Jaume Sastre Tomas

• Telecomunications Engineer
• PhD Student at University of the Balearic Islands

Hi again.

I have the following problem:

Let say  I have two random variables X, Y, discrete, each with three possible values, a,b and c.

I would like to test if in a sample of 500 objects, the two values X and Y are independent in a special way:

H_0:  Pr(X=x ^ Y=y) <= Pr(X=x) Pr(Y=y)  for each x,y = a,b,c

I use as a statistic s(T') the one in the chi squared test but in the contingency table T' of 9 cells, I add the discrepancy only if Observed_ij > Expected_ij.

In order to calculate p-values, I calculate the distribution of s(T') over a million random tables, each with the same marginal frequencies of T'.

To do so, I generate a random permutation of the values of X on the  500 objects, and another random vector with the frequencies of Y,

calculate the contingency table T' and s(T'), over a million T'

Then, the distribution of s(T') is used to calculate p-values.

But I realize that the null hypothesis H'_0 for the Montecarlo procedure is  that X and Y are independent which  is not exactly what I need.

Fortunately, for a fixed r, the number of times that there is a T' where  s(T') >= r, when H'_0: Pr(X=x ^ Y=y) = Pr(X=x) Pr(Y=y), is greater than

the number of times that there is a T' where s(T') >= r, when H_0: Pr(X=x ^ Y=y) <= Pr(X=x) Pr(Y=y), because, intuitively,  in H_0 there are more chances that the table have more cells with Observed <= Expected, that are not counted in the sum for s(T').  

Therefore, Pr(s(T')>=r | H'_0) >= Pr(s(T') >= r | H_0).

So, if I have a table T, and I calculate s(T) = r, I know that if Pr(s(T')>=r | H'_0) <= \alpha  then  Pr(s(T') >= r | H_0) <= \alpha.

Therefore, the Montecarlo distribution of s(T) under H'_0 is useful for rejecting H_0 for T. However, it is conservative.

Is this procedure correct? How can it be done better?

Thanks in advance, again.

I have a list of genes (and their interactions) and I would like to see how these genes are located in a specific network.

I would appreciate any help!

Thanks.

I have two discrete variables with three values each, A,B, and C for one, and D,E, and F, for the other (for around 500 objects).

I am only interested to reject independence in three cells of the nine cells in the table,(A,D),(A,E), (B,D).

So I have three null hypothesis:

H_AD :  Pr(AD) <= Pr(A) P(D)

H_AE :  Pr(AE) <= Pr(A) P(E)

H_BD :  Pr(BD) <= Pr(B) P(D)

which also say that for those cells, the observed values are lower or equal than the expected values according to the marginals. If the null hypothesis can be rejected, then we have that, on at least of those cells, the observed counts are greater than the expected, with some confidence.

So what I do is to use the usual chi squared test, but instead of adding over all 9 cells, I sum over only 3 cells and only if the

observation counts are above the expected ones. Then, I calculate the p-value using the chi squared distribution (one tail) with 4 degrees of freedom since all table cells still affect the answer.

Is it correct?

I have a small dataset (N=400 samples). And I want to apply Fisher exact test because I my contingency table has a lot of 0s but I have read that Fisher Exact test is only for N≤90 so I returned to Chi2-test but I read that the chi-square test is performed only if at least 80% of the cells have an expected frequency of 5 or greater, and no cell has an expected frequency smaller than 1.0, which is not my case.

Which test do I have to use?

Any response might be fine!

Thanks in advance.