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STAT 709: Exercise
Exercise 1.34. Let ν be a σ-finite measure on a σ-field F on Ω, λ be another measure with λ ø ν, and f be a nonnegative Borel function on Ω. Show that (^) ∫
f dλ =
∫ f
dλ dν
dν,
where dλdν is the Radon-Nikodym derivative. Note. Two measures λ and ν satisfying λ ø ν if ν(A) = 0 always implies λ(A) = 0, which ensures the existence of the Radon-Nikodym derivative dλdν when ν is σ-finite (see Shao, 2003, §1.1.2). Solution. By the definition of the Radon-Nikodym derivative and the lin- earity of integration, the result follows if f is a simple function. For a general nonnegative f , there is a sequence {sn} of nonnegative simple functions such that sn ≤ sn+1, n = 1, 2 , ..., and limn sn = f. Then 0 ≤ sn dλdν ≤ sn+1 dλdν and limn sn dλdν = f dλdν. By the monotone convergence theorem (e.g., Theorem 1. in Shao, 2003),
∫ f dλ = lim n
∫ sndλ = lim n
∫ sn
dλ dν
dν =
∫ f
dλ dν
dν.
Exercise 1.34. Let Fi be a σ-field on Ωi, νi be a σ-finite measure on Fi, and λi be a measure on Fi with λi ø νi, i = 1, 2. Show that λ 1 × λ 2 ø ν 1 × ν 2 and d(λ 1 × λ 2 ) d(ν 1 × ν 2 )
dλ 1 dν 1
dλ 2 dν 2
a.e. ν 1 × ν 2 ,
where ν 1 × ν 2 (or λ 1 × λ 2 ) denotes the product measure of ν 1 and ν 2 (or λ 1 and λ 2 ). Solution. Suppose that A ∈ σ(F 1 × F 2 ) and ν 1 × ν 2 (A) = 0. By Fubini’s theorem,
0 = ν 1 × ν 2 (A) =
∫ IAd(ν 1 × ν 2 ) =
∫ (∫ IAdν 1
) dν 2.
Since IA ≥ 0, this implies that there is a B ∈ F 2 such that ν 2 (Bc) = 0 and on the set B,
∫ IAdν 1 = 0. Since λ 1 ø ν 1 , on the set B ∫ IAdλ 1 =
∫ IA
dλ 1 dν 1
dν 1 = 0.
Since λ 2 ø ν 2 , λ 2 (Bc) = 0. Then
λ 1 × λ 2 (A) =
∫ IAd(λ 1 × λ 2 ) =
∫
B
(∫
A
dλ 1
) dλ 2 = 0.
Hence λ 1 × λ 2 ø ν 1 × ν 2. For the second assertion, it suffices to show that for any A ∈ σ(F 1 × F 2 ), λ(A) = ν(A), where
λ(A) =
∫
A
d(λ 1 × λ 2 ) d(ν 1 × ν 2 )
d(ν 1 × ν 2 )
and
ν(A) =
∫
A
dλ 1 dν 1
dλ 2 dν 2
d(ν 1 × ν 2 ).
Let C = F 1 × F 2. Then C satisfies the conditions specified in Exercise 2. For A 1 × A 2 ∈ F 1 × F 2 ,
λ(A) =
∫
A 1 ×A 2
d(λ 1 × λ 2 ) d(ν 1 × ν 2 )
d(ν 1 × ν 2 )
∫
A 1 ×A 2
d(λ 1 × λ 2 )
= λ 1 (A 1 )λ 2 (A 2 )
and, by Fubini’s theorem,
ν(A) =
∫
A 1 ×A 2
dλ 1 dν 1
dλ 2 dν 2
d(ν 1 × ν 2 )
∫
A 1
dλ 1 dν 1
dν 1
∫
A 2
dλ 2 dν 2
dν 2
= λ 1 (A 1 )λ 2 (A 2 ).
Hence λ(A) = ν(A) for any A ∈ C and the second assertion of this exercise follows from the result in Exercise 2.
Exercise 1.36. Let Fi be a cumulative distribution function on the real line having a Lebesgue density fi, i = 1, 2. Assume that there is a real number c such that F 1 (c) < F 2 (c). Define
F (x) =
F 1 (x) −∞ < x < c
F 2 (x) c ≤ x < ∞.
Lebesgue measure of R^2 − (A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) is 0. On each Ai, the function
(y 1 , y 2 ) = (
√ x^21 + x^22 , x 1 /x 2 ) is one-to-one with
Det
( ∂(x 1 , x 2 ) ∂(y 1 , y 2 )
)
∣∣ ∣∣ ∣∣ ∣∣
√^ y^2 1+y^22
√^ y^1 1+y^22 −^
y 1 y 22 (1+y^22 )^3 /^2 √^1 1+y^22 −^
y 1 y 2 (1+y^22 )^3 /^2
∣∣ ∣∣ ∣∣ ∣∣
y 1 1 + y 22
Since the joint Lebesgue density of (X 1 , X 2 ) is
1 2 π
e−(x
(^21) +x (^22) )/ 2
and x^21 + x^22 = y 12 , the joint Lebesgue density of (Y 1 , Y 2 ) is
∑^4 i=
2 π
e−(x
(^21) +x (^22) )/ 2 ∣∣ ∣∣Det
( ∂(x 1 , x 2 ) ∂(y 1 , y 2 )
) ∣ ∣∣ ∣ =
π
e−y
21 y 1 1 + y^22
Since the joint Lebesgue density of (Y 1 , Y 2 ) is a product of two functions that are functions of one variable, Y 1 and Y 2 are independent.
Exercise 1.51, 1.53. Let X be a random n-vector having the multivariate normal distribution Nn(μ, In). (i) Apply Cochran’s theorem to show that if A^2 = A, then Xτ^ AX has the noncentral chi-square distribution χ^2 r (δ), where A is an n × n symmetric matrix, r = rank of A, and δ = μτ^ Aμ. (ii) Let Ai be an n × n symmetric matrix satisfying A^2 i = Ai, i = 1, 2. Show that a necessary and sufficient condition that Xτ^ A 1 X and Xτ^ A 2 X are independent is A 1 A 2 = 0. Note. If X 1 , ..., Xk are independent and Xi has the normal distribution N (μi, σ^2 ), i = 1, ..., k, then the distribution of (X 12 + · · · + X k^2 )/σ^2 is called the noncentral chi-square distribution χ^2 k(δ), where δ = (μ^21 + · · · + μ^2 k)/σ^2. When δ = 0, χ^2 k is called the central chi-square distribution. Solution. (i) Since A^2 = A, i.e., A is a projection matrix,
(In − A)^2 = In − A − A + A^2 = In − A.
Hence, In − A is a projection matrix with rank tr(In − A) = tr(In) − tr(A) = n − r. The result then follows by applying Cochran’s theorem (e.g., Theorem 1.5 in Shao, 2003) to
Xτ^ X = Xτ^ AX + Xτ^ (In − A)X.
(ii) Suppose that A 1 A 2 = 0. Then
(In − A 1 − A 2 )^2 = In − A 1 − A 2 − A 1 + A^21 + A 2 A 1 − A 2 + A 1 A 2 + A^22 = In − A 1 − A 2 ,
i.e., In − A 1 − A 2 is a projection matrix with rank = tr(In − A 1 − A 2 ) = n − r 1 − r 2 , where ri = tr(Ai) is the rank of Ai, i = 1, 2. By Cochran’s theorem and
Xτ^ X = Xτ^ A 1 X + Xτ^ A 2 X + Xτ^ (In − A 1 − A 2 )X,
Xτ^ A 1 X and Xτ^ A 2 X are independent. Assume that Xτ^ A 1 X and Xτ^ A 2 X are independent. Since Xτ^ AiX has the noncentral chi-square distribution χ^2 ri (δi), where ri is the rank of Ai and δi = μτ^ Aiμ, Xτ^ (A 1 + A 2 )X has the noncentral chi-square distribution χ^2 r 1 +r 2 (δ 1 + δ 2 ). Consequently, A 1 + A 2 is a projection matrix, i.e.,
(A 1 + A 2 )^2 = A 1 + A 2 ,
which implies A 1 A 2 + A 2 A 1 = 0.
Since A^21 = A 1 , we obtain that
0 = A 1 (A 1 A 2 + A 2 A 1 ) = A 1 A 2 + A 1 A 2 A 1
and 0 = A 1 (A 1 A 2 + A 2 A 1 )A 1 = 2A 1 A 2 A 1 ,
which imply A 1 A 2 = 0.
Exercise 1.55. Let X be a random variable having a cumulative distribution function F. Show that if EX exists, then
EX =
∫ (^) ∞
0
[1 − F (x)]dx −
∫ (^0)
−∞
F (x)dx.
Solution. By Fubini’s theorem, ∫ (^) ∞
0
[1 − F (x)]dx =
∫ (^) ∞
0
∫
(x,∞)
dF (y)dx
∫ (^) ∞
0
∫
(0,y)
dxdF (y)
∫ (^) ∞
0
ydF (y).
when n 2 > 2 and
Var(F ) = E
U 12 /n^21 U 22 /n^22
− [E(F )]^2
n^22 n^21
EU 12 EU 2 − 2 −
( n 2 (n 1 + δ) n 1 (n 2 − 2)
) 2
n^22 n^21
( 2 n 1 + 4δ + (N − 1 + δ)^2 (n 2 − 2)(n 2 − 4)
(n 1 + δ)^2 (n 2 − 2)^2
)
2 n^22 [(n 1 + δ)^2 + (n 2 − 2)(n 1 + 2δ)] n^21 (n 2 − 2)^2 (n 2 − 4)
when n 2 > 4.
Exercise 1.61. Let (X, Y ) be a random 2-vector with the following Lebesgue density:
f (x, y) =
π−^1 x^2 + y^2 ≤ 1 0 x^2 + y^2 > 1.
Show that X and Y are uncorrelated, but they are not independent. Solution. Since X and Y are uniformly distributed on the Borel set {(x, y) : x^2 + y^2 ≤ 1 }, EX = EY = 0 and E(XY ) = 0. Hence Cov(X, Y ) = 0. A direct calculation shows that
P (0 < X < 1 /
2 , 0 < Y < 1 /
2 π
and
P (0 < X < 1 /
2) = P (0 < Y < 1 /
2 π
Hence,
P (0 < X < 1 /
2 , 0 < Y < 1 /
2) 6 = P (0 < X < 1 /
2)P (0 < Y < 1 /
and X and Y are not independent.
Exercise 1.48, 1.70. Let Y be a random variable having the noncentral chi-square distribution χ^2 k(δ), where k is a positive integer. Show that (i) the Lebesgue density of Y is
gδ,k(t) = e−δ/^2
∑^ ∞
j=
(δ/2)j j!
f 2 j+k(t),
where fj (t) = [Γ(j/2)2j/^2 ]−^1 tj/^2 −^1 e−t/^2 I(0,∞)(t) is the Lebesgue density of the central chi-square distribution χ^2 j , j = 1, 2 , ...;
(ii) the characteristic function of Y is (1 − 2
− 1 t)−k/^2 e
√− 1 t/(1− 2 √− 1 t) ; (iii) E(Y ) = k + δ and Var(Y ) = 2k + 4δ. Solution A. (i) Consider first k = 1. By the definition of the noncentral chi-square distribution (e.g., Shao, 2003, p. 26), the distribution of Y is the same as that of X^2 , where X has the normal distribution with mean
δ and variance 1. Since
P (Y ≤ t) = P (X ≤
t) − P (X ≤ −
t)
for t > 0, the Lebesgue density of Y is
fY (t) =
t
[fX (
t) + fX (−
t)]I(0,∞)(t),
where fX is the Lebesgue density of X. Using the fact that X has a normal distribution, we obtain that, for t > 0,
fY (t) =
2 πt
( e−(
√ t− √ δ)^2 / (^2) + e−(− √ t− √ δ)^2 / 2 )
e−δ/^2 e−t/^2 2
2 πt
( e
√ δt (^) + e √ −δt )
e−δ/^2 e−t/^2 2
2 πt
∑^ ∞ j=
δt)j j!
∑^ ∞ j=
δt)j j!
e−δ/^2 e−t/^2 √ 2 πt
∑^ ∞
j=
(δt)j (2j)!
On the other hand, for k = 1 and t > 0,
gδ, 1 (t) = e−δ/^2
∑^ ∞
j=
(δ/2)j j!
( 1 Γ(j + 1/2)2j+1/^2
tj−^1 /^2 e−t/^2
)
e−δ/^2 e−t/^2 √ 2 t
∑^ ∞
j=
(δt)j j!Γ(j + 1/2)2^2 j^
Since j!2^2 j^ Γ(j + 1/2) =
π(2j)!, fY (t) = gδ, 1 (t) holds. We then use induction. By definition, Y = X 1 + X 2 , where X 1 has the noncentral chi-square distribution χ^2 k− 1 (δ), X 2 has the central chi-square dis- tribution χ^21 , and X 1 and X 2 are independent. By the induction assumption,
Substituting t by
− 1 t in the moment generating function of Y , we obtain the characteristic function of Y as (1 − 2
− 1 t)−k/^2 e
√− 1 δt/(1− 2 √− 1 t) . (iii) Let ψY (t) be the moment generating function of Y. By the result in (ii),
ψ′(t) = ψ(t)
( δ 1 − 2 t
2 δt (1 − 2 t)^2
k 1 − 2 t
)
and
ψ′′(t) = ψ′(t)
( δ 1 − 2 t
2 δt (1 − 2 t)^2
k 1 − 2 t
)
( 4 δ (1 − 2 t)^2
2 δt (1 − 2 t)^3
2 k (1 − 2 t)^2
) .
Hence, EY = ψ′(0) = δ + k, EY 2 = ψ′′(0) = (δ + k)^2 + 4δ + 2k, and Var(Y ) = EY 2 − (EY )^2 = 4δ + 2k. Solution B. (i) We first derive result (ii). Let X be a random variable having the standard normal distribution and μ be a real number. The moment generating function of (X + μ)^2 is
ψμ(t) =
2 π
∫ e−x
(^2) / 2 et(x+μ)
2 dx
eμ (^2) t/(1− 2 t) √ 2 π
∫ e−(1−^2 t)[x−^2 μt/(1−^2 t)] (^2) / 2 dx
eμ (^2) t/(1− 2 t) √ 1 − 2 t
By definition, Y has the same distribution as X 12 + · · · + X^2 k− 1 + (Xk +
δ)^2 , where Xi’s are independent and have the standard normal distribution. From the obtained result, the moment generating function of Y is
Eet[X
(^21) +···+X k (^2) − 1 +(Xk +√δ) (^2) ] = [ψ 0 (t)]k−^1 ψ√δ(t) =
eμ (^2) t/(1− 2 t)
(1 − 2 t)k/^2
(ii) We now use the result in (ii) to prove the result in (i). From part (ii) of Solution A, the moment generating function of gδ,k is eμ (^2) t/(1− 2 t) (1 − 2 t)−k/^2 , which is the same as the moment generating function of Y derived in part (i) of this solution. By the uniqueness theorem (e.g., Theorem 1.6 in Shao,
2003), we conclude that gδ,k is the Lebesgue density of Y. (iii) Let Xi’s be as defined in (i). Then,
EY = EX 12 + · · · + EX k^2 − 1 + E(Xk +
δ)^2 = k − 1 + EX k^2 + δ + E(
δXk) = k + δ
and
Var(Y ) = Var(X 12 ) + · · · + Var(X k^2 − 1 ) + Var((Xk +
δ)^2 ) = 2(k − 1) + Var(X k^2 + 2
δXk) = 2(k − 1) + Var(X k^2 ) + Var(
δXk) + 2Cov(X k^2 , 2
δXk) = 2k + 4δ,
since Var(X i^2 ) = 2 and Cov(X k^2 , Xk) = EX k^3 − EX k^2 EXk = 0.
Exercise 1.73(g). Let φ be a characteristic function and G be a cumulative distribution function on the real line. Show that
∫ φ(ut)dG(u) is a charac- teristic function on the real line. Solution. Let F be the cumulative distribution function corresponding to φ and let X and U be independent random variables having distributions F and G, respectively. The characteristic function of U X is
Ee
√− 1 tU X
∫ ∫ e
√− 1 tux dF (x)dG(u)
=
∫ φ(ut)dG(u).
Exercise 1.74. Let φn be the characteristic function of a probability mea- sure Pn, n = 1, 2 , .... Let {an} be a sequence of nonnegative numbers with ∑∞ n=1 an^ = 1. Show that^
∑∞ n=1 anφn^ is a characteristic function and find its corresponding probability measure. Solution A. For any event A, define
P (A) =
∑^ ∞ n=
anPn(A).
Then P is a probability measure and Pn ø P for any n. Denote the Radon- Nikodym derivative of Pn with respect to P as fn, n = 1, 2 , .... By Fubini’s
