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A characterization of the arcsine distribution
Karl Michael Schmidt∗and Anatoly Zhigljavsky∗
Abstract. The following characterization of the arcsine density is established: let ξ be a r.v. supported
on (−1,1), then ξ has the arcsine density p(t) = 1/(π√1 − t2), −1 < t < 1, if and only if Elog(ξ − x)2
has the same value for almost all x ∈ [−1,1].
1 Introduction
The arcsine density on the interval (−1,1) is
p(t) =
1
π√1 − t2,
−1 < t < 1.
(1)
To define a r.v. ξ with the arcsine density (1) we can use the formula ξ = cos(πα), where α is
a r.v. with uniform distribution on (0,1). The arcsine density has several non-trivial appearances in
probability theory and statistics. For example, for a general random walk {Sn} satisfying the Lindeberg-
L´ evy condition, the limiting distribution of
n
see §11 in Billingsley (1968), Erd˝ os and Kac (1947), L´ evy (1948). The arcsine density (1) is an invariant
density for a number of maps of the interval (−1,1) onto itself, see e.g. Rivlin (1990), Theorem 4.5. This
density is the limiting density of the roots of the orthogonal polynomials which are defined on (-1,1) and
orthogonal with respect to any weight function w(·) continuous on (−1,1), see Ullman (1972), Erd˝ os and
Freud (1974), van Assche (1987).
In probability theory, the arcsine density has a number of characterizations, see Norton (1975, 1978),
Arnold and Groenveld (1980), Kemperman and Skibinsky (1982). Below we consider a characterization
of the arcsine density that is of a different nature than the ones considered in these papers.
Our main result is as follows.
1
?n
i=11[Si>0](as n → ∞) has the arcsine density on (0,1),
Theorem. Let ξ be a r.v. supported on (−1,1). This r.v. has the arcsine density (1) if and only if
Elog(ξ − x)2has the same value for almost all x ∈ [−1,1].
As a motivation for the above theorem, assume that we have a sequence of points x1,x2,... in the interval
(−1,1) that has an asymptotic c.d.f. F(·) in the sense that
1
k
j=1
?|h(x)|dF(x) < ∞. Consider an associated sequence of
the values of the normalized ratios Rk(x,y) = [Hk(x)/Hk(y)]1/ktend to 1 (as k → ∞) for almost all
x,y ∈ (−1,1) if and only if the c.d.f. F(·) has the arcsine density (1). Indeed,
logRk(x,y) = log[Hk(x)]1/k− log[Hk(y)]1/k=1
lim
k→∞
k
?
h(xj) =
?1
−1
h(t)dF(t) (2)
for any continuous function h(·) such that
polynomials Hk(x) = (x − x1)2(x − x2)2···(x − xk)2. Then the result of the Theorem implies that
k
k
?
j=1
log(x − xj)2−1
k
k
?
j=1
log(y − xj)2.
∗School of Mathematics, Cardiff University, Senghennydd Road, Cardiff, CF24 4YH, UK (SchmidtKM@cf.ac.uk,
ZhigljavskyAA@cf.ac.uk)
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Using (2), for almost all x,y ∈ [−1,1] we obtain
?
The theorem implies that the r.h.s. of (3) is zero for almost all x,y ∈ [−1,1] if and only if the c.d.f. F(·)
has the arcsine density (1).
The fact that Rk(x,y) → 1 (as k → ∞) for almost all x,y ∈ (−1,1) means that the ratios Hk(x)/Hk(y)
are almost never very large (these ratios are smaller than δkwith any δ > 1 for sufficiently large k:
k > k∗(x,y)) and very rarely are very close to 0 (they are larger than any δkwith any δ < 1 and
k > k∗(x,y)). Note that if k is fixed and xj= cos(π(2j − 1)/(2k)) for j = 1,...,k, then Hk(x) = ckT2
where ckis some constant and Tk(x) = cos[karccos(x)] is the k-th Chebyshev polynomial. In this case,
the fact that Rk(x,y)∼= 1 (as k is large) for typical x,y ∈ (−1,1) follows from the properties of the
Chebyshev polynomials.
loglim
k→∞Rk(x,y)
?
= lim
k→∞logRk(x,y) =
?1
−1
log(x − t)2dF(t) −
?1
−1
log(y − t)2dF(t).
(3)
k(t)
2Auxiliary statements and proofs
The proof of the theorem is based on three Lemmas. In Lemma 1 we observe that the expected value
of log(ξ − x)2is finite for almost all x ∈ [0,1]. In Lemma 2 we derive a specific characterization of
the uniform measure on the interval [0,π]. To prove Lemma 2 we use a general characterization of the
Lebesgue measure on the interval [0,π] established in Lemma 3.
Lemma 3 uses Fourier series, which may seem surprising but is a natural reflection of the intrinsic
relationship between the arcsine distribution and trigonometric powers, as apparent in the Chebyshev
polynomials.
Lemma 1. For any r.v. ξ supported on (−1,1), the expectation Elog(ξ − x)2is finite for almost all
x ∈ [−1,1].
Proof of Lemma 1. Let F(·) be the c.d.f. of the r.v. ξ and −1 < t < 1. The integral
?1
is bounded and continuous as a function of t, so the integral
?1
exists. By the Fubini-Tonelli theorem,
?1
so in particular it is finite for almost all x ∈ [−1,1].
−1
log(t − x)2dx = (1 + t)log(1 + t)2+ (1 − t)log(1 − t)2− 4
−1
?1
−1
log(t − x)2dxdF(t)
Elog(ξ − x)2=
−1
log(t − x)2dF(t) ∈ L1([−1,1]),
Lemma 2. Let ϕ be a r.v. distributed according to a probability measure µ(·) on [0,π]. Then the
expectation
Ex(µ) = Elog(cosϕ − x)2
is constant for almost all x ∈ [−1,1] if and only if the measure µ(·) is uniform on [0,π]; in this case, the
expectation Ex(µ) has the same value for all x ∈ [−1,1].
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Proof of Lemma 2. As x ∈ [−1,1], we can set ψ := arccosx ∈ [0,π]. Let us extend µ to [0,2π] as an
even measure (that is, we set µ(A) = µ(2π−A) for all Borel sets A ⊂ [π,2π]) and note that µ([0,2π]) = 2.
Using cosϕ = cos(2π − ϕ) for all ϕ ∈ R, we calculate
Ex(µ) =1
2
0
2
=1
2
00
2
?2π
log(cosϕ − x)2µ(dϕ) =1
?2π
?2π
0
log
?
2sinϕ − ψ
2
sinϕ + ψ
2
?2
µ(dϕ)
??2π
2 log2µ(dϕ) +log
?
sinϕ − ψ
?2
µ(dϕ) +
?2π
0
log
?
sinϕ + ψ
2
?2
µ(dϕ)
?
= 2 log2 +
?π
0
log?sin2(ϕ − ψ/2)?
˜ µ(dϕ) +
?π
0
log?sin2(ϕ + ψ/2)?
˜ µ(dϕ),
(4)
where ˜ µ(A) =1
As ˜ µ and sin2are π-periodic and even, we obtain by making the substitution ˜ ϕ = π − ϕ:
?π
This implies that the two integrals in (4) are identical and therefore
?π
Hence the expectation Ex(µ) is constant for almost all x ∈ [−1,1] if and only if
?π
The Fourier series for logsin2is not uniformly convergent, but it converges in the L2([0,π]) sense, as
logsin2∈ L2([0,π]) and {e2ikx|k ∈ Z} is an orthonormal basis of this Hilbert space. Moreover, all
(complex) Fourier coefficients of logsin2are real and non-zero. Indeed,
?π
and
?π
see formula 4.384.3 in Gradshtein and Ryzhik (1965). The statement of Lemma 2 now follows from
Lemma 3 below.
2µ(2A) for all Borel sets A ⊂ [0,π].
0
log?sin2(ϕ + ψ/2)?
˜ µ(dϕ) =
?π
0
log?sin2(π − ˜ ϕ + ψ/2)?
˜ µ(d˜ ϕ) =
?π
0
log?sin2(˜ ϕ − ψ/2)?
˜ µ(d˜ ϕ).
Ex(µ) = 2log2 + 2
0
log?sin2(ϕ − ψ/2)?
˜ µ(dϕ).
logsin2? ˜ µ(y) =
0
log?sin2(ϕ − y)?
˜ µ(dϕ) is constant for almost all y ∈ [0,π].
(5)
0
log?sin2(ϕ)?sin(2kϕ)dϕ = 0 ∀k ∈ Z
0
log?sin2(ϕ)?cos(2kϕ)dϕ = 2π
?1
0
log(sin(πt))cos(2πkt)dt =
?
−2π log2,
−π/k,
k = 0
k ∈ Z \ {0},
Lemma 3. Let ˜ µ be a probability measure on [0,π] and f ∈ L2([0,π]) be such that
f(x) = l.i.m.N→∞
N
?
k=−N
θke2ikx
(x ∈ [0,π])
where all Fourier coefficients are non-zero: θk ?= 0 ∀k ∈ Z. Then, extending f to R as a π-periodic
function, the convolution f ? ˜ µ(·) :=?π
0f(· − t) ˜ µ(dt) is constant almost everywhere if and only if ˜ µ is
the uniform measure on [0,π]; in this case, f ? ˜ µ is constant everywhere.
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Proof of Lemma 3. Assume
f ? ˜ µ(x) =
?π
0
f(x − t) ˜ µ(dt) = C = const (for almost all x ∈ [0,π]).
Then, for all k ∈ Z \ {0},
?π
0=
0
e2ikxCdx=
?π
0
e2ikx
??π
N
?
0
f(x−t) ˜ µ(dt)
?
dx=
?π
0
??π
?
0
e2ikxl.i.m.N→∞
N
?
N
?
n=−N
θne2in(x−t)dx
?
˜ µ(dt)
=
?π
0
lim
N→∞
n=−N
θne−2int
??π
0
e2i(k+n)xdx
˜ µ(dt) =
?π
0
lim
N→∞
n=−N
θne−2intπ δn,−k˜ µ(dt)
?π
= πθ−k
0
e2ikt˜ µ(dt).
As θ−k?= 0 ∀k ∈ Z \ {0} we get?π
?π
as shown above, so?π
0e2ikt˜ µ(dt) = 0 ∀k ∈ Z \ {0}.
?π
0e2iktµ?(dt) = 0 ∀k ∈ Z. As every continuous function on [0,π] can be uniformly
approximated by a linear combination of {e2ikt|k ∈ Z} and µ?is finite, this implies
?π
and hence µ?= 0. This completes the proof of the ‘only if’ part of Lemma 3. The converse is obvious,
bearing in mind that f is π-periodic and f(x − ·) ∈ L1([0,π]) for all x ∈ R.
Now set µ?= ˜ µ − ˜ µ([0,π])λ/π, where λ is the Lebesgue measure on [0,π]. Then
µ?(dt) = 0 and
00
?π
e2iktµ?(dt) =
0
e2ikt˜ µ(dt) −˜ µ([0,π])
π
?π
0
e2iktdt = 0, ∀k ∈ Z \ {0}
0
f(t)µ?(dt) = 0 ∀f ∈ C([0,π])
Proof of the Theorem. Consider the expectation
Ix= Elog(ξ − x)2=
?1
−1
log(t − x)2
π√1 − t2dt,
(6)
where r.v. ξ has the arcsine density (1). By changing the variable t = cosϕ we obtain
?π
where µ0is the uniform measure on [0,π]. Hence, by applying Lemma 2 with µ = µ0, we conclude that
Ixhas the same value for all x ∈ [−1,1]. This proves the ‘only if’ statement in the Theorem.
To complete the proof of the Theorem, we now show the converse, i.e. that if, for a random variable
ξ supported on (−1,1), Elog(ξ −x)2has the same value for almost all x ∈ [−1,1], then ξ has the arcsine
density. In view of Lemma 1 the constant value of Elog(ξ −x)2must be finite. Denote by F(·) the c.d.f.
of ξ. Then F(−1) = 0, F(1) = 1 and
?1
where t = cosϕ and˜F(ϕ) = 1 − F(cosϕ). By Lemma 2, the probability measure generated by˜F is
uniform on [0,π]; that is,˜F(ϕ) = ϕ/π ∀ϕ ∈ [0,π]. This implies
F(x) = 1 − (arccosx)/π ∀x ∈ (−1,1),
so the density of ξ is F?(x) = 1/(π√1 − x2).
Ix=
0
log(cosϕ − x)2
π sinϕ
sinϕ dϕ =1
π
?π
0
log(cosϕ − x)2dϕ = Ex(µ0),
(7)
Elog(ξ − x)2=
−1
log(t − x)2dF(t) =
?π
0
log(cosϕ − x)2d˜F(ϕ),
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3Explicit formulae for the integrals and a generalization
3.1Explicit formulae for the expectations
The value of the expectation (6) can be easily computed based on our result that it is independent of x
in the interval [−1,1].
Corollary 1. Let the r.v. ξ have density (1). Then
?
Ix= Elog(ξ − x)2=
−2log2
2log?|x| +√x2− 1?− 2log2
if |x| ≤ 1
if |x| ≥ 1.
(8)
Proof. For |x| ≤ 1 we use Ix= I0= −2log2 by evaluating the integral I0:
Ix= I0=1
π
0
?π
log?sin2(ϕ)?dϕ =2
π
?π
0
log(sinϕ) dϕ = −2log2.
(9)
Let now x ≥ 1. From (9) we have I1= −2log2. Differentiating Ixwe get
??1
(see Gradshtein and Ryzhik (1965) 3.121.2 — note that interchanging the differentiation and integration
is justified as the derivative of the integrand is bounded by an integrable function of t locally uniformly
in x, |x| > 1). Therefore, for x > 1,
?x
Combining (9) and (10) we obtain (8).
I?
x=
−1
log(x − t)2
π√1 − t2dt
??
=
?1
−1
2
π(x − t)√1 − t2dt = 2
?1
0
ds
π(x + 1 − 2s)?s(1 − s)
=
2
√x2− 1;
I−x= Ix= I1+
1
I?
zdz = −2log2 +
?x
1
2
√z2− 1dz = 2log
?
x +√x2− 1
2
?
.
(10)
3.2Arcsine density on an arbitrary interval
The arcsine density on an interval (a,b) is
p(t) =
1
π?(t − a)(b − t), a < t < b.
(11)
If a = −1 and b = 1 then (11) is reduced to (1). A simple change of variables generalizes Theorem 1 to
the following statement.
Corollary 2. Let −∞ < a < b < ∞ and let ζ be a r.v. supported on the interval (a,b). The r.v. ζ has
the arcsine density (11) if and only if Elog(ζ − z)2has the same value for almost all z ∈ [a,b].
Corollary 1 is generalized as follows.
Corollary 3. Let −∞ < a < b < ∞ and let r.v. ζ have density (11). Then
?
Elog(ζ − z)2=
2log(b − a) − 4log2
2log(b − a) + 2log
if a ≤ z ≤ b
if z < a or z > b,
?
|xz| +?x2
z− 1
?
− 4log2
(12)
where xz= −1 + 2(z − a)/(b − a).
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Proof. By changing variables t = −1 + 2(u − a)/(b − a) and x = −1 + 2(z − a)/(b − a) in the integral
?b
we obtain Elog(ζ −z)2= 2log(b−a)−2log2+Ix, where Ixis defined in (8). This immediately implies
(12).
a
log(u − z)2
π?(u − a)(b − u)du = Elog(ζ − z)2,
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