Chapter 1

Introduction and review of probability

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Problem 1.1

Let A1A_{1} and A2A_{2} be arbitrary events and show that P(A1∪A2)+P(A1A2)=P(A1)+P(A2)\mathbb {P}\left(A_{1} \cup A_{2}\right)+\mathbb {P}\left(A_{1}A_{2}\right) = \mathbb {P}\left(A_{1}\right)+\mathbb {P}\left(A_{2}\right). Explain which parts of the sample space are being double counted on both sides of this equation and which parts are being counted once.

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Problem 1.2

This exercise derives the probability of an arbitrary (non-disjoint) union of events, derives the union bound, and derives some useful limit expressions.

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(a)

For two arbitrary events A1A_{1} and A2A_{2}, show that

A1∪A2=A1∪(A2−A1),where A2−A1=A2A1c. A_{1} \cup A_{2} = A_{1} \cup (A_{2}-A_{1}), \quad \text{where } A_{2}-A_{1} = A_{2}A_{1}^{c}.

Show that A1A_{1} and A2−A1A_{2}-A_{1} are disjoint.

(b)

For an arbitrary sequence of events {An;n≥1}\left\{ A_{n}; n \geq 1\right\}, let B1=A1B_{1} = A_{1} and for each n≥2n \geq 2 define Bn=An−⋃m=1n−1AmB_{n} = A_{n}-\bigcup_{m=1}^{n-1}A_{m}. Show that B1,B2,…B_{1}, B_{2}, \ldots, are disjoint events and show that for each n≥2n \geq 2, ⋃m=1nAm=⋃m=1nBm\bigcup_{m=1}^{n}A_{m} = \bigcup_{m=1}^{n}B_{m}.

(c)

Show that

P(⋃n=1∞An)=P(⋃n=1∞Bn)=∑n=1∞P(Bn). \mathbb {P}\left(\bigcup _{n=1}^{\infty }A_{n}\right) = \mathbb {P}\left(\bigcup _{n=1}^{\infty }B_{n}\right) = \sum _{n=1}^{\infty }\mathbb {P}\left(B_{n}\right).
(d)

Show that for each nn, P(Bn)≤P(An)\mathbb {P}\left(B_{n}\right) \leq \mathbb {P}\left(A_{n}\right). Use this to show that

P(⋃n=1∞An)≤∑n=1∞P(An). \mathbb {P}\left(\bigcup _{n=1}^{\infty }A_{n}\right) \leq \sum _{n=1}^{\infty }\mathbb {P}\left(A_{n}\right).
(e)

Show that P(⋃n=1∞An)=lim⁡m→∞P(⋃n=1mAn)\mathbb {P}\left(\bigcup_{n=1}^{\infty }A_{n}\right) = \lim_{m \rightarrow \infty }\mathbb {P}\left(\bigcup_{n=1}^{m}A_{n}\right). Note that this says that the probability of a limit of unions is equal to the limit of the probabilities. This might well appear to be obvious without a proof, but you will see situations later where similar appearing interchanges cannot be made.

(f)

Show that P(⋂n=1∞An)=lim⁡n→∞P(⋂i=1nAi)\mathbb {P}\left(\bigcap_{n=1}^{\infty }A_{n}\right) = \lim_{n \rightarrow \infty }\mathbb {P}\left(\bigcap_{i=1}^{n}A_{i}\right).

Remark.
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(a)

This is what Venn diagrams were invented for.

(b)

Use induction.

(f)

Remember De Morgan's equalities.

Problem 1.3

Find the probability that a five-card poker hand, chosen randomly from a 52-card deck, contains four aces. That is, if all 52!52! arrangements of a deck of cards are equally likely, what is the probability that all four aces are in the first five cards of the deck.

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Problem 1.4

Consider a sample space of eight equiprobable sample points and let A1,A2,A3A_{1}, A_{2}, A_{3} be three events each of probability 1/21/2 such that P(A1A2A3)=P(A1)P(A2)P(A3)\mathbb {P}\left(A_{1}A_{2}A_{3}\right) = \mathbb {P}\left(A_{1}\right)\mathbb {P}\left(A_{2}\right)\mathbb {P}\left(A_{3}\right).

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(a)

Create an example where P(A1A2)=P(A1A3)=14\mathbb {P}\left(A_{1}A_{2}\right) = \mathbb {P}\left(A_{1}A_{3}\right) = \frac{1}{4} but P(A2A3)=18\mathbb {P}\left(A_{2}A_{3}\right) = \frac{1}{8}.

(b)

Show that, for your example, A2A_{2} and A3A_{3} are not independent. Note that the definition of statistical independence would be very strange if it allowed A1,A2,A3A_{1}, A_{2}, A_{3} to be independent while A2A_{2} and A3A_{3} are dependent. This illustrates why the definition of independence requires the joint factorization condition on all subsets of events, rather than just pairwise factorization of every pair of events.

Remark.
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(a)

Make a table with a row for each sample point and a column for each of the above 3 events and try different ways of assigning sample points to events (the answer is not unique).

Problem 1.5

This exercise shows that for all rv's XX, FX(x)F_{X}(x) is continuous from the right.

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(a)

For any given rv XX, any real number xx, and each integer n≥1n \geq 1, let An={ω:X>x+1/n}A_{n} = \left\{ \omega : X > x+1/n\right\}, and show that A1⊆A2⊆⋯A_{1} \subseteq A_{2} \subseteq \cdots. Use this and the corollaries to the axioms of probability to show that P(⋃n≥1An)=lim⁡n→∞P(An)\mathbb {P}\left(\bigcup_{n \geq 1}A_{n}\right) = \lim_{n \rightarrow \infty }\mathbb {P}\left(A_{n}\right).

(b)

Show that P(⋃n≥1An)=P(X>x)\mathbb {P}\left(\bigcup_{n \geq 1}A_{n}\right) = \mathbb {P}\left(X>x\right) and that P(X>x)=lim⁡n→∞P(X>x+1/n)\mathbb {P}\left(X>x\right) = \lim_{n \rightarrow \infty }\mathbb {P}\left(X>x+1/n\right).

(c)

Show that for ϵ>0\epsilon >0, lim⁡ϵ→0P(X≤x+ϵ)=P(X≤x)\lim_{\epsilon \rightarrow 0}\mathbb {P}\left(X \leq x+\epsilon \right) = \mathbb {P}\left(X \leq x\right).

(d)

Define F~X(x)=P(X<x)\widetilde{F}_{X}(x) = \mathbb {P}\left(X<x\right). Show that F~X(x)\widetilde{F}_{X}(x) is continuous from the left. In other words, the continuity from the right for the CDF arises from the almost arbitrary (but universally accepted) choice in defining the CDF as P(X≤x)\mathbb {P}\left(X \leq x\right) rather than P(X<x)\mathbb {P}\left(X<x\right).

Problem 1.6

Show that for a continuous non-negative rv XX,

∫0∞P(X>x) dx=∫0∞xfX(x) dx. \int _{0}^{\infty }\mathbb {P}\left(X>x\right) \, dx = \int _{0}^{\infty }x\mathsf{f}_{X}(x) \, dx.
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Problem 1.7

Suppose XX and YY are discrete rv's with the PMF pXY(xi,yj)\mathsf{p}_{XY}(x_{i}, y_{j}). Show (a picture will help) that this is related to the joint CDF by

pXY(xi,yj)=lim⁡δ>0, δ→0[F(xi,yj)−F(xi−δ,yj)−F(xi,yj−δ)+F(xi−δ,yj−δ)]. \mathsf{p}_{XY}(x_{i}, y_{j}) = \lim _{\delta >0, \, \delta \rightarrow 0}\left[F(x_{i}, y_{j})-F(x_{i}-\delta , y_{j})-F(x_{i}, y_{j}-\delta )+F(x_{i}-\delta , y_{j}-\delta )\right].
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Problem 1.8

A variation of Example 1.5.1 is to let MM be a rv that takes on both positive and negative values with the PMF

pM(m)=12∣m∣(∣m∣+1). \mathsf{p}_{M}(m) = \frac{1}{2\left|m\right|(\left|m\right|+1)}.

In other words, MM is symmetric around 0 and ∣M∣\left|M\right| has the same PMF as the non-negative rv NN of Example 1.5.1 (the rv with CDF FN(n)=n/(n+1)F_{N}(n) = n/(n+1) for each integer n≥1n \geq 1, PMF pN(n)=1/[n(n+1)]\mathsf{p}_{N}(n) = 1/[n(n+1)], and E[N]=∞\mathbb {E}\left[N\right] = \infty).

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(a)

Show that ∑m≥0mpM(m)=∞\sum_{m \geq 0}m\mathsf{p}_{M}(m) = \infty and ∑m<0mpM(m)=−∞\sum_{m<0}m\mathsf{p}_{M}(m) = -\infty. (Thus show that the expectation of MM not only does not exist but is undefined even as an extended real number.)

(b)

Suppose that the terms in ∑m=−∞∞mpM(m)\sum_{m=-\infty }^{\infty }m\mathsf{p}_{M}(m) are summed in the order of two positive terms for each negative term (i.e., in the order 1,2,−1,3,4,−2,5,⋯1, 2, -1, 3, 4, -2, 5, \cdots). Find the limiting value of the partial sums in this series.

(c)

Repeat (b) where, for any given integer k>0k>0, the order of summation is kk positive terms for each negative term.

Remark.
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(b)

You may find it helpful to know that

lim⁡n→∞[∑i=1n1i−∫1n1x dx]=γ, \lim _{n \rightarrow \infty }\left[\sum _{i=1}^{n}\frac{1}{i}-\int _{1}^{n}\frac{1}{x} \, dx\right] = \gamma ,

where γ\gamma is the Euler--Mascheroni constant, γ=0.57721⋯\gamma = 0.57721\cdots.

Problem 1.9

Theorem 1.4.1: Let pSn(k)\mathsf{p}_{S_{n}}(k) be the PMF of the binomial distribution for an underlying binary PMF pZ(1)=p>0\mathsf{p}_{Z}(1) = p>0, pZ(0)=1−p>0\mathsf{p}_{Z}(0) = 1-p>0 and let p~=k/n\widetilde{p} = k/n. Then for each integer p~n\widetilde{p}n, 1≤p~n≤n−11 \leq \widetilde{p}n \leq n-1,

pSn(p~n)<12πnp~(1−p~)exp⁡[−nD(p~∥p)], \mathsf{p}_{S_{n}}(\widetilde{p}n)< \sqrt{\frac{1}{2\pi n\widetilde{p}(1-\widetilde{p})}}\exp \left[-nD(\widetilde{p}\| p)\right], pSn(p~n)>(1−112np~(1−p~))12πnp~(1−p~)exp⁡[−nD(p~∥p)], \mathsf{p}_{S_{n}}(\widetilde{p}n)>\left(1-\frac{1}{12n\widetilde{p}(1-\widetilde{p})}\right)\sqrt{\frac{1}{2\pi n\widetilde{p}(1-\widetilde{p})}}\exp \left[-nD(\widetilde{p}\| p)\right],

where D(p~∥p)=p~ln⁡(p~p)+(1−p~)ln⁡(1−p~1−p)≥0D(\widetilde{p}\| p) = \widetilde{p}\ln \left(\frac{\widetilde{p}}{p}\right)+(1-\widetilde{p})\ln \left(\frac{1-\widetilde{p}}{1-p}\right) \geq 0 is the binary Kullback--Leibler divergence (relative entropy). Also, D(p~∥p)≥0D(\widetilde{p}\| p) \geq 0 with strict inequality for all p~≠p\widetilde{p} \neq p.

The bounds on the binomial in this theorem are based on the Stirling bounds. These say that for all n≥1n \geq 1, n!n! is upper and lower bounded by

2πn(ne)n<n!<2πn(ne)ne1/12n. \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}<n!<\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{1/12n}.

The ratio 2πn(n/e)n/n!\sqrt{2\pi n}(n/e)^{n}/n! of the first two terms is monotonically increasing with nn toward the limit 1, and the ratio 2πn(n/e)nexp⁡(1/12n)/n!\sqrt{2\pi n}(n/e)^{n}\exp (1/12n)/n! is monotonically decreasing toward 1. The upper bound is more accurate, but the lower bound is simpler and known as the Stirling approximation.

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(a)

Show from the Stirling bounds above and from the monotone property that

(nk)<n2πk(n−k) nnkk(n−k)n−k. \binom {n}{k}<\sqrt{\frac{n}{2\pi k(n-k)}} \, \frac{n^{n}}{k^{k}(n-k)^{n-k}}.
(b)

Use the result of (a) to upper bound pSn(k)\mathsf{p}_{S_{n}}(k) by

pSn(k)<n2πk(n−k) pk(1−p)n−knnkk(n−k)n−k. \mathsf{p}_{S_{n}}(k)<\sqrt{\frac{n}{2\pi k(n-k)}} \, \frac{p^{k}(1-p)^{n-k}n^{n}}{k^{k}(n-k)^{n-k}}.

Show that this is equivalent to the upper bound in Theorem 1.4.1.

(c)

Show that

(nk)>n2πk(n−k) nnkk(n−k)n−k[1−n12k(n−k)]. \binom {n}{k}>\sqrt{\frac{n}{2\pi k(n-k)}} \, \frac{n^{n}}{k^{k}(n-k)^{n-k}}\left[1-\frac{n}{12k(n-k)}\right].
(d)

Derive the lower bound in Theorem 1.4.1.

(e)

Show that D(p~∥p)=p~ln⁡(p~/p)+(1−p~)ln⁡[(1−p~)/(1−p)]D(\widetilde{p}\| p) = \widetilde{p}\ln (\widetilde{p}/p)+(1-\widetilde{p})\ln [(1-\widetilde{p})/(1-p)] is 0 at p~=p\widetilde{p} = p and non-negative elsewhere.

Problem 1.10

Let XX be a ternary rv taking on the three values 0, 1, 2 with probabilities p0,p1,p2p_{0}, p_{1}, p_{2} respectively. Find the median of XX for each of the cases below.

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(a)

p0=0.2p_{0} = 0.2, p1=0.4p_{1} = 0.4, p2=0.4p_{2} = 0.4.

(b)

p0=0.2p_{0} = 0.2, p1=0.2p_{1} = 0.2, p2=0.6p_{2} = 0.6.

(c)

p0=0.2p_{0} = 0.2, p1=0.3p_{1} = 0.3, p2=0.5p_{2} = 0.5.

(d)

Now suppose that XX is non-negative and continuous with the density fX(x)=1\mathsf{f}_{X}(x) = 1 for 0≤x≤0.50 \leq x \leq 0.5 and fX(x)=0\mathsf{f}_{X}(x) = 0 for 0.5<x≤10.5<x \leq 1. We know that fX(x)\mathsf{f}_{X}(x) is positive for all x>1x>1, but it is otherwise unknown. Find the median or interval of medians.

The median is sometimes (incorrectly) defined as that α\alpha for which P(X>α)=P(X<α)\mathbb {P}\left(X>\alpha \right) = \mathbb {P}\left(X<\alpha \right). Show that it is possible for no such α\alpha to exist.

Remark.
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Note 1: The median is not unique in (c). Find the interval of values that are medians. Note 2: Some people force the median to be distinct by defining it as the midpoint of the interval satisfying the definition given here.

Problem 1.11
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(a)

For any given rv YY, express E[∣Y∣]\mathbb {E}\left[\left|Y\right|\right] in terms of ∫y<0FY(y) dy\int_{y<0}F_{Y}(y) \, dy and ∫y≥0FYc(y) dy\int_{y \geq 0}F_{Y}^{c}(y) \, dy.

(b)

For some given rv XX with E[∣X∣]<∞\mathbb {E}\left[\left|X\right|\right]<\infty, let Y=X−αY = X-\alpha. Using (a), show that

E[∣X−α∣]=∫α∞FX(x) dx+∫−∞αFXc(x) dx. \mathbb {E}\left[\left|X-\alpha \right|\right] = \int _{\alpha }^{\infty }F_{X}(x) \, dx+\int _{-\infty }^{\alpha }F_{X}^{c}(x) \, dx.
(c)

Show that E[∣X−α∣]\mathbb {E}\left[\left|X-\alpha \right|\right] is minimized over α\alpha by choosing α\alpha to be a median of XX. Be careful to show that when the median is an interval, all points in this interval achieve the minimum.

Problem 1.12

Let XX be a rv with CDF FX(x)F_{X}(x). Find the CDF of the following rv's.

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(a)

The maximum of nn IID rv's, each with CDF FX(x)F_{X}(x).

(b)

The minimum of nn IID rv's, each with CDF FX(x)F_{X}(x).

(c)

The difference of the rv's defined in (a) and (b); assume XX has a density fX(x)\mathsf{f}_{X}(x).

Problem 1.13

Let XX and YY be rv's in some sample space Ω\Omega and let Z=X+YZ = X+Y, i.e., for each ω∈Ω\omega \in \Omega, Z(ω)=X(ω)+Y(ω)Z(\omega ) = X(\omega )+Y(\omega ). The purpose of this exercise is to show that ZZ is a rv. This is a mathematical fine point that many readers may prefer to simply accept without proof.

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(a)

Show that the set of ω\omega for which Z(ω)=±∞Z(\omega ) = \pm \infty has probability 0.

(b)

To show that Z=X+YZ = X+Y is a rv, we must show that for each real number α\alpha, the set {ω∈Ω:X(ω)+Y(ω)≤α}\left\{ \omega \in \Omega : X(\omega )+Y(\omega ) \leq \alpha \right\} is an event. We proceed indirectly. For an arbitrary positive integer nn and an arbitrary integer k>0k>0, let B(n,k)={ω:X(ω)≤k/n}∩{ω:Y(ω)≤α+(1−k)/n}B(n, k) = \left\{ \omega : X(\omega ) \leq k/n\right\} \cap \left\{ \omega : Y(\omega ) \leq \alpha +(1-k)/n\right\}. Let D(n)=⋃kB(n,k)D(n) = \bigcup_{k}B(n, k) and show that D(n)D(n) is an event.

(c)

On a two-dimensional sketch for a given α\alpha, show the values of X(ω)X(\omega ) and Y(ω)Y(\omega ) for which ω∈D(n)\omega \in D(n).

(d)

Show that

{ω:X(ω)+Y(ω)≤α}=⋂nD(n). \left\{ \omega : X(\omega )+Y(\omega ) \leq \alpha \right\} = \bigcap _{n}D(n).

Explain why this shows that Z=X+YZ = X+Y is a rv.

(e)

Explain why (d) implies that if Y=X1+X2+⋯+XnY = X_{1}+X_{2}+\cdots +X_{n} and if X1,X2,…,XnX_{1}, X_{2}, \ldots , X_{n} are rv's, then YY is a rv.

Remark.
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(c)

This set of values should be bounded by a staircase function.

(e)

Only one or two lines of explanation are needed.

Problem 1.14
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(a)

Let X1,X2,…,XnX_{1}, X_{2}, \ldots , X_{n} be rv's with expected values X‾1,…,X‾n\overline{X}_{1}, \ldots , \overline{X}_{n}. Show that E[X1+⋯+Xn]=X‾1+⋯+X‾n\mathbb {E}\left[X_{1}+\cdots +X_{n}\right] = \overline{X}_{1}+\cdots +\overline{X}_{n}. You may assume that the rv's have a joint density function, but do not assume that the rv's are independent.

(b)

Now assume that X1,…,XnX_{1}, \ldots , X_{n} are statistically independent and show that the expected value of the product is equal to the product of the expected values.

(c)

Again assuming that X1,…,XnX_{1}, \ldots , X_{n} are statistically independent, show that the variance of the sum is equal to the sum of the variances.

Problem 1.15
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(a)

Let h(x)=u(x)h(x) = u(x) and FX(x)=u(x)F_{X}(x) = u(x), where u(x)u(x) is the unit step, i.e., u(x)=0u(x) = 0 for −∞<x<0-\infty <x<0 and u(x)=1u(x) = 1 for x≥0x \geq 0. Using the definition of the Stieltjes integral as the limit of Riemann sums ∑nh(xn)[F(yn)−F(yn−1)]\sum_{n}h(x_{n})[F(y_{n})-F(y_{n-1})] over partitions {yn;n≥1}\left\{ y_{n}; n \geq 1\right\} with yn−1<xn≤yny_{n-1}<x_{n} \leq y_{n}, refined as the partition mesh δ→0\delta \rightarrow 0, show that ∫−11h(x) dFX(x)\int_{-1}^{1}h(x) \, dF_{X}(x) does not exist.

(b)

Let h(x)=u(x−a)h(x) = u(x-a) and FX(x)=u(x−b)F_{X}(x) = u(x-b), where aa and bb are in (−1,+1)(-1, +1). Show that ∫−11h(x) dFX(x)\int_{-1}^{1}h(x) \, dF_{X}(x) exists if and only if a≠ba \neq b. Show that the integral has the value 1 for a<ba<b and the value 0 for a>ba>b. Argue that this result is still valid in the limit of integration over (−∞,∞)(-\infty , \infty ).

(c)

Let XX and YY be independent discrete rv's, each with a finite set of possible values. Show that ∫−∞∞FX(z−y) dFY(y)\int_{-\infty }^{\infty }F_{X}(z-y) \, dF_{Y}(y), defined as a Stieltjes integral, is equal to the distribution of Z=X+YZ = X+Y at each zz other than the possible sample values of ZZ, and is undefined at each sample value of ZZ.

Remark.
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This failure of Stieltjes integration is not a serious problem; FZ(z)F_{Z}(z) is a step function, and the integral is undefined at its points of discontinuity. We automatically define FZ(z)F_{Z}(z) at those step values so that FZF_{Z} is a CDF (i.e., is continuous from the right). This problem does not arise if either XX or YY is continuous.

Problem 1.16

Let X1,X2,…,Xn,…X_{1}, X_{2}, \ldots , X_{n}, \ldots be a sequence of IID continuous rv's with the common PDF fX(x)\mathsf{f}_{X}(x); note that P(X=α)=0\mathbb {P}\left(X=\alpha \right) = 0 for all α\alpha and that P(Xi=Xj)=0\mathbb {P}\left(X_{i}=X_{j}\right) = 0 for all i≠ji \neq j. For n≥2n \geq 2, define XnX_{n} as a record-to-date of the sequence if Xn>XiX_{n}>X_{i} for all i<ni<n.

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(a)

Find the probability that X2X_{2} is a record-to-date. Use symmetry to obtain a numerical answer without computation. A one- or two-line explanation should be adequate.

(b)

Find the probability that XnX_{n} is a record-to-date, as a function of n≥1n \geq 1. Again use symmetry.

(c)

Find a simple expression for the expected number of records-to-date that occur over the first mm trials for any given integer mm. Show that this expected number is infinite in the limit m→∞m \rightarrow \infty.

Problem 1.17
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(a)

Let N1N_{1} be the index of the first record-to-date in the sequence. Find P(N1>n)\mathbb {P}\left(N_{1}>n\right) for each n≥2n \geq 2.

(b)

Show that N1N_{1} is a rv.

(c)

Show that E[N1]=∞\mathbb {E}\left[N_{1}\right] = \infty.

(d)

Let N2N_{2} be the index of the second record-to-date in the sequence. Show that N2N_{2} is a rv. You need not find the CDF of N2N_{2} here.

(e)

Contrast your result in (c) to the result from Exercise 1.16(c) saying that the expected number of records-to-date is infinite over an infinite number of trials. Note: this might be a shock to your intuition -- there is an infinite expected wait for the first of an infinite sequence of occurrences, each of which must eventually occur.

Problem 1.18
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(a)

For any given n≥2n \geq 2, find the probability that XnX_{n} and Xn+1X_{n+1} are both records-to-date.

(b)

Is the event that XnX_{n} is a record-to-date statistically independent of the event that Xn+1X_{n+1} is a record-to-date?

(c)

Find the expected number of adjacent pairs of records-to-date over the sequence X1,X2,…X_{1}, X_{2}, \ldots.

Problem 1.19
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(a)

Assume that XX is a non-negative discrete rv taking on values a1,a2,…a_{1}, a_{2}, \ldots, and let Y=h(X)Y = h(X) for some non-negative function hh. Let bi=h(ai)b_{i} = h(a_{i}), i≥1i \geq 1, be the iith value taken on by YY. Show that E[Y]=∑ibipY(bi)=∑ih(ai)pX(ai)\mathbb {E}\left[Y\right] = \sum_{i}b_{i}\mathsf{p}_{Y}(b_{i}) = \sum_{i}h(a_{i})\mathsf{p}_{X}(a_{i}). Find an example where E[X]\mathbb {E}\left[X\right] exists but E[Y]=∞\mathbb {E}\left[Y\right] = \infty.

(b)

Let XX be a non-negative continuous rv with density fX(x)\mathsf{f}_{X}(x) and let h(x)h(x) be differentiable, non-negative, and strictly increasing in xx. Let A(δ)=∑n≥1h(nδ)[F(nδ)−F(nδ−δ)]A(\delta ) = \sum_{n \geq 1}h(n\delta )[F(n\delta )-F(n\delta -\delta )], i.e., A(δ)A(\delta ) is a δ\deltath-order approximation to the Stieltjes integral ∫h(x) dF(x)\int h(x) \, dF(x). Show that if A(1)<∞A(1)<\infty, then A(2−k)≤A(2−(k−1))<∞A(2^{-k}) \leq A(2^{-(k-1)})<\infty. Show from this that ∫h(x) dF(x)\int h(x) \, dF(x) converges to a finite value. Note: This is a very special case, but it can be extended to many cases of interest. It seems better to consider these convergence questions as required rather than to consider them in general.

Problem 1.20
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(a)

Consider a positive, integer-valued rv whose CDF is given at integer values by

FY(y)=1−2(y+1)(y+2)for integer y≥0. F_{Y}(y) = 1-\frac{2}{(y+1)(y+2)} \qquad \text{for integer } y \geq 0.

Use E[Y]=∫0∞FYc(y) dy\mathbb {E}\left[Y\right] = \int_{0}^{\infty }F_{Y}^{c}(y) \, dy (equation (1.30)) to show that E[Y]=2\mathbb {E}\left[Y\right] = 2.

(b)

Find the PMF of YY and use it to check the value of E[Y]\mathbb {E}\left[Y\right].

(c)

Let XX be another positive, integer-valued rv. Assume its conditional PMF is given by

pX∣Y(x∣y)=1yfor 1≤x≤y. \mathsf{p}_{X \mid Y}(x \mid y) = \frac{1}{y} \qquad \text{for } 1 \leq x \leq y.

Find E[X∣Y=y]\mathbb {E}\left[X \mid Y=y\right] and show that E[X]=3/2\mathbb {E}\left[X\right] = 3/2. Explore finding pX(x)\mathsf{p}_{X}(x) until you are convinced that using the conditional expectation to calculate E[X]\mathbb {E}\left[X\right] is considerably easier than using pX(x)\mathsf{p}_{X}(x).

(d)

Let ZZ be another integer-valued rv with the conditional PMF

pZ∣Y(z∣y)=1y2for 1≤z≤y2. \mathsf{p}_{Z \mid Y}(z \mid y) = \frac{1}{y^{2}} \qquad \text{for } 1 \leq z \leq y^{2}.

Find E[Z∣Y=y]\mathbb {E}\left[Z \mid Y=y\right] for each integer y≥1y \geq 1 and find E[Z]\mathbb {E}\left[Z\right].

Problem 1.21
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(a)

Show that, for uncorrelated rv's, the expected value of the product is equal to the product of the expected values (by definition, XX and YY are uncorrelated if E[(X−X‾)(Y−Y‾)]=0\mathbb {E}\left[(X-\overline{X})(Y-\overline{Y})\right] = 0).

(b)

Show that if XX and YY are uncorrelated, then the variance of X+YX+Y is equal to the variance of XX plus the variance of YY.

(c)

Show that if X1,…,XnX_{1}, \ldots , X_{n} are uncorrelated, then the variance of the sum is equal to the sum of the variances.

(d)

Show that independent rv's are uncorrelated.

(e)

Let X,YX, Y be identically distributed ternary valued rv's with the PMF pX(−1)=pX(1)=1/4\mathsf{p}_{X}(-1) = \mathsf{p}_{X}(1) = 1/4; pX(0)=1/2\mathsf{p}_{X}(0) = 1/2. Find a simple joint probability assignment such that XX and YY are uncorrelated but dependent.

(f)

You have seen that the MGF of a sum of independent rv's is equal to the product of the individual MGFs. Give an example where this is false if the variables are uncorrelated but dependent.

Problem 1.22

Suppose XX has the Poisson PMF, pX(n)=λnexp⁡(−λ)/n!\mathsf{p}_{X}(n) = \lambda^{n}\exp (-\lambda )/n! for n≥0n \geq 0 and YY has the Poisson PMF, pY(m)=μnexp⁡(−μ)/n!\mathsf{p}_{Y}(m) = \mu^{n}\exp (-\mu )/n! for n≥0n \geq 0. Assume that XX and YY are independent. Show that Z=X+YZ = X+Y is a Poisson rv and find the conditional distribution of YY conditional on Z=nZ=n.

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Problem 1.23
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(a)

Suppose X,YX, Y and ZZ are binary rv's, each taking on the value 0 with probability 1/21/2 and the value 1 with probability 1/21/2. Find a simple example in which X,Y,ZX, Y, Z are statistically dependent but are pairwise statistically independent (i.e., X,YX, Y are statistically independent, X,ZX, Z are statistically independent, and Y,ZY, Z are statistically independent). Give pXYZ(x,y,z)\mathsf{p}_{XYZ}(x, y, z) for your example.

(b)

Is pairwise statistical independence enough to ensure that

E[∏i=1nXi]=∏i=1nE[Xi] \mathbb {E}\left[\prod _{i=1}^{n}X_{i}\right] = \prod _{i=1}^{n}\mathbb {E}\left[X_{i}\right]

for a set of rv's X1,…,XnX_{1}, \ldots , X_{n}?

Problem 1.24

Show that E[X]\mathbb {E}\left[X\right] is the value of α\alpha that minimizes E[(X−α)2]\mathbb {E}\left[(X-\alpha )^{2}\right].

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Problem 1.25

For each of the following random variables, find the endpoints r−r_{-} and r+r_{+} of the interval for which the moment generating function g(r)\mathsf{g}(r) exists. Determine in each case whether g(r)\mathsf{g}(r) exists at r−r_{-} and r+r_{+}. For parts (a) and (b) you should also find and sketch g(r)\mathsf{g}(r). For parts (c) and (d), g(r)\mathsf{g}(r) has no closed form.

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(a)

Let λ,θ\lambda , \theta, be positive numbers and let XX have the density

fX(x)=12λexp⁡(−λx); x≥0;fX(x)=12θexp⁡(θx); x<0. \mathsf{f}_{X}(x) = \frac{1}{2}\lambda \exp (-\lambda x); \, x \geq 0; \qquad \mathsf{f}_{X}(x) = \frac{1}{2}\theta \exp (\theta x); \, x<0.
(b)

Let YY be a Gaussian rv with mean mm and variance σ2\sigma^{2}.

(c)

Let ZZ be a non-negative rv with density

fZ(z)=k(1+z)−2exp⁡(−λz)z≥0, \mathsf{f}_{Z}(z) = k(1+z)^{-2}\exp (-\lambda z) \qquad z \geq 0,

where λ>0\lambda >0 and k=[∫z≥0(1+z)−2exp⁡(−λz) dz]−1k = \left[\int_{z \geq 0}(1+z)^{-2}\exp (-\lambda z) \, dz\right]^{-1}.

(d)

For the ZZ of (c), find the limit of γ′(r)\gamma '(r) as rr approaches λ\lambda from below. Then replace (1+z)2(1+z)^{2} with ∣1+z∣3\left|1+z\right|^{3} in the definition of fZ(z)\mathsf{f}_{Z}(z) and kk and show whether the above limit is then finite or not.

Problem 1.26
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(a)

Assume that the MGF of the rv XX exists (i.e., is finite) in the interval (r−,r+)(r_{-}, r_{+}), r−<0<r+r_{-}<0<r_{+}, and assume r−<r<r+r_{-}<r<r_{+} throughout. For any finite constant cc, express the MGF of X−cX-c, i.e., g(X−c)(r)\mathsf{g}_{(X-c)}(r), in terms of the MGF of XX and show that g(X−c)′′(r)≥0\mathsf{g}_{(X-c)}''(r) \geq 0. Explain why g(X−c)′′(r)≥0\mathsf{g}_{(X-c)}''(r) \geq 0.

(b)

Show that g(X−c)′′(r)=[gX′′(r)−2cgX′(r)+c2gX(r)]e−rc\mathsf{g}_{(X-c)}''(r) = [\mathsf{g}_{X}''(r)-2c\mathsf{g}_{X}'(r)+c^{2}\mathsf{g}_{X}(r)]e^{-rc}.

(c)

Use (a) and (b) to show that gX′′(r)gX(r)−[gX′(r)]2≥0\mathsf{g}_{X}''(r)\mathsf{g}_{X}(r)-[\mathsf{g}_{X}'(r)]^{2} \geq 0. Let γX(r)=ln⁡gX(r)\gamma_{X}(r) = \ln \mathsf{g}_{X}(r) and show that γX′′(r)≥0\gamma_{X}''(r) \geq 0.

(d)

Assume that XX is non-deterministic, i.e., there is no value of α\alpha such that P(X=α)=1\mathbb {P}\left(X=\alpha \right) = 1. Show that the inequality sign $\geq$' may be replaced by >>' everywhere in (a), (b) and (c).

Problem 1.27

A computer system has nn users, each with a unique name and password. Due to a software error, the nn passwords are randomly permuted internally (i.e., each of the n!n! possible permutations are equally likely). Only those users lucky enough to have had their passwords unchanged in the permutation are able to continue using the system.

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(a)

What is the probability that a particular user, say user 1, is able to continue using the system?

(b)

What is the expected number of users able to continue using the system?

Problem 1.28

Suppose the rv XX is continuous and has the CDF FX(x)F_{X}(x). Consider another rv Y=FX(X)Y = F_{X}(X). That is, for each sample point ω\omega such that X(ω)=xX(\omega ) = x, we have Y(ω)=FX(x)Y(\omega ) = F_{X}(x). Show that YY is uniformly distributed in the interval 0 to 1.

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Problem 1.29

Let ZZ be an integer-valued rv with the PMF pZ(n)=1/k\mathsf{p}_{Z}(n) = 1/k for 0≤n≤k−10 \leq n \leq k-1. Find the mean, variance, and moment generating function of ZZ.

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Problem 1.30
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(a)

Let YY be a non-negative rv and y>0y>0 be some fixed number. Let AA be the event that Y≥yY \geq y. Show that yIA≤Yy\mathbb {I}_{A} \leq Y (i.e., that this inequality is satisfied for every ω∈Ω\omega \in \Omega).

(b)

Use your result in (a) to prove the Markov inequality.

Problem 1.31
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(a)

Minimize E[Y]\mathbb {E}\left[Y\right] over all non-negative rv's such that P(Y≥b)=β\mathbb {P}\left(Y \geq b\right) = \beta for some given b>0b>0 and 0<β<10<\beta <1. What is the rv that achieves the minimum?

(b)

Use (a) to prove the Markov inequality and also point out the distribution that meets the inequality with equality.

Problem 1.32

This inequality states that if a zero-mean rv XX has a variance σ2\sigma^{2}, then it satisfies the inequality

P(X≥b)≤σ2σ2+b2for every b>0, \mathbb {P}\left(X \geq b\right) \leq \frac{\sigma ^{2}}{\sigma ^{2}+b^{2}} \qquad \text{for every } b>0,

with equality for some bb only if XX is binary and P(X=b)=σ2/(σ2+b2)\mathbb {P}\left(X=b\right) = \sigma^{2}/(\sigma^{2}+b^{2}). We prove this here using the same approach as in Exercise 1.31. Let XX be a zero-mean rv that satisfies P(X≥b)=β\mathbb {P}\left(X \geq b\right) = \beta for some given b>0b>0 and 0<β<10<\beta <1. The variance σ2\sigma^{2} of XX can be expressed as

σ2=∫−∞b−x2fX(x) dx+∫b∞x2fX(x) dx. \sigma ^{2} = \int _{-\infty }^{b^{-}}x^{2}\mathsf{f}_{X}(x) \, dx+\int _{b}^{\infty }x^{2}\mathsf{f}_{X}(x) \, dx.

We will first minimize σ2\sigma^{2} over all zero-mean XX satisfying P(X≥b)=β\mathbb {P}\left(X \geq b\right) = \beta.

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(a)

Show that the second integral above satisfies ∫b∞x2fX(x) dx≥b2β\int_{b}^{\infty }x^{2}\mathsf{f}_{X}(x) \, dx \geq b^{2}\beta.

(b)

Show that the first integral above is constrained by

∫−∞b−fX(x) dx=1−βand∫−∞b−xfX(x) dx≤−bβ. \int _{-\infty }^{b^{-}}\mathsf{f}_{X}(x) \, dx = 1-\beta \qquad \text{and} \qquad \int _{-\infty }^{b^{-}}x\mathsf{f}_{X}(x) \, dx \leq -b\beta .
(c)

Minimize the first integral subject to the constraints in (b).

(d)

Combine the results in (a) and (c) to show that σ2≥b2β/(1−β)\sigma^{2} \geq b^{2}\beta /(1-\beta ). Find the minimizing distribution.

(e)

Use (d) to establish the one-sided Chebyshev inequality above. Also show (trivially) that if YY has a mean Y‾\overline{Y} and variance σ2\sigma^{2}, then P(Y−Y‾≥b)≤σ2/(σ2+b2)\mathbb {P}\left(Y-\overline{Y} \geq b\right) \leq \sigma^{2}/(\sigma^{2}+b^{2}).

Problem 1.33

Here we show that if XX is a zero-mean rv with a variance σ2\sigma^{2}, then the median α\alpha satisfies ∣α∣≤σ\left|\alpha \right| \leq \sigma.

The one-sided Chebyshev inequality (Exercise 1.32) states that for a zero-mean rv XX with variance σ2\sigma^{2}, P(X≥b)≤σ2/(σ2+b2)\mathbb {P}\left(X \geq b\right) \leq \sigma^{2}/(\sigma^{2}+b^{2}) for every b>0b>0, with equality for some bb only if XX is binary and P(X=b)=σ2/(σ2+b2)\mathbb {P}\left(X=b\right) = \sigma^{2}/(\sigma^{2}+b^{2}).

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(a)

First show that ∣α∣≤σ\left|\alpha \right| \leq \sigma for the special case where XX is binary with equiprobable values at ±σ\pm \sigma.

(b)

For all zero-mean rv's XX with variance σ2\sigma^{2} other than the special case in (a), show that

P(X≥σ)<0.5. \mathbb {P}\left(X \geq \sigma \right)<0.5.
(c)

Show that P(X≥α)≥0.5\mathbb {P}\left(X \geq \alpha \right) \geq 0.5. Other than the special case in (a), show that this implies that α<σ\alpha <\sigma.

(d)

Other than the special case in (a), show that ∣α∣<σ\left|\alpha \right|<\sigma. You have then shown that ∣α∣≤σ\left|\alpha \right| \leq \sigma with equality only for the binary case with values ±σ\pm \sigma. For rv's YY with a non-zero mean, this shows that ∣α−Y‾∣≤σ\left|\alpha -\overline{Y}\right| \leq \sigma.

Problem 1.34

We stressed the importance of the mean of a rv XX in terms of its association with the sample average via the WLLN. Here we show that there is a form of WLLN for the median and for the entire CDF, say FX(x)F_{X}(x) of XX via sufficiently many independent sample values of XX.

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(a)

For any given xx, let Ij(x)\mathbb {I}_{j}(x) be the indicator function of the event {Xj≤x}\left\{ X_{j} \leq x\right\} where X1,X2,…,Xj,…X_{1}, X_{2}, \ldots , X_{j}, \ldots are IID rv's with the CDF FX(x)F_{X}(x). State the WLLN for the IID rv's {I1(x),I2(x),…}\left\{ \mathbb {I}_{1}(x), \mathbb {I}_{2}(x), \ldots \right\}.

(b)

Does the answer to (a) require XX to have a mean or variance?

(c)

Suggest a procedure for evaluating the median of XX from the sample values of X1,X2,…X_{1}, X_{2}, \ldots. Assume that XX is a continuous rv and that its PDF is positive in an open interval around the median. You need not be precise, but try to think the issue through carefully.

What you have seen here, without stating it precisely or proving it is that the median has a law of large numbers associated with it, saying that the sample median of nn IID samples of a rv is close to the true median with high probability.

Problem 1.35
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(a)

Show that for any integers 0<k<n0<k<n,

(nk+1)≤(nk)n−kk. \binom {n}{k+1} \leq \binom {n}{k}\frac{n-k}{k}.
(b)

Extend (a) to show that, for all ℓ≤n−k\ell \leq n-k,

(nk+ℓ)≤(nk)[n−kk]ℓ. \binom {n}{k+\ell } \leq \binom {n}{k}\left[\frac{n-k}{k}\right]^{\ell }.
(c)

Let p~=k/n\widetilde{p} = k/n and q~=1−p~\widetilde{q} = 1-\widetilde{p}. Let SnS_{n} be the sum of nn binary IID rv's with pX(0)=q\mathsf{p}_{X}(0) = q and pX(1)=p\mathsf{p}_{X}(1) = p. Show that for all ℓ≤n−k\ell \leq n-k,

pSn(k+ℓ)≤pSn(k)(q~pp~q)ℓ. \mathsf{p}_{S_{n}}(k+\ell ) \leq \mathsf{p}_{S_{n}}(k)\left(\frac{\widetilde{q}p}{\widetilde{p}q}\right)^{\ell }.
(d)

For k/n>pk/n>p, show that

P(Sn≥kn)≤p~qp~−ppSn(k). \mathbb {P}\left(S_{n} \geq kn\right) \leq \frac{\widetilde{p}q}{\widetilde{p}-p}\mathsf{p}_{S_{n}}(k).
(e)

Now let ℓ\ell be fixed and k=⌈np~⌉k = \lceil n\widetilde{p}\rceil for fixed p~\widetilde{p} such that 1>p~>p1>\widetilde{p}>p. Argue that as n→∞n \rightarrow \infty,

pSn(k+ℓ)∼pSn(k)(q~pp~q)ℓandP(Sn≥kn)∼p~qp~−ppSn(k), \mathsf{p}_{S_{n}}(k+\ell ) \sim \mathsf{p}_{S_{n}}(k)\left(\frac{\widetilde{q}p}{\widetilde{p}q}\right)^{\ell } \qquad \text{and} \qquad \mathbb {P}\left(S_{n} \geq kn\right) \sim \frac{\widetilde{p}q}{\widetilde{p}-p}\mathsf{p}_{S_{n}}(k),

where a(n)∼b(n)a(n) \sim b(n) means that lim⁡n→∞a(n)/b(n)=1\lim_{n \rightarrow \infty }a(n)/b(n) = 1.

Problem 1.36

A sequence {an;n≥1}\left\{ a_{n}; n \geq 1\right\} of real numbers has the limit 0 if, for all ϵ>0\epsilon >0, there is an m(ϵ)m(\epsilon ) such that ∣an∣≤ϵ\left|a_{n}\right| \leq \epsilon for all n≥m(ϵ)n \geq m(\epsilon ). Show that the sequences in (a) and (b) below satisfy lim⁡n→∞an=0\lim_{n \rightarrow \infty }a_{n} = 0 but the sequence in (c) does not have a limit.

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(a)

an=1ln⁡(ln⁡(n+1))a_{n} = \dfrac {1}{\ln (\ln (n+1))}

(b)

an=n10exp⁡(−n)a_{n} = n^{10}\exp (-n)

(c)

an=1a_{n} = 1 for n=10ℓn = 10^{\ell } for each positive integer ℓ\ell and an=0a_{n} = 0 otherwise.

(d)

Show that the definition can be changed (with no change in meaning) by replacing ϵ\epsilon with either 1/k1/k or 2−k2^{-k} for every positive integer kk.

Problem 1.37

Represent the MGF of a rv XX by

gX(r)=∫−∞0erx dF(x)+∫0∞erx dF(x). \mathsf{g}_{X}(r) = \int _{-\infty }^{0}e^{rx} \, dF(x)+\int _{0}^{\infty }e^{rx} \, dF(x).

In each of the following parts, you are welcome to restrict XX to be either discrete or continuous.

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(a)

Show that the first integral always exists (i.e., is finite) for r≥0r \geq 0 and that the second integral always exists for r≤0r \leq 0.

(b)

Show that if the second integral exists for a given r1>0r_{1}>0, then it also exists for all rr in the range 0≤r≤r10 \leq r \leq r_{1}.

(c)

Show that if the first integral exists for a given r2<0r_{2}<0, then it also exists for all rr in the range r2≤r≤0r_{2} \leq r \leq 0.

(d)

Show that the range of rr over which gX(r)\mathsf{g}_{X}(r) exists is an interval from some r−≤0r_{-} \leq 0 to some r+≥0r_{+} \geq 0 (the interval might or might not include each endpoint, and the magnitude of either or both endpoints might be 0, ∞\infty, or any point between).

(e)

Find an example where r+=1r_{+}=1 and the MGF does not exist for r=1r=1. Find another example where r+=1r_{+}=1 and the MGF does exist for r=1r=1.

Problem 1.38

Let {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\} be a sequence of independent but not identically distributed rv's. We say that the weak law of large numbers (WLLN) holds for this sequence if for all ϵ>0\epsilon >0

lim⁡n→∞P(∣Snn−E[Sn]n∣≥ϵ)=0where Sn=X1+X2+⋯+Xn. \lim _{n \rightarrow \infty }\mathbb {P}\left(\left|\frac{S_{n}}{n}-\frac{\mathbb {E}\left[S_{n}\right]}{n}\right| \geq \epsilon \right) = 0 \qquad \text{where } S_{n} = X_{1}+X_{2}+\cdots +X_{n}.
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(a)

Show that the WLLN holds if there is some constant AA such that σXn2≤A\sigma_{X_{n}}^{2} \leq A for all nn.

(b)

Suppose that σXn2≤An1−α\sigma_{X_{n}}^{2} \leq An^{1-\alpha } for some α\alpha, 0<α<10<\alpha <1 and for all n≥1n \geq 1. Show that the WLLN holds in this case.

Problem 1.39

Let {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} be IID binary rv's. Let P(Xi=1)=δ\mathbb {P}\left(X_{i}=1\right) = \delta, P(Xi=0)=1−δ\mathbb {P}\left(X_{i}=0\right) = 1-\delta. Let Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n}. Let mm be an arbitrary but fixed positive integer. Think! Then evaluate the following and explain your answers:

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(a)

lim⁡n→∞∑i : nδ−m≤i≤nδ+mP(Sn=i)\lim_{n \rightarrow \infty }\sum_{i \, : \, n\delta -m \leq i \leq n\delta +m}\mathbb {P}\left(S_{n}=i\right).

(b)

lim⁡n→∞∑i : 0≤i≤nδ+mP(Sn=i)\lim_{n \rightarrow \infty }\sum_{i \, : \, 0 \leq i \leq n\delta +m}\mathbb {P}\left(S_{n}=i\right).

(c)

lim⁡n→∞∑i : n(δ−1/m)≤i≤n(δ+1/m)P(Sn=i)\lim_{n \rightarrow \infty }\sum_{i \, : \, n(\delta -1/m) \leq i \leq n(\delta +1/m)}\mathbb {P}\left(S_{n}=i\right).

Problem 1.40

The WLLN is said to hold for a zero-mean sequence {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\} if

lim⁡n→∞P(∣Snn∣>ϵ)=0for every ϵ>0. \lim _{n \rightarrow \infty }\mathbb {P}\left(\left|\frac{S_{n}}{n}\right|>\epsilon \right) = 0 \qquad \text{for every } \epsilon >0.

The CLT is said to hold for {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\} if for some σ>0\sigma >0 and all z∈Rz \in \mathbb {R},

lim⁡n→∞P(Snσn≤z)=Φ(z), \lim _{n \rightarrow \infty }\mathbb {P}\left(\frac{S_{n}}{\sigma \sqrt{n}} \leq z\right) = \Phi (z),

where Φ(z)\Phi (z) is the normal CDF. Show that if the CLT holds, then the WLLN also holds.

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Remark.
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Note 1: If you hate ϵ,δ\epsilon , \delta arguments, you will hate this. Note 2: It will probably ease the pain if you convert the WLLN statement to: For every ϵ>0,δ>0\epsilon >0, \delta >0 there exists an n(ϵ,δ)n(\epsilon , \delta ) such that for every n≥n(ϵ,δ)n \geq n(\epsilon , \delta ),

P(Snn<−ϵ)≤δandP(Snn>ϵ)≤δ. \mathbb {P}\left(\frac{S_{n}}{n}<-\epsilon \right) \leq \delta \qquad \text{and} \qquad \mathbb {P}\left(\frac{S_{n}}{n}>\epsilon \right) \leq \delta .
Problem 1.41
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(a)

Show that if X1,X2,…X_{1}, X_{2}, \ldots are IID, then the truncated versions X˘1,X˘2,…\breve{X}_{1}, \breve{X}_{2}, \ldots are also IID.

(b)

Show that each X˘i\breve{X}_{i} has a finite mean E[X˘]\mathbb {E}\left[\breve{X}\right] and finite variance σX˘2\sigma_{\breve{X}}^{2}. Show that the variance is upper bounded by the second moment around the original mean X‾\overline{X}, i.e., show that σX˘2≤E[∣X˘−E[X]∣2]\sigma_{\breve{X}}^{2} \leq \mathbb {E}\left[\left|\breve{X}-\mathbb {E}\left[X\right]\right|^{2}\right].

(c)

Assume that X˘i\breve{X}_{i} is XiX_{i} truncated to X‾±b\overline{X} \pm b. Show that ∣X˘−X‾∣≤b\left|\breve{X}-\overline{X}\right| \leq b and that ∣X˘−X‾∣≤∣X−X‾∣\left|\breve{X}-\overline{X}\right| \leq \left|X-\overline{X}\right|. Use this to show that σX˘2≤bE[∣X˘−X‾∣]≤2bE[∣X∣]\sigma_{\breve{X}}^{2} \leq b\mathbb {E}\left[\left|\breve{X}-\overline{X}\right|\right] \leq 2b\mathbb {E}\left[\left|X\right|\right].

(d)

Let S˘n=X˘1+⋯+X˘n\breve{S}_{n} = \breve{X}_{1}+\cdots +\breve{X}_{n} and show that for any ϵ>0\epsilon >0,

P(∣S˘nn−E[X˘]∣≥ϵ2)≤8bE[∣X∣]nϵ2. \mathbb {P}\left(\left|\frac{\breve{S}_{n}}{n}-\mathbb {E}\left[\breve{X}\right]\right| \geq \frac{\epsilon }{2}\right) \leq \frac{8b\mathbb {E}\left[\left|X\right|\right]}{n\epsilon ^{2}}.
(e)

Use the fact that XX and X˘\breve{X} have the same CDF from X‾−b\overline{X}-b to X‾+b\overline{X}+b, along with E[X−X˘]=−∫−∞X‾−bFX(x) dx+∫X‾+b∞Fc(x) dx\mathbb {E}\left[X-\breve{X}\right] = -\int_{-\infty }^{\overline{X}-b}F_{X}(x) \, dx+\int_{\overline{X}+b}^{\infty }F^{c}(x) \, dx, to show that for all sufficiently large bb, ∣E[X−X˘]∣≤ϵ/2\left|\mathbb {E}\left[X-\breve{X}\right]\right| \leq \epsilon /2. Use this to show that

P(∣S˘nn−E[X]∣≥ϵ)≤8bE[∣X∣]nϵ2for all large enough b. \mathbb {P}\left(\left|\frac{\breve{S}_{n}}{n}-\mathbb {E}\left[X\right]\right| \geq \epsilon \right) \leq \frac{8b\mathbb {E}\left[\left|X\right|\right]}{n\epsilon ^{2}} \qquad \text{for all large enough } b.
(f)

Use the following equation to justify the truncation argument that Sn/nS_{n}/n and S˘n/n\breve{S}_{n}/n have the same limiting behavior.

P(∣Snn−E[X]∣>ϵ)=P({∣Snn−E[X]∣>ϵ}∩{Sn=S˘n})+P({∣Snn−E[X]∣>ϵ}∩{Sn≠S˘n}). \begin{aligned} \mathbb {P}\left(\left|\frac{S_{n}}{n}-\mathbb {E}\left[X\right]\right|>\epsilon \right) & = \mathbb {P}\left(\left\{ \left|\frac{S_{n}}{n}-\mathbb {E}\left[X\right]\right|>\epsilon \right\} \cap \left\{ S_{n}=\breve{S}_{n}\right\} \right) \\ & \quad +\mathbb {P}\left(\left\{ \left|\frac{S_{n}}{n}-\mathbb {E}\left[X\right]\right|>\epsilon \right\} \cap \left\{ S_{n} \neq \breve{S}_{n}\right\} \right). \end{aligned}
Problem 1.42

Let {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} be IID rv's with mean 0 and infinite variance. Assume that E[∣Xi∣1+h]=β\mathbb {E}\left[\left|X_{i}\right|^{1+h}\right] = \beta for some given hh, 0<h<10<h<1 and some finite β\beta. Let Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n}.

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(a)

Show that P(∣Xi∣≥y)≤βy−1−h\mathbb {P}\left(\left|X_{i}\right| \geq y\right) \leq \beta y^{-1-h}.

(b)

Let {X˘i;i≥1}\left\{ \breve{X}_{i}; i \geq 1\right\} be truncated variables defined as

X˘i={b:Xi≥b,Xi:−b≤Xi≤b,−b:Xi≤−b. \breve{X}_{i} = \begin{cases} b & : X_{i} \geq b, \\ X_{i} & : -b \leq X_{i} \leq b, \\ -b & : X_{i} \leq -b. \end{cases}

Show that E[X˘2]≤2βb1−h/(1−h)\mathbb {E}\left[\breve{X}^{2}\right] \leq 2\beta b^{1-h}/(1-h).

(c)

Let S˘n=X˘1+⋯+X˘n\breve{S}_{n} = \breve{X}_{1}+\cdots +\breve{X}_{n}. Show that P(Sn≠S˘n)≤nβb−1−h\mathbb {P}\left(S_{n} \neq \breve{S}_{n}\right) \leq n\beta b^{-1-h}.

(d)

Show that

P(∣Snn∣≥ϵ)≤β[2b1−h(1−h)nϵ2+nb1+h]. \mathbb {P}\left(\left|\frac{S_{n}}{n}\right| \geq \epsilon \right) \leq \beta \left[\frac{2b^{1-h}}{(1-h)n\epsilon ^{2}}+\frac{n}{b^{1+h}}\right].
(e)

Optimize your bound with respect to bb. How fast does this optimized bound approach 0 with increasing nn?

Problem 1.43

Assume that {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a sequence of rv's and α\alpha is a number with the property that lim⁡n→∞E[(Zn−α)2]=0\lim_{n \rightarrow \infty }\mathbb {E}\left[(Z_{n}-\alpha )^{2}\right] = 0.

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(a)

Let ϵ>0\epsilon >0 be arbitrary and show that for each n≥0n \geq 0,

P(∣Zn−α∣≥ϵ)≤E[(Zn−α)2]ϵ2. \mathbb {P}\left(\left|Z_{n}-\alpha \right| \geq \epsilon \right) \leq \frac{\mathbb {E}\left[(Z_{n}-\alpha )^{2}\right]}{\epsilon ^{2}}.
(b)

For the ϵ\epsilon above, let δ>0\delta >0 be arbitrary. Show that there is an integer mm such that E[(Zn−α)2]≤ϵ2δ\mathbb {E}\left[(Z_{n}-\alpha )^{2}\right] \leq \epsilon^{2}\delta for all n≥mn \geq m.

(c)

Show that this implies convergence in probability.

Problem 1.44

Let X1,X2…X_{1}, X_{2}\ldots be a sequence of IID rv's each with mean 0 and variance σ2\sigma^{2}. Let Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n} for all nn and consider the random variable Sn/σn−S2n/σ2nS_{n}/\sigma \sqrt{n}-S_{2n}/\sigma \sqrt{2n}. Find the limiting CDF for this sequence of rv's as n→∞n \rightarrow \infty. The point of this exercise is to see clearly that the CDF of Sn/σn−S2n/σ2nS_{n}/\sigma \sqrt{n}-S_{2n}/\sigma \sqrt{2n} is converging in nn but that the sequence of rv's is not converging in any reasonable sense.

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Problem 1.45

Verify that lim⁡y→∞yP(Y≥y)=0\lim_{y \rightarrow \infty }y\mathbb {P}\left(Y \geq y\right) = 0 if Y≥0Y \geq 0 and Y‾<∞\overline{Y}<\infty.

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Problem 1.46

Show that ∏m≥n(1−1/m)=0\prod_{m \geq n}(1-1/m) = 0.

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Problem 1.47

Consider a discrete rv XX with the PMF

pX(−1)=(1−10−10)/2,pX(1)=(1−10−10)/2,pX(1012)=10−10. \begin{aligned} \mathsf{p}_{X}(-1) & = (1-10^{-10})/2, \\ \mathsf{p}_{X}(1) & = (1-10^{-10})/2, \\ \mathsf{p}_{X}(10^{12}) & = 10^{-10}. \end{aligned}
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(a)

Find the mean and variance of XX. Assuming that {Xm;m≥1}\left\{ X_{m}; m \geq 1\right\} is an IID sequence with the distribution of XX and that Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n} for each nn, find the mean and variance of SnS_{n}. (No explanations needed.)

(b)

Let n=106n = 10^{6} and describe the event {Sn≤106}\left\{ S_{n} \leq 10^{6}\right\} in words. Find an exact expression for P(Sn≤106)=FSn(106)\mathbb {P}\left(S_{n} \leq 10^{6}\right) = F_{S_{n}}(10^{6}).

(c)

Find a way to use the union bound to get a simple upper bound and approximation of 1−FSn(106)1-F_{S_{n}}(10^{6}).

(d)

Sketch the CDF of SnS_{n} for n=106n = 10^{6}. You can choose the horizontal axis for your sketch to go from −1-1 to +1+1 or from −3×103-3 \times 10^{3} to 3×1033 \times 10^{3} or from −106-10^{6} to 10610^{6} or from 0 to 101210^{12}, whichever you think will best describe this CDF.

(e)

Now let n=1010n = 10^{10}. Give an exact expression for P(Sn≤1010)\mathbb {P}\left(S_{n} \leq 10^{10}\right) and show that this can be approximated by e−1e^{-1}. Sketch the CDF of SnS_{n} for n=1010n = 10^{10}, using a horizontal axis going from slightly below 0 to slightly more than 2×10122 \times 10^{12}.

(f)

Can you make a qualitative statement about how the CDF of a rv XX affects the required size of nn before the WLLN and the CLT provide much of an indication about SnS_{n}.

Problem 1.48

Let {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} be a sequence of rv's and assume that lim⁡n→∞E[∣Yn∣]=0\lim_{n \rightarrow \infty }\mathbb {E}\left[\left|Y_{n}\right|\right] = 0. Show that {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} converges to 0 in probability.

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