Chapter 11

Topics from the Theory of Characteristic Functions

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Problem 11.1

Let gg and hh be real-valued functions defined on R\mathbb {R} for which ∫Rg(x)dμ\int_{\mathbb {R}} g(x) d \mu and ∫Rh(x)dμ\int_{\mathbb {R}} h(x) d \mu are finite, where μ\mu is a (σ\sigma-finite) measure in R\mathbb {R}. Then show that:

∣∫R[g(x)+ih(x)]dμ∣≤∫R∣g(x)+ih(x)∣dμ. \left|\int _{\mathbb {R}}[g(x)+i h(x)] d \mu \right| \leq \int _{\mathbb {R}}\left|g(x)+i h(x)\right| d \mu .

In particular, ∣E[Z]∣≤E[∣Z∣]\left|\mathbb {E}\left[Z\right]\right| \leq \mathbb {E}\left[\left|Z\right|\right], where ZZ is a complex-valued r.v.; i.e., Z=X+iYZ = X+i Y with XX and YY real-valued r.v.s.

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Problem 11.2

In reference to the proof of Theorem 2, show that lim⁡t→0g(t)=b−a\lim_{t \rightarrow 0} g(t) = b-a as claimed there.

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Problem 11.3

In reference to the proof of Proposition 1 (see also proof of Proposition 3), provide the details of the convergence

1t∫0tg(v)dv→g(0) as t↓0 \frac{1}{t} \int _{0}^{t} g(v) d v \rightarrow g(0) \text{ as } t \downarrow 0
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Problem 11.4

Show that ∣eix−1∣≤∣x∣\left|e^{i x}-1\right| \leq \left|x\right| for all x∈Rx \in \mathbb {R}.

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Problem 11.5

Show that (1+cnn)n→n→∞ec\left(1+\frac{c_{n}}{n}\right)^{n} \underset {n \rightarrow \infty }{\rightarrow } e^{c}, when cn→∞c_{n} \rightarrow \infty, where cn∈Cc_{n} \in \mathbb {C}, the complex plane, n≥1n \geq 1.

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Problem 11.6

Let XX be a r.v. having the Cauchy distribution with parameters μ=0\mu = 0 and σ=1\sigma = 1 (i.e., the p.d.f. of XX is given by p(x)=1π×11+x2,x∈Rp(x) = \frac{1}{\pi } \times \frac{1}{1+x^{2}}, x \in \mathbb {R}). Then show that:

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(i)

The E[X]\mathbb {E}\left[X\right] does not exist.

(ii)

The ch.f. fX(t)=e−∣t∣,t∈Rf_{X}(t) = e^{-\left|t\right|}, t \in \mathbb {R}. Next, let X1,…,XnX_{1}, \ldots , X_{n} be independent r.v.s distributed as XX and set Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n}. Then

(iii)

Identify the ch.f. fSn/n(t)f_{S_{n} / n}(t).

(iv)

Show that Snn→n→∞P0\frac{S_{n}}{n} \xrightarrow [n \rightarrow \infty ]{P} 0 by showing that Snn→n→∞d0\frac{S_{n}}{n} \xrightarrow [n \rightarrow \infty ]{d} 0. (Although, by intuition, one would expect such a convergence, because of symmetry about 0 of the Cauchy distribution!)

Problem 11.7

Show that: [(2n)!]1/2n[(2n−1)!]1/(2n−1)→n→∞→1\frac{[(2 n)!]^{1 / 2 n}}{[(2 n-1)!]^{1 /(2 n-1)}} \xrightarrow [n \rightarrow \infty ]{\rightarrow } 1.

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Problem 11.8

Recall the following lemma (Lemma 6). Let XX be a r.v. such that μ(n)=E[∣X∣n]<∞\mu^{(n)} = \mathbb {E}\left[\left|X\right|^{n}\right] < \infty for all n=1,2,…n = 1, 2, \ldots (so that m(n)=E[Xn]m^{(n)} = \mathbb {E}\left[X^{n}\right] are also finite). Then the series

S(t)=∑n=0∞m(n)n!(it)nandS∗(t)=∑n=0∞μ(n)n!∣t∣n S(t) = \sum _{n=0}^{\infty } \frac{m^{(n)}}{n!} (it)^{n} \quad \text{and} \quad S^{*}(t) = \sum _{n=0}^{\infty } \frac{\mu ^{(n)}}{n!} \left|t\right|^{n}

have the same radii of convergence.

Establish the validity of the relations claimed in the following corollary (the Corollary to Lemma 6). A sufficient condition for the series S(t)S(t) to converge for some t0≠0t_{0} \neq 0 is that

lim‾⁡n→∞[m(n)n!]1/n<∞orlim‾⁡n→∞[μ(n)]1/nn<∞. \varlimsup _{n \rightarrow \infty } \left[\frac{m^{(n)}}{n!}\right]^{1/n} < \infty \quad \text{or} \quad \varlimsup _{n \rightarrow \infty } \frac{\left[\mu ^{(n)}\right]^{1/n}}{n} < \infty .
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Problem 11.9

If X1,…,XnX_{1}, \ldots , X_{n} are i.i.d. r.v.s with E[X1]=μ∈R\mathbb {E}\left[X_{1}\right] = \mu \in \mathbb {R} and σ2(X1)=σ2∈(0,∞)\sigma^{2}\left(X_{1}\right) = \sigma^{2} \in (0, \infty ), then (by Application 2 to Theorem 9 in this chapter) it follows that Sn−nμσn⇒n→∞Z∼N(0,1)\frac{S_{n}-n \mu }{\sigma \sqrt{n}} \underset {n \rightarrow \infty }{\Rightarrow } Z \sim N(0,1), where Sn=∑j=1nXjS_{n} = \sum_{j = 1}^{n} X_{j}. Show that {Sn−nμσn}\left\{ \frac{S_{n}-n \mu }{\sigma \sqrt{n}}\right\} does not converge in probability as n→∞n \rightarrow \infty.

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Problem 11.10

According to the WLLN (Application 1 to Theorem 9), if the r.v.s X1,…,XnX_{1}, \ldots , X_{n} are i.i.d. with finite E[X1]\mathbb {E}\left[X_{1}\right], then

Xˉn=SnnPn→∞E[X1], where Sn=∑j=1nXj. \bar{X}_{n} = \frac{S_{n}}{n} \underset {n \rightarrow \infty }{P} \mathbb {E}\left[X_{1}\right] \text{, where } S_{n} = \sum _{j = 1}^{n} X_{j} .

The following example shows that it is possible for {Snn}\left\{ \frac{S_{n}}{n}\right\} to converge in probability to a finite constant, as n→∞n \rightarrow \infty, even if the E[X1]\mathbb {E}\left[X_{1}\right] does not exist. To this effect, for j=1,2,…j = 1,2, \ldots, let XjX_{j} be i.i.d. r.v.s such that P(Xj=−n)=P(Xj=n)=c/n2log⁡n,n≥3\mathbb {P}\left(X_{j} = -n\right) = \mathbb {P}\left(X_{j} = n\right) = c / n^{2} \log n, n \geq 3, where c=12∑n=3∞1n2log⁡nc = \frac{1}{2} \sum_{n = 3}^{\infty } \frac{1}{n^{2}} \log n. Then show that E[X1]\mathbb {E}\left[X_{1}\right] does not exist, but Snn→n→∞P0\frac{S_{n}}{n} \xrightarrow [n \rightarrow \infty ]{P} 0, where Sn=∑j=1nXjS_{n} = \sum_{j = 1}^{n} X_{j}.

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Problem 11.11

If X∼B(n,p)X \sim B(n, p), show that

fX(t)=(peit+q)n,q=1−p f_{X}(t) = \left(p e^{i t}+q\right)^{n}, \quad q = 1-p
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Problem 11.12

If X∼Pois⁡(λ)X \sim \operatorname {Pois}\left(\lambda \right), show that

fX(t)=eλeit−λ f_{X}(t) = e^{\lambda e^{i t}-\lambda }
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Problem 11.13
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(i)

If Z∼N(0,1)Z \sim N(0,1), show that fZ(t)=e−t2/2f_{Z}(t) = e^{-t^{2} / 2}.

(ii)

If X∼N(μ,σ2)X \sim N\left(\mu , \sigma^{2}\right), use the fact that Z=X−μσ∼N(0,1)Z = \frac{X-\mu }{\sigma } \sim N(0,1) and part (i) in order to show that

fX(t)=eiμt−σ2t22 f_{X}(t) = e^{i \mu t-\frac{\sigma ^{2} t^{2}}{2}}
Problem 11.14

If XX has the Gamma distribution with parameters α\alpha and β\beta; i.e., if its p.d.f. is given by

p(x;α,β)=1Γ(α)βαxα−1e−x/β,x>0( and 0 for x≤0) p(x ; \alpha , \beta ) = \frac{1}{\Gamma (\alpha ) \beta ^{\alpha }} x^{\alpha -1} e^{-x / \beta }, \quad x > 0(\text{ and } 0 \text{ for } x \leq 0)

where the Gamma function Γ(α)\Gamma (\alpha ) is given by Γ(α)=∫0∞yα−1e−ydy(α,β>0)\Gamma (\alpha ) = \int_{0}^{\infty } y^{\alpha -1} e^{-y} d y(\alpha , \beta > 0), then show that

fX(t)=1(1−iβt)α f_{X}(t) = \frac{1}{(1-i \beta t)^{\alpha }}

In particular, for α=1\alpha = 1 and β=1/λ\beta = 1 / \lambda, we get the ch.f. of the Negative Exponential distribution with parameter λ\lambda; i.e., fX(t)=1/(1−itλ)f_{X}(t) = 1 /\left(1-\frac{i t}{\lambda }\right); and for α=r2(r>0\alpha = \frac{r}{2}(r > 0 integer) and β=2\beta = 2, we get the ch.f. of the chi-square distribution with rr degrees of freedom; i.e., fX(t)=1/(1−2it)r/2f_{X}(t) = 1 /(1-2 i t)^{r / 2}.

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Problem 11.15

If the r.v.s XX and YY have the Bivariate Normal distribution with parameters μ1,μ2\mu_{1}, \mu_{2} in R,0<σ1,σ2<∞\mathbb {R}, 0 < \sigma_{1}, \sigma_{2} < \infty, and ρ∈[−1,1]\rho \in [-1,1], show that their joint ch.f. is given by

fX,Y(t1,t2)=exp⁡[iμ1t1+iμ2t2−12(σ12t12+2ρσ1σ2t1t2+σ22t22)] \begin{aligned} f_{X, Y}\left(t_{1}, t_{2}\right) = \exp \left[i \mu _{1} t_{1}+i \mu _{2} t_{2}-\frac{1}{2}\left(\sigma _{1}^{2} t_{1}^{2}+2 \rho \sigma _{1} \sigma _{2} t_{1} t_{2}+\sigma _{2}^{2} t_{2}^{2}\right)\right] \end{aligned}

For this purpose, do the following:

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(i)

Assume first that μ1=μ2=0\mu_{1} = \mu_{2} = 0 and σ1=σ2=1\sigma_{1} = \sigma_{2} = 1, and use Exercises 12 (ii) in Chapter 9 and 13 (ii) in this chapter to show that:

fX,Y(t1,t2)=exp⁡[−12(t12+2ρt1t2+t22)]. f_{X, Y}\left(t_{1}, t_{2}\right) = \exp \left[-\frac{1}{2}\left(t_{1}^{2}+2 \rho t_{1} t_{2}+t_{2}^{2}\right)\right] .
(ii)

For the general case, use the transformations U=(X−μ1)/σ1,V=(Y−μ2)/σ2U = \left(X-\mu_{1}\right) / \sigma_{1}, V = \left(Y-\mu_{2}\right) / \sigma_{2} and verify that E[U]=E[V]=0,Var⁡[U]=Var⁡[V]=1,ρ(U,V)=ρ(X,Y)=ρ\mathbb {E}\left[U\right] = \mathbb {E}\left[V\right] = 0, \operatorname {Var}\left[U\right] = \operatorname {Var}\left[V\right] = 1, \rho (U, V) = \rho (X, Y) = \rho. Then use Exercise 15 in Chapter 9 and part (i) here to arrive at the desired expression for the ch.f. fX,Yf_{X, Y}.

Problem 11.16

Let the r.v.s XX and YY have the Bivariate Normal distribution with parameters μ1,μ2∈R,0<σ1,σ2<∞\mu_{1}, \mu_{2} \in \mathbb {R}, 0 < \sigma_{1}, \sigma_{2} < \infty, and ρ∈[−1,1]\rho \in [-1,1], and set U=X+Y,V=X−YU = X+Y, V = X-Y.

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(i)

Verify that E[U]=μ1+μ2,Var⁡[U]=σ12+σ22+2ρσ1σ2,E[V]=μ1−μ2,Var⁡[V]=σ12+σ22−2ρσ1σ2\mathbb {E}\left[U\right] = \mu_{1}+\mu_{2}, \operatorname {Var}\left[U\right] = \sigma_{1}^{2}+\sigma_{2}^{2}+2 \rho \sigma_{1} \sigma_{2}, \mathbb {E}\left[V\right] = \mu_{1}-\mu_{2}, \operatorname {Var}\left[V\right] = \sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2}, and Cov⁡[U,V]=σ12−σ22\operatorname {Cov}\left[ U, V \right] = \sigma_{1}^{2}-\sigma_{2}^{2} (by using Exercises 12 (ii) and 14 (ii) in Chapter 9).

(ii)

Since

fU,V(t1,t2)=E[eit1U+it2V]=E[eit1(X+Y)+it2(X−Y)]=E[ei(t1+t2)X+i(t1−t2)Y]=fX,Y(t1+t2,t1−t2) \begin{aligned} f_{U, V}\left(t_{1}, t_{2}\right) & = \mathbb {E}\left[e^{i t_{1} U+i t_{2} V}\right] = \mathbb {E}\left[e^{i t_{1}(X+Y)+i t_{2}(X-Y)}\right] \\ & = \mathbb {E}\left[e^{i\left(t_{1}+t_{2}\right) X+i\left(t_{1}-t_{2}\right) Y}\right] \\ & = f_{X, Y}\left(t_{1}+t_{2}, t_{1}-t_{2}\right) \end{aligned}

use Exercise 15 in order to conclude that

fU,V(t1,t2)=exp⁡[i(μ1+μ2)t1+i(μ1−μ2)t2−12((σ12+σ22+2ρσ1σ2)t12+2(σ12−σ22)t1t2+(σ12+σ22−2ρσ1σ2)t22)]. \begin{aligned} & f_{U, V}\left(t_{1}, t_{2}\right) = \exp \left[i\left(\mu _{1}+\mu _{2}\right) t_{1}+i\left(\mu _{1}-\mu _{2}\right) t_{2}-\frac{1}{2}\left(\left(\sigma _{1}^{2}+\sigma _{2}^{2}+2 \rho \sigma _{1} \sigma _{2}\right) t_{1}^{2}+2\left(\sigma _{1}^{2}-\sigma _{2}^{2}\right) t_{1} t_{2}+\left(\sigma _{1}^{2}+\sigma _{2}^{2}-2 \rho \sigma _{1} \sigma _{2}\right) t_{2}^{2}\right)\right] . \end{aligned}
(iii)

From part (ii) and Exercise 15, conclude that the r.v.s UU and VV have the Bivariate Normal distribution with parameters μ1+μ2,μ1−μ2,σ12+σ22+2ρσ1σ2=τ12,σ12+σ22−2ρσ1σ2=τ22\mu_{1}+\mu_{2}, \mu_{1}-\mu_{2}, \sigma_{1}^{2}+\sigma_{2}^{2}+ 2 \rho \sigma_{1} \sigma_{2} = \tau_{1}^{2}, \sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2} = \tau_{2}^{2}, and ρ(U,V)=(σ12−σ22)/τ1τ2\rho (U, V) = \left(\sigma_{1}^{2}-\sigma_{2}^{2}\right) / \tau_{1} \tau_{2}.

(iv)

From part (iii) and Exercise 12 (ii) in Chapter 10, conclude that UU and VV are independent if and only if σ1=σ2\sigma_{1} = \sigma_{2}.

Problem 11.17

In this exercise, the r.v.s X1,…,XkX_{1}, \ldots , X_{k} are independent with distributions as indicated, and X=X1+⋯+XkX = X_{1}+\cdots +X_{k}. Then use Exercise 13 in Chapter 10, the inversion formula (Theorem 2), and the appropriate ch.f.s in order to show that:

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(i)

If Xj∼B(nj,p),j=1,…,kX_{j} \sim B\left(n_{j}, p\right), j = 1, \ldots , k, then X∼B(n,p)X \sim B(n, p), where n=n1+⋯+nkn = n_{1}+\cdots + n_{k}.

(ii)

If Xj∼Pois⁡(λj),j=1,…,kX_{j} \sim \operatorname {Pois}\left(\lambda_{j}\right), j = 1, \ldots , k, then X∼Pois⁡(λ)X \sim \operatorname {Pois}\left(\lambda \right), where λ=λ1+⋯+λk\lambda = \lambda_{1}+\cdots +\lambda_{k}.

(iii)

If Xj∼N(μj,σj2),j=1,…,kX_{j} \sim N\left(\mu_{j}, \sigma_{j}^{2}\right), j = 1, \ldots , k, then X∼N(μ,σ2)X \sim N\left(\mu , \sigma^{2}\right), where μ=μ1+⋯+μk\mu = \mu_{1}+\cdots +\mu_{k} and σ2=σ12+⋯+σk2\sigma^{2} = \sigma_{1}^{2}+\cdots +\sigma_{k}^{2}. Also, c1X1+⋯+ckXk∼N(c1μ1+⋯+ckμk,c12σ12+⋯+ck2σk2)c_{1} X_{1}+\cdots +c_{k} X_{k} \sim N\left(c_{1} \mu_{1}+\cdots +c_{k} \mu_{k}, c_{1}^{2} \sigma_{1}^{2}+\cdots +c_{k}^{2} \sigma_{k}^{2}\right), where c1,…,ckc_{1}, \ldots , c_{k} are constants.

(iv)

If Xj∼X_{j} \sim Gamma with parameters αj\alpha_{j} and β,j=1,…,k\beta , j = 1, \ldots , k, then X∼X \sim Gamma with parameters α=α1+⋯+αk\alpha = \alpha_{1}+\cdots +\alpha_{k} and β\beta. In particular, if XjX_{j} has the Negative Exponential distribution with parameter λ,j=1,…,k\lambda , j = 1, \ldots , k, then X∼X \sim Gamma with parameters α=k\alpha = k and β=1/λ\beta = 1 / \lambda, whereas, if Xj∼χrj2,j=1,…,kX_{j} \sim \chi_{r_{j}}^{2}, j = 1, \ldots , k, then X∼χr2X \sim \chi_{r}^{2}, where r=r1+⋯+rkr = r_{1}+\cdots +r_{k}.

Problem 11.18

Show that the r.v.s X1,…,XkX_{1}, \ldots , X_{k} are independent if and only if, for all t1,…,tkt_{1}, \ldots , t_{k} in R,fX1,…,Xk(t1,…,tk)=fX1(t1)×⋯×fXk(tk)\mathbb {R}, f_{X_{1}, \ldots , X_{k}}\left(t_{1}, \ldots , t_{k}\right) = f_{X_{1}}\left(t_{1}\right) \times \cdots \times f_{X_{k}}\left(t_{k}\right).

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Problem 11.19

If ff is a ch.f., then show that ff is positive definite; i.e., for all tk,tlt_{k}, t_{l} reals and all complex numbers zk,zlz_{k}, z_{l}, we have

∑k=1n∑l=1nf(tk−tl)zkzˉl≥0 \sum _{k = 1}^{n} \sum _{l = 1}^{n} f\left(t_{k}-t_{l}\right) z_{k} \bar{z}_{l} \geq 0

for every integer n≥1n \geq 1.

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Problem 11.20

For n=1,2,…n = 1,2, \ldots, let XnX_{n} be a r.v. distributed as Poisson with parameter n,Xn∼Pois⁡(n)n, X_{n} \sim \operatorname {Pois}\left(n\right) (so that E[Xn]=Var⁡[Xn]=n\mathbb {E}\left[X_{n}\right] = \operatorname {Var}\left[X_{n}\right] = n), and set Yn=(Xn−n)/nY_{n} = \left(X_{n}-n\right) / \sqrt{n}. Then show that Yn→n→∞dZ∼N(0,1)Y_{n} \xrightarrow [n \rightarrow \infty ]{d} Z \sim N(0,1).

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Problem 11.21

Let F1F_{1} and F2F_{2} be d.f.s, and let GG be their convolution, G=F1∗F2G = F_{1} * F_{2}. Then:

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(i)

If F1F_{1} is absolutely continuous with respect to Lebesgue measure with p.d.f. p1p_{1}, it follows that GG is absolutely continuous with respect to Lebesgue measure with p.d.f. pp given by

p(u)=∫−∞∞p1(u−y)dF2(y) p(u) = \int _{-\infty }^{\infty } p_{1}(u-y) d F_{2}(y)
(ii)

Furthermore, if F2F_{2} is also absolutely continuous with respect to Lebesgue measure with p.d.f. p2p_{2}, then

p(u)=∫−∞∞p1(u−y)p2(y)dy. p(u) = \int _{-\infty }^{\infty } p_{1}(u-y) p_{2}(y) d y .
Problem 11.22

If the r.v.s XX and YY are i.i.d. with ch.f. ff, then the ch.f. of the r.v. X−YX-Y is ∣f(t)∣2\left|f(t)\right|^{2}.

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Problem 11.23

Let X1,…,XnX_{1}, \ldots , X_{n} be independent r.v.s, each of which is symmetric about 0. Then the r.v. X1+⋯+XnX_{1}+\cdots +X_{n} is also symmetric about 0.

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Problem 11.24

For n≥1n \geq 1, let gng_{n} and gg be functions defined on E⊆RkE \subseteq \mathbb {R}^{k} into R\mathbb {R}, and recall that {gn}\left\{ g_{n}\right\} is said to converge continuously to gg on EE, if for every x∈E,gn(xn)→g(x)x \in E, g_{n}\left(x_{n}\right) \rightarrow g(x) whenever xn→xx_{n} \rightarrow x, as n→∞n \rightarrow \infty. Then show that if {gn}\left\{ g_{n}\right\} converges continuously to gg on EE, it follows that gg is continuous on EE.

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Problem 11.25

For n≥1n \geq 1, let gn,gg_{n}, g and EE be as in Exercise 24. Then:

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(i)

If {gn}\left\{ g_{n}\right\} converges to gg uniformly on EE and gg is continuous on EE, it follows that {gn}\left\{ g_{n}\right\} converges continuously to gg on EE.

(ii)

If EE is compact and {gn}\left\{ g_{n}\right\} converges continuously to gg on EE, then the convergence is uniform.

Problem 11.26

For n≥1n \geq 1, let Xn,YnX_{n}, Y_{n} and X,YX, Y be r.v.s defined on the probability space (Ω,A,P)(\Omega , \mathcal{A}, P), let d1,d2d_{1}, d_{2} and cnc_{n} be constants with 0≠cn→∞0 \neq c_{n} \rightarrow \infty as n→∞n \rightarrow \infty, and suppose that cn(Xn−d1,Yn−d2)→d(X,Y)c_{n}\left(X_{n}-d_{1}, Y_{n}-d_{2}\right) \xrightarrow {d}(X, Y) as n→∞n \rightarrow \infty. Let g:R2→Rg: \mathbb {R}^{2} \rightarrow \mathbb {R} be differentiable, and suppose its (first-order) partial derivatives g˙x,g˙y\dot{g}_{x}, \dot{g}_{y} are continuous at (d1,d2d_{1}, d_{2}). Then show that, as n→∞n \rightarrow \infty,

cn[g(Xn,Yn)−g(d1,d2)]→d[g˙x(d1,d2)g˙y(d1,d2)](XY)′=g˙x(d1,d2)X+g˙y(d1,d2)Y. \begin{aligned} c_{n}\left[g\left(X_{n}, Y_{n}\right)-g\left(d_{1}, d_{2}\right)\right] \xrightarrow {d} & {\left[\dot{g}_{x}\left(d_{1}, d_{2}\right) \quad \dot{g}_{y}\left(d_{1}, d_{2}\right)\right](X \quad Y)^{\prime } } \\ & = \dot{g}_{x}\left(d_{1}, d_{2}\right) X+\dot{g}_{y}\left(d_{1}, d_{2}\right) Y . \end{aligned}
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