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Probability Theory - Science topic

The analysis of random phenomena and variable, stochastic processes and modeling.
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Hi everyone,
In engineering design, there are usually only a few data points or low order moments, so it is meaningful to fit a relatively accurate probability density function to guide engineering design. What are the methods of fitting probability density functions through small amounts of data or low order statistical moments?
Best regards
Tao Wang
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Good explanation is performed by Chao Dang,
Best regards
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I Will be more than happy if somebody help me in this case. Does it has an specific function in R? or we should utilize quantile -copula methods...? or other???
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Send me equations.
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Dear colleagues,
I would appreciate if you give comments on the following question.
Best regards
Ali Taghavi
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George Stoica
Thank you for the suggestion, George
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We need to prepare a weighted average multi-model ensemble of projected future daily precipitation by assigning weights to individual CMIP6 models based on past performance. For this purpose, We want to use Bayesian Model Averaging. Since the distribution of precipitation is highly skewed with large number of zeros in it, a mixed (discrete-gamma) distribution is preferred as the conditional PDF as per Sloughter et al., (2007).
Considering 'y' as the reference (observed ) data and 'fk' as the modelled data of kth model,
The conditional PDF consists of two parts. The first part estimates P(y=0|fk) using a logistic regression model. The second part consists the following the term P(y>0|fk)*g(y|fk).
Since the computation of P(y>0|fk) is not mentioned in the referred manuscript, If I can compute P(y=0|fk), Can I compute P(y>0|fk) as 1-P(y=0|fk) in this case?
If not, Can someone help in computing P(y>0|fk)?
You can find the the referred paper here https://doi.org/10.1175/MWR3441.1
Thanks
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Yes. You can proceed with that formula as you deal with Precipitation data, which contains only non-negative values. according to axioms of probability P(y≠0|fk)=1-P(y=0|fk).
You can find a worked example in the book titled “Statistical Methods in Hydrology and Hydroclimatology(DOI: 10.1007/978-981-10-8779-0)”
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It is said Bell's inequality is a consequence of probability theory, which has nothing to do with quantum or not quantum. There are many papers discuss this issue, but I don't know which one is the original? Where can I find such material? Thanks.
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Attached for your kind perusal.
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The birth and death probabilities are p_i and q_i respectively and (1-(p_i+q_i)) is the probability for no change in the process. zero ({0}) is an absorbing state and sate space is {0,1,2, ...}. What are the conditions for {0} to be recurrence (positive or null)? Is the set {1,2,3,...} transient? What we can say about duration of process until absorption and stationary distribution if it exists and etc?
Every comment is appreciated.
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There is no logical (reasonable) condition that {0} is not absorbing, so it is always a recurrence state. {1,2,...} is always transient.
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I created a new test for uniformity, but so far, I've had no luck finding its critical values analytically, I could only obtain them by Monte Carlo simulation. What's worse is that histograms show that the null distribution does not approach normal distribution even at large n, so I cannot approximate it with mean and standard deviation.
Is there any sort of "standard procedure" for deriving null distribution of a test statistic? Or at least approximating it with an analytical expression?
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Dear Igor Yegin
You might start by looking at how the distributions for existing tests of uniformity are obtained. This seems to be via theory for Weiner Processes (also called Empirical Processses). A book that contains all this is the following:
Empirical Processes with Applications to Statistics
G.R. Shorak & J.A. Wellner
Wiley , 1986 ISBN 0-471-86725-X
(there might be possibly later editions)
This paper:
Conference Paper A review of the properties of tests for uniformity
notes a transformed version of the Sherman statistic that supposedly has improved convergence to normality. It looks to have been obtained empirically, but the papers referenced for this may make it clear.
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How to calculate the sum and the subtraction of many random variables that follow exponential distributions and have different parameters ?
(The value of Lambda is different for all or some variables).
example :
L(t) = f(t) + g(t) - h(t)
with
f(t) = a.expo(-a.t)
g(t) = b.expo(-b.t)
h(t) = c.expo(-c.t)
st:
a = Lambda_1
b = Lambda_2
c = Lambda_3.
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(continued)
In case of more terms (all with different means m_j>0, j=1,2,...,n) the formulas are as follows (ti replaced by -s)
ch.f.(X_1+X_2+...+X_n)(t) = 1/ [ (1+m_1 s) (1 + m_2 s) ... (1 + m_n s)]
= \sum_{j=1}^n A_j / (1+m_j s),
where A_j = \prod_{k\ne j} [ 1 - m_k / m_j]^{-1}
Therefore, in such cases the density of the sum is equal to
\sum_{j=1}^n A_j / m_j \exp( - x/m_j ), for x>0.
If X_j in the sum is preceded by sign -, then the first two formulas remain valid after replacing m_j by - m_j. The last requires replacing the exponential density for positive variable by the opposite one.
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In discussing Quantum Mechanics (QM), I shall restrict myself here to Schroedinger's Non-Relativistic Wave Mechanics (WM), as Dirac showed (in his 1930 text) [using Hilbert State Vectors] that Heisenberg's Matrix Mechanics (MM) was simply mathematically equivalent.
WM was invented in 1925 when Schroedinger adopted de Broglie's radical proposal that a quantum particle, like an electron, could "have" both contradictory point particle properties (like momentum, P) and wave properties, like a wave-length or wave-number K) by: K = h P; where h is Planck's constant (smuggling in quantization). Next he ASSUMED that a free electron could be represented as a spherical wave described by the Wave Equation. Then, he "joined the QM Club" by restricting his math approach to an isolated hydrogen atom, with its one orbital electron moving around the single proton (each with only one electron charge,e) at a spatial separation r at time t (i.e. x;t). He then linearized out the time by assuming a harmonic form: Exp{i w t) along with Einstein's EM frequency (photon) rule: E = h w. This gave him his famous Wave Equation [using W instead of Greek letter, psi]: H W = E W where H was the classical particle Hamiltonian H =K+U with K the kinetic energy [K= p2/2m] and U the Coulomb potential energy [U = e2/r]. Replacing the quadratic momentum term gave the Laplacian in 3D spherical polar co-ordinates [r, theta, phi]. He then remembered this resembled the 19th century oscillating sphere model with its known complete (infinite series) solution for n=1 to N=infinity for W=Y(l:cos theta) exp[i m phi] introducing the integer parameters l [angular momentum] and m [rotation around the Z axis]. By assuming the math solution is separable, he was left with the linear radial equation that could be solved [with difficulty] but approximated to Bohr's 1913 2D circular [planetary] model E values.
The "TRICK" was to isolate out from the infinite sums, all terms that only included EACH of the finite n terms [measured from n=1 to 6]. This was Dirac's key to match the nth wave function W(n:x,t) with his own Hilbert ket vector: W(n:x,t) = |n, x, t>.
So, I maintain that QM has failed to map its mathematics to a SINGLE hydrogen atom [the physical assumptions used therein] but to the full [almost infinite] collection of atoms present in real experiments. This then results in multiple epistemological nonsense such as Born's probability theory, wave function collapse and the multiverse theory.
This is NOT needed IF we reject the Continuum Hypothesis [imported from Classical Mechanics] and stick to finite difference mathematics.
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QM, like CM, imagines time as a "Fourth Dimension" (orthogonal to the 3 space directions). Physics then adopts Newton's Timeless calculus to write a set of equations that are VALID across all space at the same single time [t].
Apart from great simplifications, what is the physical justification for this model?
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In probability theory, fractional Brownian motion (fBm), also called a fractal Brownian motion, is a generalization of Brownian motion. Unlike classical Brownian motion, the increments of fBm need not be independent. fBm is a continuous-time Gaussian process BH(t) on [0, T], that starts at zero, has expectation zero for all t in [0, T],
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Thanks for your useful answers.
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I am working in statistical seismology and we are running into a HIGHLY controversial topic. What can we say about the largest possible event (earthquake) that could happen in an area based on data? We make estimates, but what reliability do these estimates carry? There are epistemic and random uncertainties involved. There are many theoretical estimators for this quantity but many scientist doubt that they are of any practical value. I do not believe we seismologists are qualified to do more than "rambling" about the problem and I think some input from philosophers would be extremely enlightening.
I refer to papers:
Pisarenko VF (1991). Statistical evaluation of maximum possible magnitude. Izvestiya Earth Phys 27:757–763
Zöller, G. & Holschneider, M. (2016). The Maximum Possible and the Maximum Expected
Earthquake Magnitude for Production-Induced Earthquakes at the Gas Field in Groningen, The
Netherlands. Bull. Seismol. Soc. Am. 106, 2917-2921.
Zöller, G. (2017) Comment on “Estimation of Earthquake Hazard Parameters from Incomplete Data
Files. Part III. Incorporation of Uncertainty of Earthquake‐ Occurrence Model” by Andrzej
Kijko, Ansie Smit, and Markvard A. Sellevoll. Bull. Seismol. Soc. Am. 107: 1975-1978.
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and Albania ...
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It is known that the FPE gives the time evolution of the probability density function of the stochastic differential equation.
I could not see any reference that relates the PDF obtain by the FPE with trajectories of the SDE.
for instance, consider the solution of corresponding FPE of an SDE converges to pdf=\delta{x0} asymptotically in time.
does it mean that all the trajectories of the SDE will converge to x0 asymptotically in time?
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The Fokker-Plank equation can be treated as a so-called forward Kolmogorov equation for a certain diffusion process.
To derive a stochastic equation for this diffusion process it is very useful if you know a generator of this process. Finally, to find out a form of the generator you have to consider a PDE, dual to the Fokker-Plank equation which is called the backward Kolmogorov equation. The elliptic operator in the backward Kolmogorov equation coincides with the generator of the required disffusion process. Let me give you an example.
Assume that you consider the Cauchy problem for the Fokker-Plank type equation
u_t=Lu, u(0,x)=u_0(x),
where Lu(t,x)=[A^2(x)u(t,x)]_{xx}-[a(x)u(t,x)]_x.
The dual equation is h_t+L^*h=0, where L^*h= A^2(x)h_{xx}+a(x)h_x.
As a result the required diffusion process x(t) satisfies the SDE
dx(t)=a(x(t))dt+A(x(t))dw(t), x(0)= \xi,
where w(t) is a Wiener process and \xi is a random variable independent on w(t) with the distribution density u_0(x).
You may see the book Bogachev V.I., Krylov N.V., Röckner M., Shaposhnikov S.V. "Fokker-Planck-Kolmogorov equations"
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Dear all,
I measured a variable that takes values between 0 and 0.1 (with a minimum of 0.00053). This variable will be used in a regression analysis, but it has values of skewness and kurtosis of 3.8 and 14.3, respectively, hence requiring a transformation in order to reduce those values.
I first thought about a log transformation. However, in this way, the resulting values of the variable will be negative, and I would avoid this. Another option is multiplying all values for 1,000 and then use a log transformation. But, how can I justify this choice to referees?
Have you ever experienced this problem? How have you solved it?
Thank you for your attention to this matter.
Best
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Can i delete some of variables which has more than ±2 skewness and kurtosis to get better scores?
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suggest with probable theories and examples
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Difficult to define, this concept of spiritual intelligence.
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Consider the following cases, where i have put my understanding
Notation - 0+ = tending to zero
{b} = singleton set
  1. lim (n-->infinity) (1/n) = 0+
  2. lim (n-->infinity) (n/n) = lim (n-->infinity) (1) = 1
  3. Step 2 can also be looked as lim (n-->infinity) ((1/n)/(1/n)) = 0+/0+= 1 (Here both 0+ are same and they are not exact 0)
  4. lim (n-->infinity) (n2/n) = infinity
  5. step 4 can also be viewed as lim (n-->infinity) ((1/n)/(1/n2)) = 0+/0+= infinity (here both 0+ are not same and one 0+ is like infinite times the other. Which is again a conclusion that 1/n or 1/n2 with limit n goes to infinity is exact zero)
Now the real question is this from probability theory or set theory.
I found this description of singleton as
{b} = infinite intersection over 'n' of (b-1/n , b+1/n]
but according to my understanding(as above), it still should represent a range of real number and not single point. For that intersection to result in a point, 1/n should be exact zero.
These two descriptions, one from probability theory and other from calculus doesn't seem to agree to each other according to my understanding.
Can you please tell where i am doing wrong ?
I might have used some terminologies carelessly, but i hope you got the point what i am trying to ask.
Thanks
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Many of the engineer's face the same problem about the exact meaning of the concept (Limit of sequence, limit f(x), etc, ). This because the short methods for computations they learned in calculus, without focusing on the analysis of the method. So, you need to read more about REAL ANALYSIS, in particular, the Archimedes Property.
In fact, the two concepts in your question are the same.
Using Lim(1/n)=0 ,we obtain the infinite ∩(b - 1/n, b+1/n ] ={ b}.
To see this,
Let x ∈ ∩(b - 1/n, b+1/n ] for any n, then
b - 1/n < x ≤ b + 1/n for any n.
As n goes to infinity (1/n) goes to zero, we obtain
b - 0 < x ≤ b + 0 hence, x = b the only possible value.
we deduce that ∩(b - 1/n, b+1/n ] ={ b}.
You may ask, are there another point a which is very close to b
and belongs to the intersection?
Assume that b - (1/n) < a ≤ b + (1/n) for any n,
- (1/n) < a - b ≤ (1/n) ( by subtracting b )
as n goes to infinity, we obtain 0 < a - b ≤ 0,
provides a = b.
Therefore, ∩(b - 1/n, b+1/n ] = { b}.
Best regards
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If X is an asymptotic normal variate then we can find the asymptotic distribution of h(X) by using delta method if h(X) \in R. But if h(X) is not Real valued function (e.g., h(X) could be a positive function), what is the asymptotic distribution of h(X)?
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I'm having a problem with one of my works. I need to use the central limit theorem, but I first need to prove that my variables are weakly dependent. Does someone have an example that I can use as a base for my work? Thanks in advance.
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The question is unclear. Does "I need to prove that my variables are weakly dependent" mean that you want to prove that the variables are:
(a) not independent;
(b) not strongly dependent.?
In the context of the central limit theorem,there needs to be some idea of an increasing data-set. What is this?
In some types of practical application, it may not be possible to "prove" anything in the sense of a mathematical proof starting from some standard model (that then itself needs to be justified), but you be able to justify (from the practical context and experience in that context) making the direct assumption that weak dependence applies. I hesitate to point you to my own publication , as it may not be relevant to you,
Again, when you say "I need to use the central limit theorem", are you wanting to use it to get a formal asymptotic distributional result without actually wanting to apply it , or do you need a practically useful approximate result? If the latter, it may be useful to concentrate on the variance, and its asymptotic behaviour, as this should give some clue about the effects of various levels of dependence (where, for studying the variance, interest is limited to the correlation properties only).
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according to the probability theory,
suppose that we calculate the experimental probability of students who prefer mathematics and it was .70%from a sample of 20 students (14/20), is that correct to use these percentage (70% to calculate the probability of prefer mathematics in case of applying the same survey on a sample of 200 students?
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14/20 is a proportion or percentage. It's a relative frequency, not a probability. One may (!) use the relative frequeny as an estimate for the probability, that a student sampled under similar conditions will prefer mathematics (and this also only in the case that there is a countable set of possible alternatives).
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I've been thinking about this topic for a while. I admit that I still need to do more work to fully understand the full implications of the problem, but suggests that under certain conditions, Bayesian inference may have pathological results. Does it matter for science? Can we just avoid theories that generate those pathologies? If not, what can we do about it?
I have provided a more detailed commentary here:
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My practical experience suggests that that different priors do not usually make much difference unless there is very liitle information in the data. indded different results could be seen as a feature and not a problem so that you could do a sesnitivity analysis with different priors. There is a nice eaxmple of this in Clayton and Hills Statsitical Methods in epidemiology
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Dear researchers , I'm a student in Master 1 (EDP) , and am a beginner in research , I have one international paper entitled " A new special function and it's application in probability " , I want people here to Give me comments to improve that research for the futur contribution in mathematics ? , Now I want theorist in probability and numerical analyis to give us any constrictive opinion about that research in all needed sides , For checking that paper via the journal webpage , just to check this link , Thanks som much for any comments or any kind of help.
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If you have developed a new probability density function, this first suggests that you have taken into account a new phenomenon that needs to be estimated and that is not adaptable to classical probability laws.
If so, how would it facilitate a way of life, phenomena that are difficult to measure with certainty in the present, and that merits the trouble of evaluating its chances of being realized?
Or, what information or role does it assume in other disciplines: in medicine, physics, statistics, demography, risks, etc.?
Science is a kind of molecule, that is to say, which attaches, although it is specific, to other external notions. In this case, how to adapt this theory in others to facilitate its application in the real world.
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Most multivariate techniques, such as Linear Discriminant Analysis (LDA), Factor Analysis, MANOVA and Multivariate Regression are based on an assumption of multivariate normality. On occasion when you report such an application, the Editor or Reviewer will challenge whether you have established the applicability of that assumption to your data. How does one do that and what sample size do you need relative to the number of variables? You can check for certain properties of the multivariate normal distribution, such as marginal normality, linearity of all relationships between variables and normality of all linear combinations. But is there a definitive test or battery of tests?
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you can test multivariate normality by assessing the multivariate skewness and kurtosis through the software in the following link
for more information about multivariate normality please refer to the attached papers
  • Cain, M. K., Zhang, Z., & Yuan, K. H. (2017). Univariate and multivariate skewness and kurtosis for measuring nonnormality: Prevalence, influence and estimation. Behavior research methods, 49(5), 1716-1735.
  • Ramayah, T., Yeap, J. A., Ahmad, N. H., Halim, H. A., & Rahman, S. A. (2017). Testing a confirmatory model of facebook usage in smartpls using consistent PLS. International Journal of Business and Innovation, 3(2), 1-14.
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Is the canonical unit 2 standard probability simplex, the convex hull of the equilateral triangle in the three dimensional Cartesian plane whose vertices are (1,0,0) , (0,1,0) and (0,0,1) in euclidean coordinates, closed under all and only all convex combinations of  probability vector 
that is the set of all non negative triples/vectors of three real numbers that are non negative and sum to 1, ?
Do any  unit probability vectors, set of three non negative three numbers at each pt, if conceveid as a probability vector space,  go missing; for example <p1=0.3, p2=0.2, p3=0.5>may not be an element of the domain if the probability  simplex in barry-centric /probabilty coordinate s a function of p1, p2, p3 .
where y denotes p2, and z denotes p3,  is not  constructed appropriately?
and the pi entries of each vector,  p1, p2 p3 in <p1, p2,p3> p1+p2+p3=1 pi>=0
in the  x,y,z plane where x =m=1/3 for example, denotes the set of probability vectors whose first entry is 1/3 ie < p1=1/3, p2, p3> p2+p3=2/3 p1, p2, p3>=0; p1=1/3 +p2+p3=1?
p1=1/3, the coordinates value of all vectors whose first entry is x=p1=m =1/3 ie
Does using absolute barry-centric coordinates rule out this possibility? That vector going missing?
where <p1=0.3, p2=0.2, p3=0.5> is the vector located at p1, p2 ,p3 in absolute barycentric coordinates.
Given that its convex hull, - it is the smallest such set such that inscribed in the equilateral such that any subset of its not closed under all convex combinations of the vertices (I presume that this means all and only triples of non negative pi that sum to 1 are included, and because any subset may not include the vertices etc). so that the there are no vectors with negative entries
every go missing in the domain  in the , when its traditionally described in three coordinates, as the convex hull of three standard unit vectors  (1,0 0) (0,0,1 and (0,1,0), the equilateral triangle in Cartesian coordinates x,y,z in three dimensional euclidean spaces whose vertices are    Or can this only be guaranteed by representing in this fashion.
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Of course it is, its what it is by definition. I probably should have thought more about this, back when I originally posted. If it isnt, then nothing is. Its the "closed" convex hull of its vertices, or all points in [0,1]^3, that can be expressed by convex combinations of its vertices:, (1,0,0), (0,0,1), (0,1,0).
So that any vector (x,y,z), x+y+z=1 , or rather point, (x,y,z), in [0,1]^3, where x+y+z=1,and x >=0,y,>=0z>=0 can be simply expressed by x* (1,0,0)+(0,1,0)+z*(0,0,1)=(x,y,z), as closed under all convex combinations of the vertices convexity (of this form) simply means/ requires that its contains all points in [0,1]^3, that can be expressed by-negative c1, c2,c3 ; c1*(1,0,0)+c2*(0,1,0) +c3*(0,0,1) where c1+c2+c3 =1, it just is the set of 3 all and only non-negative coordinates, that sum to one . And clearly we can set c1=x, c2=y, and c3=y and these are non-negative and sum to one. So its all, and "only" probably triples (x,y,z),x+y+z=1 .
I should have thought this through.
The main questions are (1) and (2):
(1): does the canonical 2 probability simplex, contain the canonical 2 simplex of the sums at each point x+y, x+z, z+y, clearly at each point (x,y,z) in the canonical 2 probability simplex, x+y,in [0,1] x+z\in [0,1], z+yin [0,1] and their sum (x+y+x+z+z+y=2(x+y+z)=2*1=1, but does it contain at "every" point, every such convex combination of (x+y=l, x+z=g, z+y=h), 1>=h>=0,1>=g>=0, 1>=l>=0 & l+g, g+h, h+l in where h+g+l=2 at some point (x,y,z)?
Id say yes, because if there were a (l,g,h), l+g+h=2, 1>=h>=0,1>=g>=0, 1>=l>=0, but with no accompanying (x,y,z) x+y+z=1, (x,y,z)>=0, l= x+y, g=x+z, h=z+y
, g=x+h-y, y=x+h-g, so l=x+x+h-g, l=2x+h-: x=1/2*(l-h+g), which is uniquely determined where l-h+g<=l+g+h=2, as h, g, l>=0, and clearly x+y+z=1/2(2x+2y+2x)1/2(l+g+h)=1, the only way that point could not be in the simplex is if, for example x<0
which in this case of l+g<h, as h<=1, g<=1, l<l, as per the conditions above, that we have a contradiction as (l+g)+h=2 where, l+g<h gives (l+g)+(l+g)<(l+g)+h=2, to 2(l+g)<2, L+g<1 but h<=1, so 1+h<=1+1=2, so, 1+h<=2,
L+g<1 gives (l+g)+h< 1+h
and (l+g)+h<1+h<=2, L+g+h<2, ie l+g+h<1
1/2*(1-g+h)<=1/2*2=1
And moreover, (2):
(3)Does it contain all such h+g+l=1, where the l, g, h, are the sums of the first 2, the first and third, second and third coordinates, respectively of 3 "distinct" cartesian points in the simplex, (x1,y1,z1), (x2,y2,z2), and (x3, z3, y3), l=x1+y1, g=x2+z2, h= y3+z3, h+g+l=1, where none of the x1,x2,x3, x2,y2,z2, x3,z3, y3 are 0?.
Clearly for any m, in [0,1] there is a point in the simplex <x,y,z>, x=m.
So as x is an element of a point, in the simplex, then at those points, where the x coordinate is m, m=x>=0, and clearly x=m<=1.
As, 1>=m>=0 entails that 0<=1-m<=1, then (1-m) is in [0,1],. Then as, for any real in [0,1], there are pts in the simplex, where the x coordinate attains that real, then there is a pt, whose x coordinate is x1, where, (the x coordinate), assumes the value, x1=1-x=1-m.
<x1,y1,z1>, x1+y1+z1=1 so, y1+z1=1-x1=1-(1-m)=m, so x+z assumes the value m,
As the y and z coordinates can assume any value in [0,1] as well, then we have all non negative coordinate of three points, in [0,1]^3:, p1 =(x1,y1,z1), p2=(x2,y2,z2), and p3=(x3, z3, y3), where x1+y2+y3<=3 where x1,y2,y3, are all in [0,1], and thus, the subset, comprising all, sets of three points, p1, p2, p3 in [0,1]^3 where x1+y2+z3=2 , where x1,y2, z3 are non-negative and in [0,1,] are in the simplex.
As, at any pt, in the simplex, and at all pts , x+y+z=1, so: (x1+z1+y1)+)x2+y2+z2)+(x3+y3+z3)=3, and as x1+y2+z3=2 ,
so( x1+y1)+(y2+z2)+(z3+x3)= 3-(x1+y2+z3)= 1,
ie for any real in [0,1] x=m at(x,y,z) if and only there is a point (x1,y1,z1), x1+z1= m
(ie not on the edges of the simplex, where, aeprt from the vertices, only two of the coordinate of a point are non-zero and not 1 ?
At the edges we set y1=0, x1=x1+y1=0+x1= l,
x2=0, so as x2+y2+z2=1, x2+z2 in [0,1], so we set z2=g, z2+x2=z2+0=z2=h so,
z2+x2=g, and at the third point. Let z3=0, h=y3 so y3+z3=0+y3=y3=h
I suppose so, there was a counter-example there wou as if it did not there would have to be no point in the simplex (x,y,z), where 0<=y+z=l=1-g-h<=1 for some, (h,g, l), h+g+l=1 h>=0, g>=0, l>=0, where of course, this gives that 0=1-1<=1-(g+h)=l>=1-0=0 , It also has to express all such combinations because if did not for some l, g, h.
So as long as the the set of all probability triples, 2 canonical probability simplex , contain on its edges, (the 2 positive entries,), the set of call of all probability doubles, that is, the "canonical 1 probability simplex/ the set of 2 non-negative points in R^2 whose sum is 1 (all convex combinations of 2 non-negatives that sum to 1, then the it will contain some such three such edge points, meeting the above sum constraintm(x1,y1,z1), (x2,y2,z2), and (x3, z3, y3), l=x1+y1, g=x2+z2, h= y3+z3, h+g+l=1.
Where along the edges, the double in the canonical 1 probability simplex, which any triple on the edges of canonical 2 probability simplex, identifies, are those doubles, formed by the, 2 elements, of the 3 elements at that point (x,y,z), which happen to be positive (non-zero). SO, setting the first positive coordinate, in said triple (which will be x, or y) to be the x coordinate of the double, and the second positive entry (which will be z or y) to y, ie (x=1/6, y=0, z=5/6) goes to (x=1/6, y=5/6) which will be unique except at the vertices (where one can use both (1,0) or (0,1), and x+y=1 as the other point z will be zero, as the point was taken from the edge of the set of canonical 2 simplex (x,y, z), where precisely one of (x,y,z), is 0, (except at the vertices, themselves, where 2 of them are of course)
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If f(t) represents the probability density of failure rate, then how it it possible that f(t) will follow exponential distribution whereas the failure rate is constant?
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Dear Parag Sen,
your problem concerns the relationship between Poisson distribution and exponential distribution - namely:
If the random variable X represents the number of errors (system failures) in a given time period and has the Poisson distribution, then the intervals between every two consecutive errors have the exponential distribution.
For example, see Wikipedia:
If for every t > 0 the number of arrivals in the time interval [0, t] follows the Poisson distribution with mean value λt, then the sequence of inter-arrival times are independent and identically distributed exponential random variables having mean 1/λ.
In general, the exponential distribution describes the distribution of time intervals between every two subsequent Poisson events.
The answer to your question can be found at the following addresses:
Best regards
Anatol Badach
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I would like to find the probability distribution of log[U/(1-U)] when U~u(0,1). How to derive this?
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I do not quite agree with the answer of Wulf Rehder. He has computed only the cumulative distribution function(CDF). The question wants the probability distribution function(PDF), so we need to take a derivative over the CDF.
In the first case, PDF = 1/(1+x)^2 = (1-u)^2
In the second case, PDF = e^x/(1+e^x)^2 = u(1-u)
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Suppose we have n samples (x1, x2, …, xn) independently taken from a normal distribution, where known variance σ2 and unknown mean μ.
Considering non-informative prior distributions, the posterior distribution of the mean p(μ/D) follows normal distribution with μn and σn2, where μn is the sample mean of the n samples (i.e., μn=(x1+x2+…+xn)/n), σn2 is σ2/n, and D = {x1, x2, …, xn} (i.e., p(μ/D) ~ N(μn, σ2/n)).
Let the new data D’ be {x1, x2, …, xn, x1new, x2new, …, xknew}. That is, we take additional k (k<n) samples independently from the original distribution N(μ, σ2). However, before taking the additional samples, we can know the posterior predictive distribution for the additional sample. According to Bayesian statistics, the posterior predictive distribution p(xnew/D) follows normal distribution with μn and σn2+ σ2 (i.e., p(xnew/D) ~ N(μn, σ2/n+ σ2)). Namely, the variance becomes higher to reflect the uncertainty of μ. So far, this is what I know.
My question is, if we know p(xnew/D) for the additional samples, can we predict the posterior distribution p(μ/D’) before taking the additional k samples? I think that p(μ/D’) seems to be calculated based on p(xnew/D), but I have not gotten the answer yet. So, I need help. Please borrow your wisdom. Thanks in advance.
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I am not sure that I am right but lets add something to the discussion.
As likelihod is the joint prob (prodct of pdfs)of the sample and also prior/posterior predictive is the prob fnctn of xnew, therefore, to me, the liklihood fntn of x and xnew is the product of likelihood of x and prior/posterior predictive of xnew. U have the likelihood fnctn u can derive the posterior p(mu/D’).
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At the end of the nineteenth century, many researchers concentrated on various alternative methods based on the theory of infinite series. These methods have been combined under a single heading which is called \textit{Summability Methods}.  In recent years, these methods  have been used in approximation by linear positive operators. Also, in connection with the concept of statistical convergence and statistical summability, many useful developments have been used in various contexts, for example, approximation theory, probability theory, quantum mechanics, analytic continuation, Fourier analysis, the behaviors of dynamical systems, the theory of orthogonal series, and fixed point theory
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I think that it is almost completely correct and useful (especially for young researchers), each of the previous Responders' replies to the question of Professor Ugur Kadak: "What do you think about the future of summability and its applications?".
With a cordial greetings,
Jasmina Fatkić
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I have an upcoming exam, with 8 questions that may cover 10 fields. Suppose the pass criteria is 70% then how many fields i have to study to clear the exam?
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Following
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Slutsky theorem is commonly used to prove the consistency of estimators in Econometrics.
The theorem is stated as:
For a continuous function g(X_k) that is not a function of k,
plim g(X_k) = g (plim X_k)
where X_k is the sequence of random variables.
Could anyone suggest any literature on how to prove this theorem?
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The question of whether Quantum Mechanics is a complete science had sparked a historic debate led by Albert Einstein on one side and Niels Bohr on the other side. It is interesting that quantum physicists from the school of Copenhagen had to resort to philosophical arguments to defend the soundness of quantum mechanics in terms of its ability to faithfully interpret dynamic systems. The fuzziness of the central notion of the quantum wavefunction seems to have never been resolved to this day, a problem that prompted Richard Feynman in no small part to assert that “Nobody understands quantum mechanics”. I offer the view that the very mathematical tool at work in QM, the Theory of Probability (ToP), might be the first element responsible for the weaknesses of QM as a science. In Chapter 7 of the Title Quanto-Geometry: Overture of Cosmic Consciousness or Universal Knowledge for All, I discuss its limits and show the necessary extensions required for causal categories of interpretation in ToP, thus leading to completeness of QM. Downloadable here:
What do you think? Is QM obscure in its soul as formulated or are its limits attributable to Statistical Theory? Do you think the proposed extensions contribute any further intelligibility at all? What aspect of QM do you find particularly challenging?
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Hello Vikram.
Wish you well equally in the needed works of unblocking the avenues of theoretical physics.
Cordially.
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This is an example in Durrett's book "Probability theory: theory and examples", it's about the coupling time in Markov chains, but I can't see the reason behind it.
The trick is played by two persons A and B. A writes 100 digits from 0-9 randomly, B choose one of the first 10 numbers and does not tell A. If B has chosen 7,say, he counts 7 places along the list, notes the digits at the location, and continue the process. If the digit is 0 he counts 10. A possible sequence is underlined in the list:
3 4 7 8 2 3 7 5 6 1 6 4 6 5 7 8 3 1 5 3 0 7 9  2  3 .........
The trick is that, without knowing B's first digit, A can point to B's final stopping location. He just starts the process from any one of the first 10 places, and conclude that he's stopping location is the same as B's. The probability of making an error is less than 3%.
I'm puzzled by the reasoning behind the example, can anyone explain it to me ?
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The key point is that if A and B ever use the same position in their respective sequences, then from that point on their sequences are identical.  The reason that the trick works is that there is a fairly high probability that this will happen. 
Coupling in Markov chains works in a similar way.
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I saw a claim in some paper without proof. Because the formula is complicated, I wrote it in the attached file. I can not prove one part of the assertion. Please tell me how to prove it if you can understand.
Thank you
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Dear  Alejandro Santoyo
Thanks for your reply.
The attached file is my reply.
Best regard.
Masahiro Fujimoto
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Extinction probability and expected number of progeny are to be calculated at the end
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Choose the branch at random, depending on its probability, each time the branch comes up.
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State dependent additivity and state independent additivity? ;
akin to more to cauchy additivity versus local  kolmorgov additivity/normalization of subjective credence/utility,  in a simplex representation of subjective probability or utility ranked by objective probability  distinction? Ie in the 2 or more unit simplex (at least three atomic outcomes on each unit probability vector, finitely additive space) where every events is ranked globally within vectors and between distinct vectors by < > and especially '='
i [resume that one is mere representability and the other unique-ness
the distinction between the trival
(1)x+y+z x,y,z mutually exclusive and exhaustive F(x)+F(y)+F(z)=1
(2)or F(x u y) = F(x)+F(y) xu y ie F(A V B)=F(A)+F(B)=F(A)+ F(B) A, B disjoint
disjoint on samevector)
(3)F(A)+F(AC)=1 disjoint and mutually excluisve on the same unit vector
and more like this or the properties below something more these
to these (3a, 3b, 3C) which are uniqueness properties
forall x,y events in the simplex
(3.A) F(x+y)=F(x)+F(y) cauchy addivity(same vector or probability state, or not)
This needs no explaining
aritrarily in the simplex of interest (ie whether they are on the same vector or not)
or(B)  x+y =z+m=F(z)+F(m) (any arbitary two or more events with teh same objective sum must have the same credence sum, same vector or not) disjoint or not (almost jensens equality)
or (C)F(1-x-y)+F(x)+F(y)=1 *(any arbitrary three events in the simplex, same vector or not, must to one in credence if they sum to one in objective chance)
(D) F(1-x)+F(x)=1 any arbitary two events whose sum is one,in chance must sum to 1 in credence same probability space,/state/vector or not
global symmetry (distinct from complement additivity) it applies to non disjoint events on disitnct vectors to the equalities in the rank. 'rank equalities, plus complement addivitity' gives rise to this in a two outcome system, a
. It seems to be entailed by global modal cross world rank, so long as there at least three outcome, without use of mixtures, unions or tradeoffs. Iff ones domain is the entire simplex
that is adding up function values of sums of evenst on distinct vectors to the value of some other event on some non commutting (arguably) probability vector
 F(x+y)=F(x)+F(y)
In the context of certain probabilistic and/or utility unique-ness theorems,where one takes one objective probability function  and tries to show that any other probability function, given ones' constraints, must be the same function.
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and F(x+y)=F(x)+F(y)
In the context of certain probabilistic and/or utility unique-ness theorems,where one takes one objective probability function  and tries to show that any other probability function, given ones' constraints, must be the same function.
what is meant by state dependent additivity does that mean that instead of F(x U y)=F(x)+F(y) x,y disjoint (lie on the same vector) that (same finite probability triple) ; or instead of F(x)+F(y)+F(z)=1 iff x,y,z are elements of the very same vector (same triple)
in the simplex one literally instead has that F(x+y)=F(x)+F(y) over the entire domain of the function. ie adding up (arguably non commuting) elements of distinct vectors,   or F(x)+F(y)+F(1-x-y)=1 arbitarily over the simplex; or domain of interest, where the only restriction is that one can only add up elements as many times as they are present in the domain. Whilst with cauchy additivity one if one domain is merely a single vector. <1,3, 1,6, 1,2, unit event=1 > one so long 1/6 is in the domain (supposing that entire domain, probability vector space, just is that vector dom(F)={1,3 1,6, 1,2 1),  where if F(1)=1 one can arbitrarily add up 1=F(1)=F(1/6+1/6+1/6 ) six times= 6F(1/6), so F(1/6)=1/6.
I presume however, that if ones domain is the entire simplex there would not be any relevant difference, between outright Cauchy additivity and state independent additivity; and thus to presume would be outright presumptuous.  Or this a name for cross world global rank which entails its long as there at least three atomic events on each simplex (so long as the simplex is well constructed, and the rank is global and modal), even if finite local standard additivity is presumed). As one can transfer values of equiprobable events onto other vectors where they are disjoint?
if by state independent addivity  this means one can arbitrary add up the function values of F(1/6) for objective probability six times, to F(1)=1 the function value lets say at chance=1 to attain that F(1/6)=6 so long as those events are ranked equal and are present at least six times somewhere or other, even if in distinct state, or vectors (in the same system).
or does this apply to local additivity, where one has a global modally transitive rank over the simplex where n>=2 (ie the number of elements in each triple is at least three) because a cross world rank, with equalities, will entail this in any case, if justified. So if one can derive that cross world additivity must hold given finite additivity and global modal rank including equalities cross world/vector, on pain of either either local addivitity failing (probabilism)  or ones justified global and local  total rank  is violated, justified for whatever reason is violated)  including equalities must hold (for whatever reason) does this count as presumptious.
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In order to get a homogeneous population by inspecting two conditions and filtering  the entire population (all possible members) according these two conditions, then used the all remaining filtered members in the research, Is it still population? or it is a sample ( what is called?).
working on mathematical equation by adding other part to it then find the solution and applying it on the real world. can we generalize its result to other real world?
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Rula -
I am not sure I understand your process, but if your 'sample' is really just a census of a special part of your population, then you can get descriptive statistics on it, but you cannot do inference to the entire population from it. 
You might find the following instructive and entertaining.  I think it is quite good.
Ken Brewer's Waksberg Award article: 
 Brewer, K.R.W. (2014), “Three controversies in the history of survey sampling,” Survey Methodology,
(December 2013/January 2014), Vol 39, No 2, pp. 249-262. Statistics Canada, Catalogue No. 12-001-X.
Cheers - Jim
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Is there a distinction between strong or complete qualitative probability orders which are considered to be strong representation or total probability relations neither of which involve in-com parables events, use the stronger form of Scott axiom (not cases of weak, partial or intermediate agreements) and both of  whose representation is considered 'strong '
of type (1)P>=Q iff P(x)>= P(y)
versus type
(2) x=y iff P(x)=P(x)
y>x iff P(y)>Pr(x) and
y<x  iff P(y)<Pr(x)
The last link speaks about by worry about some total orders that only use
totallity A<=B or B<=A without a trichotomy https://antimeta.wordpress.com/category
/probability/page/3/
where they refer to:
f≥g, but we don’t know whether f>g or f≈g. S
However, as it stands, this dominance principle leaves some preference relations among actions underspecified. That is, if f and g are actions such that f strictly dominates g in some states, but they have the same (or equipreferable) outcomes in the others, then we know that f≥g, but we don’t know whether f>g or f≈g. So the axioms for a partial ordering on the outcomes, together with the dominance principle, don’t suffice to uniquely specify an induced partial ordering on the acti
.
 The both uses a total order  over
totality
A <=B or B >=A
l definition of equality and anti-symmetry,  A=B iff A<=B and B>=A
A<= B iff [A< B or A=B] iff not A>B
A>=B iff [A>B or A=B]iff not A<B
where A>B equiv B<A,and
A>=B  equiv B<=A iff (A<B)
where = is an equivalence relation, symmetric, transitive and reflexive
<=.=> are reflexive transitive, negative transitive,complementary and total
, whilst <, > are irreflexive and ass-ymetric,
transitive
A<B , B<C implies A>C
A<B B=C implies A>C
A<B, A<=B implies A>C
and negatively transitive
and complementary
A>B iff ~A<~B
<|=|>, are mutually exclusive.
and where equality s, is an equivalence class not denoting identity or in-comparability but generally equality in rank (in probability) whilst the second kind uses negatively transitive weakly connected strict weak orders,r <|=|>,
weak connected-ness not  (A=B) then A<B or A> B
whilst the second kind uses both trichotomous strongly connected strict total orders,  for <|=|>,.
(2) trichotomoy A<B or A=B or A>B are made explicit,  where the relations are mutually exclusive and exhaustive in (2(
(3) strong connectected. not  (A=B) iff A<B or A> B, and
and satisfy the axioms of A>= emptyset, \Omega > emptyset , \Omega >=  A
scotts conditions and the separability and archimedean axioms and monotone continuity if required
In the first kind <= |>= is primitive which makes me suspect, whilst in the second <|=|> are primitive.
Please see the attached document.And whether trich-otomoy fails
in the first type, which appears a bit fuzzier yet totality holds in both case A>=B or B<=B  where
What is unclear is whether there is any canonical meaning to weak orders (as opposed total pre-orders, or strict weak orders) .
In the context of qualitative probability this is sometimes seen as synonymous with a complete or total order. , as opposed to a partial order which allows for incomparable s, its generally a partial order, which allows for comparable equalities but between non identical events usually put in the same equivalence class (ie A is as probable as B, when A=B, as opposed, one and the same event, or 'who knows/ for in-comparability) Fihsburn hints at a second distinction where A may not be as likely as B, and it must be the case
not A>B and not  A< B  yet not A=B is possible in the second yet
A>= B or A<=B must hold
which appears to say that you can quasi -compare the events (can say that A less then or more probable,  than B ,but not which of the two A<B, A=B, , that is which relation it specifically stands in
but yet one cannot say that A>B  or A<B
)
and satisfy definitions
and A<=B iff A<B or A=B iff B>=A, iff ~A>=~B, where this mutually exclusive to A<B equiv ~B>~A
A>=B iff A>B or A<=B
iff iff B>=A where this mutually exclusive to A>B equiv ~B<~A
and both (1) and (2) using as a total ordering over >= |<=
(1)totalityA<= B or B<=A
(2)equality in rank and anti-symmetric biconditional  A=B iff A<=B and B>=A where = is an equivalence relation, symmetric, transitive and reflexive
(2) A<=B iff A<B or A=B, A>=B iff A>B or A<=B
(3) and satisfy the criterion that >|<|>=|<=,  are
complementary, A>B iff ~B<~A
transitive and negatively transitive,
where A<B iff B<A and where , =, <|> are mutually exclusive,
The difference between the two seem to be whether A>=B and A<= B is equivalent to A=B; or where in the first kind, it counts as strongly respresenting the structure even if A>=B comes out A>B because one could not specify whether A>B or A=B yet you could compare them in the sense that under <= one can say that its either less or equal in probability or more than or equal, but not precisely which of the two it is.
either some weakening of anti-symmetry of the both and the fact that the first kind use
whilst the less ambiguous orders trich-otomous orders use not  (A=B) iff A<B or A> B; generally trichotomy is not considered, when it comes to using  satisfying scotts axiom , in its strongest sense, for strict aggreement
and I am wondering whether the trich-otomous forms which appear to be required for real valued or strictly increasing probability functions are slightly stronger, when it comes to dense order but require a stronger form of scotts axiom, that involves <. > and not just <=.
but where in (1) these <=|>= relation is primitive and trich-otomoy is not explicit, nor is strong connected-ness whilst in (2)A neq B iff A>B or A<B
>|=|< is primitive and both
(1) totality A<= B or B<=A
(2) A<B or A=B or A>B are made explicit,  where the relations are mutually exclusive and exhaustive in (2(
and (2) trichotomy hold and are modelled as strict total trichotomous orders,
as opposed to a weakly connected strict weak order, with an associated total pre-order, or what may be a total order,
, or at least are made explicit.  I get the impression that the first kind as deccribed by FIshburn 1970 considers a weird relation that does not involve incomparables, and is consided total but A>=B and B<=A but one cannot that A is as likely as B, or that its fuzzy in the sense
that one can say that B is either less than or equal in probability to A, or conversely, but if B<= A one cannot /need not  whether say A=B or A<B,
not A=B] iff A<B or A>B
and strongly connected in the second.
where A=B iff A<=B and B>=A in both cases
where <= is transitive , negative transitive, complementary, total, and reflexive
A>=B or B<=A
are considered complete
and
y
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You are way too dispersed.
Try to undestand well the more basic stuff, one topic at a time.
Then move on.
Do not start from Gleason or weak measurements  for eample.
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What is general, is meant by the orthogonality relation x⊥y 
in the functional equation: orthogonal additivity below (1)
(1)∀(x,y)∈(dom(F)∩[x⊥y]):F(x+y)=F(x)+F(y)
Particularly as used in the following two references listed below (by De Zelah and Ratz, in the context of quantum spin half born rule derivations
What is general, is meant by the orthogonality relation x⊥y 
denotes x is orthogonal to y.
.
(1)∀(x,y)∈(dom(F)∩[x⊥y]):F(x+y)=F(x)+F(y)
 
See Rätz, Jürg, On orthogonally additive mappings, Aequationes Math. 28, 35-49 (1985). ZBL0569.39006.
2.See 'Comment on `Gleason-type theorem including Qubits and pro-jective measurements: the Born rule beyond quantum physics' ", by Michael J. W. Hall ">https://www.researchgate.net/profile/Francisco_De_Zela
where x⊥y
denote: -'logical/set theoretical Disjointedness' of events, on the same basis.That is mutual exclusivity? Or - or geometrical orthogonality events . That is, on perpendicular/possibly non-commuting vectors/bases in spin 12 system? (for example, spin up x direction versus spin up ydirection)? - OR something else?
That is, within QM, in the Hilbert inner product metric, what does it mean for the frame function two events to be explicitly allowed to add as per x⊥y
, in functional equation, Orthogonal additivity.
.
Does z⊥m
denote events mean events whose amplitude modul-i squared (P(A),PR(B))=(z,m)respectively and which lie on the same basis/vector in a spin system, for example
:
A spin up on basis in x direction,P(A)=||amplitude(spin up_x)||2=z
and¬A spin down on basis in x direction;P(¬A)=||amplitude(spin down_x)||2=m
 
2.Versus Non commuting bases:
A spin up in x direction
andA spin down in y direction
 
Corresponding to the logical and geometric notions roughly, respectively
How does one distinguish this, from events that are non-commuting from events& orthogonal events in the geometric sense from events, that are orthogonal in the logical sense and disjoint events on the same basis?
Is there a distinct operator, inner product in quantum mechanics (which equals zero) which determines when the frame function probabilities of events can explicitly be assumed to add?
That is disjoint, in the sense of **Kolmogorov that is disjoint or mutually exclusive, or its analog in quantum mechanics as in (1)?
(1)F(X∪Y)=F(X)+F(Y)where in (1);X∩Y=∅,∧X,Y∈Ω,,X,Y are mutually exclusive and lie on the same basis
(1a)P(A∪AC)=1,A∪AC=Ω=⊤∧A∪AC=∅
(1.b)∀(Ai)∈Ω;where,Ai∈F,the singletons, atoms, of which are in the algebra of events F;P(∪n=|Ω|i=1Ai)=[∑i=1n=|Ω|P(Ai)]=1
where, in 1(b)[∪n=|Ω|i=1Ai]=⊤∧∀(j);j≠i⟺Ai∩Aj=∅
 
This being opposed to: when the events on distinct vectors, that are geometrically orthogonal or non commuting/complementary bases/distinct bases, which are not explicitly specified to add, they may happen to, as was derived for n≥3
, where this is really a form of the much stronger global ,Cauchy additivity equation (2) which relate to unique-ness , when one has (1), in addition ? (2)(2)∀(x,y)∈dom(unitsphere);F(x+y)=F(x)+F(y)
where x+y,,x,y, can be translated as ||x+y||2,||x|2,||y||2 Of these two notions, to which does orthogonal additivity relate to, is probabilism (or quantum probabilism) with some further richness properties that make it closer to (2)in a restricted form
That is the sense of being on bases that are at right angles or being at 90
degrees from one another, not simultaneously measurable such as (A),(B):
as
(A)spin-up at angle y with||spin−up−y||2=P(spin up, measured@ angle,y)12
(B)spin-up at angle x with||spin−upxy||2=P(spin up, measured@ angle,x)=12
 
Both seem to use the Hilbert space inner product? I presume one is for bases, and another is for events.
Does this relate to when the events commute, are on vectors at right angles from each other, or rather do commute and lie on the same basis. If the former is just standard probabilism.
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An arbitrary wave functionin Hilbert space may be written as
psi) = sum over i of i)(i  psi) = sum over i of P(i) psi)
where P(i)=i)(i  is a projector to the state i)
Each P(i) squared is equal to itself, The product of any two is zero for different índices.
As a ressult of a measurment, one of the P(i) has eigenvalue 1, all the rest zero, and it is
determined that the particle is in state i, mutually exclusive to all other meauement posible results.
There is then a kind of additivity, and mutually exclusive posible results
I =sum over all i of P(i)
is the unit operator.
Im using ) and ( for the ket an bra rspectively in Dirac language.
Read Diracs Q. M.
as for orthogonality it is in (i,j) = 1 if i=j and 0 if i not equal to j
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 I was wondering if someone knows whether Mathematica allows one to plot another 'probability function over the unit 2 simplex (it can be expressed as a function of two arguments subject to certain contrainsts function .
Where I am taking the domain to be of the function to be over vectors in  the 2- standard simplex  itself as it were,subject to certain contrain'ts. 
Does it actually have a closed form expression as a function x,y coordinates. as a function of two arguments. I presume  mathematica can allows you plot it .and optimize certain functions over  tenary plot, ternary graph, triangle plot, simplex plot, G? Is that correct
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If you plot is assigned to G, then type FullForm[G]. Pay attention to the head.If you see Graphics[stuff], then you can use the Show command to combine plots, Its you see Graph[stuff], it might still be possible, but is harder. If you code is not too long, I recommend attaching it to your question.
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Is there any other ideas than "probability theory" to derive the intervals from the responses in intervall-type Type-2 Takagi-Sugeno Systems?
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Dear Prof. Mohamed-Mourad Lafifi, I would like to thank you for your valuble information and time.
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Is the ortho-complement of a proposition, in quantum logic/probability or hilbert space, the logical comple-ment of a proposition
such as 'spin up at direction x and 'spin down in direction y'
(mutually exhaustive, on the) same prepared spin system at the as same angle, ie on the same basis) and thus one can add them the probabilities to one, as in the usual probabilistic sense to .when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).
Or is this a geometric notion that that allows one to add probabilities of events together that lie on distinct bases (bases/vectors on distinct angles of the same prepared spin system). Which is stronger (closer to a restricted form of cauchy additivity)
the on the very same vector in an orth-onormal basis; the very same basis.
, in the sense that they are disjoint A\cap B emptyset , and mutually exclusive. Or is its geometric, and relates to some kind of relation between events on distinct bases of a spin system.
When one speaks of an orthonormal bases or vectors that are mutually orthogonal , and are unit vectors that are mutually orthogonal, is this where ||u||=sqrt(x^2+y^2)=1 and ||v|=sqrt(x_i^2+y_i2)=1
u.v=x_2.v_2+ y_2.y_2=0;
does this denote, complementarity/non commutting, geometric orthonality, or logical orthogonality, that is events are disjoints in the logical sense (lie in the same basis or a commutting bases, and can be added) or are on distinct bases..
In other words
Is is there of way of using the inner product to determine whether the events are lie on non commutting and cannot be explicitely assume to add. I presume that the hilbert space inner product denoted logical orthogonality
x is amplitude moduli of spin up in direction of the vector u, with probability ||x||^2, and
y is the same but for spin down down in the same direction vector u with probability ||y||2
and  x_i, y_i are the corresponding amplitude,moduli of spin up and spin down  for the direction of the vector v,and probabilities,
||x_i||^2
||y_i||^2.
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presume that he did not merely say that we are presuming Given a normalization of the frame function, presumed to some such probability function, that on any specific bases,
(1)where any given amplitudes moduli squared sum to one, the frame function values do,three which lie on the very same basis sum to 1, as they are mutually exhaustive and exclusive; when the amplitudes mod squared do). A\cap,B=emptyset,\capC ,=emptyset A, B C, are mutually exclusive and exhaustive \Omega=A\vee B\vee C0, so P(A)+PR(B)+PR(C)=1
(2) and when he applied that the function is odd F(-x)+F(x)=0, or any given two, amplutdues mod squared which lies on the same basis, and are mutually exhaustive and exclusive sum to one , will have frame function values which sum to one, as instance of complement additivity PR(A)+PR(A_C)=1 A\cap A_c =\emptyset, A\vee A_c is mutually exhaustic, where A_c is teh set theoric complement . ie if Omega={A,B, C}, A_C=B \vee C
and some other property for instance complement additivty, any event and its mutually exclusive, exhaustive complement which sum to one, must have frame function values which do) but that he from this, and given the rank and symmetry of invoked by quantum non-contextuality and quantum logic, he meant this as a form of global addititity.
with probability
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There are traditional ways of getting the cut-off value that maximize the sensitivity and specificity of a given test from the ROC curve. But I think those ways posses some arbitrariness in doing the job. How about this;
Can we use Bayes formula to calculate the value of the variable which equates the probability of the disease to the probability of the control, assuming that both classes are normally distributed? This should be the ideal cut-off, without considering any factors related to the cost,,,etc that can affect the conventional method. So from a set of observations, we can use referential statistics to calculate the mean and the standard deviation of both classes, if not already known. Then by letting the cut-off to be the value that makes the conditional probability of the disease given the value equal to the conditional probability of the control given the value, we can set the cut-off value as a function of the difference between the two means and a function of SD. Moreover, if we consider the prior probabilities of both classes not to be equal to each other, we can even see how the cut-off values moves toward one side of the variable scale in a very natural way.  This can also help if we have more than one variable and in this case it would work similar to discriminant analysis methods.
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Bayes is still considered one of the highly popular  methods, although using support vector machines provides better  support  for fixing cut off values in a more natural way then delimiting them through baye's classifier. 
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Is the following function F:[0,1] to [0,1]
F strictly monotonic increasing F(1)=1,(i presume this unnecessary as its specified by the first two
(1)ie x+y=1 if and only iof F(x)+F(y)=1 F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1
(2)F homomorphic(2) x+y+z=1 if and only F(x)+F(y)+F(m)= 1
(3) x+y+z+m=1; if and only ifF(x)+F(y)+F(z)+F(M)=1
Give F(x)=x and F continuous (it appears to entail cauchys equation with F(1)=1 over ethe unit triangle due tot he common term). F(x+y)+F(z)=1, F(x)+F(y)+F(z)=1, etc, F(x+y)=1-F(z)=F(x)+F(y)
.
I
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t appears that the third equation is redundant and I presuming any further equation where they sum to one (any given four, any given five, which sum to one is derivable from the first two) at least given that the first two entail some form of cauchy additivity. Is that correct
e
the above equation for any (4)given four points in the domain which sum to one
Does this generalize to Cauchy equation over the unit triangle. Ithough that one had to show that it held for any arbitary number of components not just 2 and three (although it appeas that the equations just generalize
F:[0,1] to [0,1]
F(1)=1 F(0)=1
F strictly monotonic increasing
(1)ie F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1
(2)F homomorphic(2) x+y+z if and only F(x)+F(y)+F(m)= 1
(3)x+y+z=2 if and only if F(x)+F(y)+F(z)=2, and th
(4) x+y+m+z=1 if and only F(x)+F(y)+F(z)+F(M)=1
Are the latter two redundant, ie do the first two generalize to any arbitrary factor; as the first two appears to entail cauchy equations of the unit trial
F(x+y+M)+F(Z)=1 F(Z)+F(x+y)+F((M)=1
so F(X+y+M)=F(x+y)+F(M)
but also x+y+z+m=1 entails
((x+y)+(1-x+y)) entails
F(x)+F(Y)+F(1-x+y) by (1) and (2)
F(x+y)+F(1-x-y)=1, so  by cancelling out common terms F(X)+F(Y)=F(x+Y)
and so
F(x+y+M)=F(x+y)+F(M)=F(X)+F(y)+F(M)
z+ (x+y+M)=1 entials F(z)+F(x+y+M)=1 so F(x+y+m)=1-F(z)
so 1-F(Z)=F(X+Y)+M)=F(X)+F(Y)+F(M), and so F(X)+F(Y)+F(Z)+F(M)=1 x+y+z+m=1, appears in addition to a form of cauchy equation s
These appear to entail those first two fixed pints
These appear to entail F(x+y)=F(x)+F(y), over x,y,(x +y)\in in [0,1],;
I thought one required that any for any n=2.........infinity which sum to one. But I presume this is entailed by these first both collectively working, as any given four example which sum to one can always be express as (x+y) +z+m = 1F(x+y)+F(m)+F(z)=1;
F(x+y+M)+F(Z)=1 so F(x+y+M)=F(x+y)+F(M) so -F(x+y)=-F(X+y+
F(M)+F(Z)=1-F(x+y)
F(x+y+m,)+F(Z)=1,
F(x+y+M)=F(x+y)+F(M)
yet 1-F(1-(X+y+M)=F(Z) as  so 1-F(1-x+
x+y+M+z=1 z=1-(x+y+M), then F((x+y+M))+F(z)=1, F(z)=1-F(x+y+z) so as z=1-x-m-y F(z)=F(1-(x+y-+M)=1-F(x+y+M)=F(z)
so 1-F(x+y+M)=1-F(x+y)-F(M), F(1-x+y+=F(M_M= ;1-F(x+y)=F(M)+F(Z)
(2) appears to just say this ie F(1-x-y)=1-F(x)-F(y)
which given 2 entails F(0)=0, F(1)=1; ie (1) specifies F(0.5)=0.5, and F(0)+F(1)=1 and F(0.5)+F(0.5)=1
(2) entails that 0 +0.5+0.5=1 so that F(0.5)+F(0.5)+F(0)=1,
so that 2 times 0.5+F(0)=1; 1+F(0)=1,which gives, F(0)=0
which by subsitution into into one gives F(1)=
and that (2)  F(1/3)=1/3 however when one has both
Moreover if(A) (x+y) +z =1 then (B)x+y+z =1 so clearly by (1) on A, (2) on (B)
F(x+y)+F(z)=1 and F(x)+F(y)+F(z)=1,
F(x+y)+F(z)=F(x)+F(y)+F(z) but F(z) is common
so \forall (x)(y)\in [0,1] so long as x+y is in [0,1]F(x+y)=F(x)+F(y)
where F(1)=1 and F non negative and strictly monotone increasing. Is this a pair wise result. It appears to generalize to F(x)=x for all rationals as far i can see
ie F(10)=2F(20), as 10+10+80=1 so F(10)+F(10)+F(80)=1, so 2F(10)+F(80)=1 and by the first equation 20 +80=1 so F(20)+F(80)=1 so that
F(20)=2F(10)
and F(30)=F(10)+F(20) use 10+20+70=1=F(10)+F(20)+F(70)=1 and F(30)+F(70)=1 and one can 10 F(10)=1=0.1 and one can seeiming building this standard sequence, It appears to entail F(x)=x for all rations even x>0..5 and F(x+y) generalizes to arbitarily many components if x+y in [0,1[
Or is just a cauchy function on the restricted interval with with strictly monononitc and positive, on [0,1] to [0,1] then F(x)=x
forall(x,y)
They appear t
and otherwise x+y+z>1 then F(x)+F(y)+F(z)>1 and x+y+z<1 if and only if F(x)+F(z)+F(m)>1
x+y =1 iff F(x)+F(y)=1
where otherwise x+y>1 iff F(x+y)>1, and x+y<1 if and only F(x)+F(y)<1
x+y+z=1 if and only if F(x)+F(y)+F(z)=1
x+y+z>1 if and only F(x)+F(y)+F(z)>1; x+y+z+m<1 iff and only F(x)+F(y)+F(z)+F(M)<1
x+y+z+m iff F(x)+F(y)+F(z)+F(m)=1
otherwise x+y+m +z>1 if and only if F(x)+F(y)+F(M)>1
x+y+z=2 if and only F(x)+F(y)+F(z)=2
where x+y+z>2 if and only if F(x)+F(y)+F(z)>2
x+y+z<2 F(x)+F(z)+F(m)<2
where other x+y+z>z
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source modeling
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Some comments:
1. If  m  is a not random constant, than one can use the name  Poisson pd on the ordered set {0=0^m, 1=1^m, 2^m, 3^m, 4^m,. . . .}
2. The name  log-Poisson should be accordingly preferred for    e^x, instead;  indeed, y=e^x  is of log-normal distribution if x is normal.
3. If   m  is random, no particular names can be assignedalso, not any particular  distribution can be implied without information on the joint pd of  n  and  m
4. For the limit pd of sums like  a logx+b logx+c logx+d logx +...,  the CLT is applicable in very special circumstnces only, which in particular  have to take into account non-zero correlation between  the coefficients a,b,c,d,...and the rv  x in such a way, that the terms are iid;  this is due to the fact, that if  a,b,c,d,... are iid, then   a logx, b logx, c logx, d logx , ...  are not independent
Further comments are in prepartion:)
Best regards, Joachim
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I did analyzed these counts (about 100 C/S) by poisson function and gaussian function and i observed that, nearly there is no difference between using of a gaussian and poisson functions. 
Although, i know that: 1- gaussian function is used if the count rates are greater than 20 count 2- poisson function is used if the count rates are smaller than 20 count.
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@Deepak
Good catch. @Mahmoud said analyzed counts, but discussed count rates. One must be very careful with count rates. It is always better to do the analysis in counts. Many count rate formulas are incorrect or used incorrectly. The problem is combining count rates taken over different time periods.
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Bottleneck analysis software offers three models (IAM, TPM and SMM) and three tests (Sign, Standardized difference and Wilcoxon). IAM is known for use with allozymes, and TPM and SMM with microsatellites. The latter models under different tests above show presence of bottleneck while not so under another test.
For example; 
SIGN TEST
Assumptions: all loci fit T.P.M., mutation-drift equilibrium.
Expected number of loci with heterozygosity excess: 18.06
8 loci with heterozygosity deficiency and 22 loci with heterozygosity excess.
Probability: 0.09738
Assumptions: all loci fit S.M.M., mutation-drift equilibrium.
Expected number of loci with heterozygosity excess: 17.80
13 loci with heterozygosity deficiency and 17 loci with heterozygosity excess.
Probability: 0.45129
______________________________________________________________
STANDARDIZED DIFFERENCES TEST
Assumptions: all loci fit T.P.M., mutation-drift equilibrium.
T2: 2.072 Probability: 0.01913
Assumptions: all loci fit S.M.M., mutation-drift equilibrium.
T2: -0.802 Probability: 0.21140
_____________________________________________________________
WILCOXON TEST
Assumptions: all loci fit T.P.M., mutation-drift equilibrium.
Probability (one tail for H deficiency): 0.99017
Probability (one tail for H excess): 0.01042
Probability (two tails for H excess or deficiency): 0.02085
Assumptions: all loci fit S.M.M., mutation-drift equilibrium.
Probability (one tail for H deficiency): 0.59608
Probability (one tail for H excess): 0.41179
Probability (two tails for H excess or deficiency): 0.82358
___________________________________________________________
MODE-SHIFT
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
| 0.410 | 0.312 | 0.195 | 0.044 | 0.029 | 0.000 | 0.010 | 0.000 | 0.000 | 0.000 |
When different models under different tests produces different results on presence bottleneck as in Standardized difference and Wilcoxon test above. What is the bottom line of drawing the conclusion on presence or absence of bottleneck in a population? 
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Dear Dragos, Thanks for the article. Interesting and helpful indeed. 
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Dear all,
I am searching a result concerning the weak convergence of signed measures defined on the Borel algebra of a given metric space. The result that i find so far is established for weak convergence of probability measures (please see Portmanteau theorem page 11; book attached).
Does the same result apply for signed measures? If not, what could be the possible differences in the equivalent conditions?
(providing me with good references would be appreciated)
Sincerely,
Nathalie
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Dear All Followers of this question,
let me give answer to my subproblem by presenting a counterexample, where $\delta(a)$  stands for the p.d. concentrated at $a$, for $a \in R$:
For  $n \in Z_0=\{...,-2,-1, 1,2,3, . . . \}$  let  the coefficients be
$b_n : =  sign(n) \sqrt{ \pi |n|}$ and let the signed measures be
$Q_n := ( -\delta(b_{-2 n-1} +\delta(b_{-2 n} + \delta(b_{2 n} - \delta(b_{2 n+1} $.
Their characteristic functions equal  
$ f_n(t)  = 2 [  \cos(t \sqrt{2n\pi} - \cos(t \sqrt{(2n+1)\pi} ] $
By elementary trigonometry, $f_n(t) \to 0$,  for all $t \in R$.
On the other hand the integrals of the obviously continuous and bounded function $\cos(x^2), x \in \R$ with respect to $Q_n$ are all equal to
 $ \int_R \, cos(x^2) \,  dQ_n(x) = 2  [ \cos ((b_{2n})^2) -  \cos ((b_{2n+1})^2) $,
which equals 4, for every  $n \in  N$. Thus the measures $Q_n$ are not weakly convergent to the zero measure, although absolute variations of the measures are bounded (equal 4).
Thus, thank you Natalie for your question! Without it we could  miss this special feature of signed measures.
Joachim
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I was wondering if there are any set of n; n>=3continuous or somewhat smooth functions (certain polynomials), all of which have the same domain [0,1]
f^n(min,max):[0,1] \to[min,max]; where max>min, min, max\in [0,1] that have these properties
1.\forall v \in [0,1]\sum_n=1\toN(f^n(v)=1; for all values in the domain [0,1], the sum of the 'function values' =1at all points in this domain, and for all possible min and max values of the ranges of the n functions
2. The n functions have   range continuously between [min,max], max>=min, both in [0,1]; and will all reach their maximum at some point, but only once (if possible).
3. These functions must reach their global maximum value only  at the same point on the domain st,t the other distinct functions attain their global minimum value. These global minima for each fi,  may occur at multiple points in the domain [0,1] .
4. There is, for each point in the domain, under which any distinct function fi\neqfj,( fj j\neqi) reaches a global maxima, at  least one point (the same point) in the domain  s.t that  any fi attains its global  minima. Correspondingly, each function fi has (or at least) n-1 points (in the domain) under which it attains its global minima, if each of the n functions has a singular point in the domain under which it attains its global maxima (the n-1 points corresponding to the distinct points in the domain under which the n-1 distinct functions, fj\neqfi attain their global maxima .
That is, for each, of the (n-1) global maximum values of the range of the other functions, f^i;kneqi, where there are n functions  ; the  value of the domain in [0,1] when one function reaches its maximum, is the same value of the domain wherein other functions at which they all reach their absolute minimum . W
5. The max values of the range of said functions,  f^n= 1-\sum_{i;i\neqn}(min(ranf^i)). Ie the maximum value of the functions, f^i for all i\in{1,n} for n functions, is set by  1- sum (of the min function value/range of the other 'distinct' n-1 functions)
5. It must be such that whenever the minimum values are set so as to sum to one. ie min of any function= max of that same function for all such functions; that all  the functions values become a flat line (ie, ie min=max, for such functions)
6.This should be possible for all possible combinations of min values of all 'n' f's (functiions), So each combination of n, non negative values in [0,1] inclusive that sum to one. So one should be able to set the min value of any given value of a function to any value in [0,1], so long as the conjoint sum, sums to one. Likewise presumable if any such function is set such that min=max, all of them are, and these will mins will add to one.
It must be be such that these functions should have to change, to make this work.So that for all min, max range values for the n functions and all elements in their domain (which is common to all and is [0,1])  these function values \forall(n, min function values)\forall(v in [0,1])\sumf^i(v) sum to one
6. Moreover and most importantly they must also allow for the non trivial case where the function minimums do not sum to one (ie are not all flat lines); but such that the sum of the values of the function of all such functions sum to one for all values in their common domain ,v=[0,1](which just is their domain, as their domains;f^i:[0,1]->[min,max] are the same).
Ie so that the functions will range between a maximum and minimum values (which are not equal) in a continuous and smooth fashion (no gaps, steps or spikes).
In this case it must be such that it possible for the the sum of the minimum values, to sum to some positive value smaller then one, although not greater than one, or smaller then zero; for all or 'some' possible combinations (obviously values may not be possible. although, not in virtue of the sum of the functions at some point not being one, but because the max has been set to be smaller then the min for some function; that is the sum of the mins is a particular combination greater then one; (ie 7 function min=0.16....., so that max of fi=1-1=0<0.166=min fi; that is if they sum to more then one;
7. I was wondering if there for any n>3,4,5,6,7,,,,four sets of such sets of 3,4,5,6,7 functions that will do this.
And then given this; it must also be such that the function can be modified so that a function only ever has a minimum of 0 if the maximum is 0; without having to set th sum of the mins of the function to 1.
And likewise such that the function only ever reaches a max of one if its min is one, unless under certain circumstancess; unless that is, it is possible to do this,for the same reason; without doing this having to set the sum of the mins to one. If indeed one does want some of the functions to range between. If possible, Although, What is most important, is that if any function has a min and max set to one, all of the rest of the functions values sit at zero for all v in [0,1].
It also must be such that if the maximum range value for any given function f1 is larger than that of another function f2 in the set, then f1's minimum value will be larger then that of f2's minimum range value; and conversely.
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I agree peter, I apologize. Nonetheless, However,  i must insist,logic entails that one can leave A>T out of the conjunction; you already contradicted yourself (you already said 'can' ,NOT 'must').
It does not entail that one must leave it. Consequentially it does not entail it. What is necessarily possible, is not 'necessarily, necessarily necessary, and that which is entailed or necessary, are only statements which are Necessarily Necessarily
Secondly, I never said that I did not originally say it max>=min, when I later said A>B that was a correction.
Secondly I repeat, with reference even back to the original statement, max>=Min; that I never said it was NOT Tautologous, despite the fact that I DID SAY THAT IT SOUNDED TAUTOLOGOUS.
I did not say this for example,
" it only " merely" "sounded tautologous', so in what sense did I contradict what you said? ."
  I agree, that it is tautologous, and I meant that back then, then just as do now. With relation to what I said originally. The original claim 'It sounds tautologous' does not contradict it being nor being interpreted as saying 'it is tautological' in the strict sense.  technically speaking does not contradict the possible interpretation that it is, and that perhaps I  meantthat IT IS TAUTOLOGOUS.
In the  the literal sense, at least.
It only contradicts that claim if you apply an interpretation involving common sense and the maxim of pleasant  communication. Nonetheless I apologize i wont be like this in the future. I am just pointing out again and again, that back then, nor now, did I ever say, that I was not tautologous. Common sense, appears to give that implication though so I wont do that in the future
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Are there sets of three, or rather for any N>3  sets of  N sur-jective uniformly continuous functions, for all N>3, where n denotes the number of function in the sets, such that each function in the set has the same domain [0,1] and the same range [0,1],such that these  functions, in a given set, sum to one on all points in the domain [0,1]. ie \forallv\in[0,1] [\sum_{i=1}^{i=1n)fi(v)} =1 in [0,1] .Moro-ever, Are there arbitrarily many such functions for all n>=3.
By non trivial, I mean, -non linear, (and presumably not quadratic) functions which just so happen to be that that their sum give value 1, in [0,1] or perhaps for any domain, and not by way of the algebraic sum cancelling out to a constant 1 as in the case of x, 1-x (ie error correcting) .Presumably due to the nature of the derivatives
That is, so that the functions, would sum to one, regardless of the domain [0,1] or least for if the  domain [0,1] is held fixed, and the functions are weighted;  without it being a mere artefact that the algebraic sum to cancels out to give a constant,1,for any x; that is error-correcting functions. That is three (or n>3) functions all of which have a maximum range value of one, miniimum range value of zero, and which sum to one for all points in the domain [0,1]; and which are surjective and uniformly continuous, ie for every value in the range ri\in [0,1], these functions have some (at least one)  value, ci in the domain [0,1] such that f(ci)=ri such that takes on that .
Where these maximum values and min values for all fi coincide (the same element of the domain for which fi corresponds to 1, is such that other n-1/2 functions correspond to zero etc); where obviously these three/n max points such that fi(c)=1,(one, and only one for each function, fi,i1<=i<=n for n>=3, such functions) are distinct (correspond to distinct elements of the domain), although they correspond to the same element of the domain for for which each of the other 2/n-1 are at their minimum value (ie 0)
So for n=3 there is one one max point for each function, and at least , and presumably only, two/n-1 min points), s.t, when one functions fi(c) hits 1, fi(c)=1,the other 2/n-1 functions fj(c),jneqi take the value 0, fj(c)=0 for all j.And are there at least two such such sets for all sets of N>=3 such functions. So there are  three distinct  domain points in [0,1] corresponding to a n/3-tuple of function values,f1,f2,f3(c)= <1,0,0>,,f1,f2,f3(c1)=<0,1,0>f1,f2,f3(c2)=,<0,0,1>; corresponding the three distinct elements/points in the domain [0,1] , c\neqc1\neqc2, c,c1,c2,\in  [0,1] . Where the first function is at its maximum value here 1, and the other two are their minimum, zero at c, the second is at its maximum at c1 (1), and first and third are a their mimimum (zero here) at c1 etc
And are continuous  (no gaps, and uniformly continuous, ie no spikes).
As one could not make use of these such functions, if one wanted them to be weighted otherwise; if said functions have sums which either (A) as just cancel out to be constant, or two (are such that it their sums are do not cancel out to a constant, but just so happen to line up because the domain is [0,1]
Likwise, the functions a similar form; so that one does not want, one having two maximums whilst the other two for example have one maxima, and two minima. Perhaps Berstein polynomials could be so weighted, but I do not know; the linear forms cancel out but their weighted bezauir forms seem to be a little unstable from what I have read.
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I passed through
  1. Pythagorean theorem in 2D,3D,...nD [or norm of unit vectors in Rn]
  2. unit circle, unit sphere, unit sphere in Rn
  3. transformation to generalized spherical coordinates in Rn
A set of uniformly continuous surjective functions f1,...,fn from [0,1] to [0,1] is enclosed.
In particular, for n=3:  
f1(x) = cos2(2pi.x).cos2(4pi.x),
f2(x) = cos2(2pi.x).sin2(4pi.x),
f3(x) = sin2(2pi.x).
The construction shows that many other convenient sets of functions exist.
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Let's say the initial population in NSGA ii is  i.i.d and uniformly distributed. Has anyone done research about what we can say about the distribution after k iterations in NSGA ii? The individuals are surely no longer i.i.d but are there asymptotic results for large populations sizes?
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Dear Ernst,
You can utilize the following codes if you use MATLAB as follows:
before merging the population at the end of algorithm, please type "Stop time;"
and save the initial population in another Pop, e.g. Pop2.
Yours,
Niknamfar
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Dear researchers,
How can be calculated the failure probability for a component with λ=1*E-6 failure per hour and a month (τ=1 month) proof test interval at one year؟
Please see the following files.
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Dear Dr Limon,
Thanks.
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Which equation (14 or 15) in the paper (attached in the link) gives us the average outage probability?
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Outage probability is the probability that a system does not function as per the expectations, such as the probability when the instantaneous SNR falls below a threshold SNR. But if the SNR considered is received SNR, then that received SNR which depends on the wireless channel could be considered random. So, we need to average out the received SNRs for various channel realizations. Hope this helps. 
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I have read that a drawback with Edgeworth series expansion is that "... they can be inaccurate, especially in the tails, due to mainly two reasons: (1) They are obtained under a Taylor series around the mean. (2) They guarantee (asymptotically) an absolute error, not a relative one. This is an issue when one wants to approximate very small quantities, for which the absolute error might be small, but the relative error important."
So my resulting question is if there are any attractive alternative ways of approximating stochastic variables with some corresponding method that still is useful in the tails of the distribution, and does not (for example) result in negative probabilities (which is mentioned as another drawback with the approach).
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Another criticism of the Edgeworth (and Gram-Charlier) series is that they do not always (usually?) converge--they are similar to asymptotic expansions).
Instead of expanding the density, you could expand a function that transforms it to normality. The Cornish-Fisher expansions (for such a transformation and its inverse) achieve this. These are based on the Edgeworth expansion. They might also fail to converge but still give approximations.
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Let {A_n} be a sequence of sets, limsup A_n=, liminf A_n and lim A_n are defined in probability theory.
I would like to know who introduced this notions and find some papers on this context!
Thanks so much!
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Hello,
I think there is not a unique answer to your question. However :
- Fréchet in his thesis in 1906 introduced the bases for topology over general sets and went on with several works on abstract sets in the following years
- the liminf definition may be due to Borel in his 1909 paper introducing the strong law of large numbers by means of (today's name of course) Borel-Cantelli lemma expressing that only a finite number of events in a sequence take place.
Sincerely,
Laurent
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If I have two independent variables (time series) for two different sources(let's say from Observation and from Model).
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Thank You Juan and Shameek.
Juan your chapter may help in my work . Actually I want to use this technique for rainfall forecasting for monthly and Seasonally.
Thank you again 
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I would like to study the problem of insolvency in the insurance companies and what I thought about is classical probability models where I can examine the probability of the ultimate ruin ?
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Bear in mind, however, that in the real world ruin often occurs not because a threshold of extreme value was exceeded, but because the model used to estimate the capital requirements was insufficient.  For example, a model may assume that risks in different regions or different sectors are independent and can be used to hedge each other, but situations will always exist where the risks are correlated, with one event then resulting in losses over many sectors or regions.
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significance of gender as dummy variable
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Additionally, although this might be practically unimportant in your case, if you have no other IV, regressing the outcome variable only on gender provides you with the average outcome of the base category of your gender variable with the intercept of the model, and the difference to that intercept with the regression coefficient of your gender variable.
y = 3  - 1*Gender would mean that for your base category of gender (Gender = 0), the average of your outcome variable is 3, and for Gender = 1, the average of the outcome variable is 2 (3 - 1). Regression tells you whether this difference is significant.
If you have other control variables, this would be a bit more complex. See here for a very easy and basic introduction: http://www.theanalysisfactor.com/interpreting-regression-coefficients/
Hope this helps.
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Hi everybody,
I'm going to apply multiDE, an R package for detection of Differentially Expressed Genes in multiple treatment conditions, on some RNA-seq time series data... (I want to assume each time point as a treatment condition)
Let's assume Yidg denotes the normalized read counts for sample "i", in condition "d" for gene "g".
We also assume that Yidg marginally follows the negative binomial distribution with expectation "μdg" and dispersion parameter "ϕg" (i.e., the variance of Yidg is μdg+ϕgμ2dg).
The statistical methodology behind this package is a two factorial log linear model : logμdg = μ + αd + βg + γdg = μ + αd + βg + UdVg,
where μ is the grand mean, αd is the main effect for condition d, βg is the main effect for gene g, and γdg:=UdVg is the interaction effect between gene g and condition d.
My professor has asked me to estimate the main effect for condition (α), the main effect for gene (β) and the effect of interaction between gene and condition (γ). While the package can only show "Ud" in its output...
I'm in grave need of help to find out how I can estimate those effects please...
My main problem is I don't know how I can calculate μdg. Maybe if I can calculate it, then applying a regression strategy would be helpful to estimate those effects...
here it is the link to the full paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4917940/
Thanks in advance
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To reply in detail to your question would require that I download your paper and examine in detail what you would like to do.
In general the application of log-linear models is well know and excellent books are available, so just chose one of them  and pursue your research.
But, it seems, in my experience see in my directory the paper
Article: Solving large protein secondary structure classification problems by a nonlinear complementarity algorithm with {0, 1} variables
C. Cifarelli · G. Patrizi
Full-text Article · Feb 2007  and the  Optimization Methods and Software  that you need a log-linear model to estimate categorical effects such as the ones that you mention.
In this case, there are three excellent books, a bit dated but surely the best, written by Shelby Haberman, especially the 2 volume work entitled " Analysis of Qualitative Data" , which your University Library can borrow from my department if it is not available in your nearby university libraries.  Perhaps you can use also books on categorical analysis by Alan Agresti and there should be the necessary software also available in R.
I will look into your problem very willingly if I am given some time to study your paper supposing that this research probably concerns your thesis or a joint paper.
Looking forward to your answer,
Giacomo Patrizi
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Well! Generalized (G-) classes with more than 2 parameters are around. The questions is "the model with more than 3 shape parameters is worthy?" Many statisticians have on objection on fitting models with more than 3 parameters. Well! we all should think about it. Especially the recent work of Morad Alizadeh, Ahmed Afify and Laba Handique. The special models contain 4 or more parameters . Look what is happening with modern distribution theory! You can fit an elephant with 4 or 5 parameters. A very serious concern.
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In regression analysis we consider a large number of parameters to limited size data and nobody bothers about it.Why so much concern when a distribution contains three shape parameters?.The larger question in both cases is how successfully we can deal with the estimation problem
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Given a set of values of a variable x= 40, 25, 20, 15, 12, and 5:
The probability of exceedance of a value x=20 written as  P (x>=20)  can be got by arranging data is descending order thus giving a value of 0.5
The probability of non-exceedance of a value x=20 written as  P (x<=20) and can be got by arranging data is ascending order thus giving a value of 0.67
On checking P(x>=1) = 1 - P(x<=1) gives
0.5 = 1 - 0.67 which is not correct. Is this error creating by the estimations made using the probability formulae?
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First of all the Correct fact is: P(x>=1) = 1 - P(x<1)
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I would like to capture the dependence between returns using regime switching copulas and I'd like to know if there is any code currently available.
More in details, I would like to estimate the maximum likelihood estimates using the EM algorithm displayed in Hamilton in particular. In the framework, we consider two states of the economy, each one characterized by a certain copula and dependence parameter.
Thank you very much in advance.
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Code is available for R software. You need to edit for your specific use.
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When I use adequacy Model to fit the new probability model by use appropriate data sets.
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Are you running some sort of nonlinear optimization?
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I have considered Hydraulic conductivity (Ks) of clayey soil as random variable with log normal distribution. I have got negative mean (lambda) after the determination of measures of variation. Logically, I should not have negative mean for physical parameter Ks.. Find the attached excel document.. Kindly provide solution as soon as possible..
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Yes, it is possible to have a negative value for lognormal mean. The main purpose of using a lognormal distribution for probabilistic analysis is to have only positive values assigned to the variables (engineering properties like hydraulic conductivity). The lognormal of the variables is normally distributed however, you would be interested in assigning the corresponding normally distributed value to the hydraulic conductivity, and in that case you will happen to raise the lognormal mean to the exponent function rendering positive values only (exponent function is always positive). Hope this helps.
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Hello! I am only starting to study GRP (General Renewal Process), so sorry, if my question will be too simple -:). Main expression for GRP is A[i] - A[i-1] = q*(S[i] - S[i-1]), where A is Virtual Age, S is Real Time and q is restoration factor. It is clear (???), that this formula also proved for non-constant restoration factor q[i] – e.g., q[i] = q_CM for Corrective Maintenance and q[i] = q_PM for Preventive Maintenance. But I see some collision, if sometimes q[i] = 0 (e.g., q_PM = 0). On the one hand, q[i] = 0 means, that it is replacement (Good as New) and in this case the Aging is absent, i.e. A[i] = 0. On the other hand, according above formula, A[i] = A[i-1] and it isn't 0, because q[i-1] =/= 0. What is your opinion?
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Why we are using the (characteristics function) CHF for evaluating the (probability density function) PDF of any random variable, why not directly evaluate PDF for random variable..
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It is not necessary to use CHF .  For example if you know the distribution function (DF) then  its derivative is the PDF.
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I am working on simply supported beam, please let me know any hint or idea?
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MATLAB can help you building a PDF graph. You can do the same with Python using commonly known libraries; although there many software choices. You can read about statistics and PDF functions in the following books:
J. P. Marques de Sá, "Applied Statistics using SPSS, STATISTICA, MATLAB and R", 2007.
Andrew N. O’Connor,"Probability Distributions Used in Reliability Engineering", 2011.
Good Luck!
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If the homogeneity of regression slopes assumption for ANCOVA and levene’s test was violated (significance ), what is the alternative test for ANCOVA? I have post-test mean scores as dependent, pre-test scores as covariates, one independent variables with two treatment modes, and one moderator variables. Is it Welch’s test suitable for ANCOVA?
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It is not an alternative actually but you could run a linear regression and check for assumptions of it.
Or you could make use of a mixed model allowing for non-homogenous variances.
Another one might be going on with ANCOVA despite violations.