Section 5.2

Some applications

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Problem 5.2.1

Let XX be the number of events in the sequence A1,A2,…,AnA_{1}, A_{2}, \ldots , A_{n} which occur. Let Sm=E[(Xm)]S_{m} = \mathbb {E}\left[\binom {X}{m}\right], the mean value of the random binomial coefficient (Xm)\binom {X}{m}, and show that

P(X≥i)=∑j=in(−1)j−i(j−1i−1)Sj, for 1≤i≤n, where Sm=∑j=mn(j−1m−1)P(X≥j), for 1≤m≤n. \begin{aligned} & \mathbb {P}\left(X \geq i\right) = \sum _{j = i}^{n}(-1)^{j-i}\binom {j-1}{i-1} S_{j}, \quad \text{ for } 1 \leq i \leq n, \\ & \text{ where } \quad S_{m} = \sum _{j = m}^{n}\binom {j-1}{m-1} \mathbb {P}\left(X \geq j\right), \quad \text{ for } 1 \leq m \leq n. \end{aligned}
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Problem 5.2.2

Each person in a group of nn people chooses another at random. Find the probability:

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(a)

that exactly kk people are chosen by nobody,

(b)

that at least kk people are chosen by nobody.

Problem 5.2.3
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(a)

Let XX have the Poisson distribution with parameter YY, where YY has the Poisson distribution with parameter μ\mu. Show that GX+Y(x)=exp⁡{μ(xex−1−1)}G_{X+Y}(x) = \exp \left\{ \mu \left(x e^{x-1}-1\right)\right\}.

(b)

Let X1,X2,…X_{1}, X_{2}, \ldots be independent identically distributed random variables with the logarithmic mass function

f(k)=(1−p)kklog⁡(1/p),k≥1 f(k) = \frac{(1-p)^{k}}{k \log (1 / p)}, \quad k \geq 1

where 0<p<10 < p < 1. If NN is independent of the XiX_{i} and has the Poisson distribution with parameter μ\mu, show that Y=∑i=1NXiY = \sum_{i = 1}^{N} X_{i} has a type of negative binomial distribution.

Problem 5.2.4

Let XX have the binomial distribution with parameters nn and pp, and show that

E[11+X]=1−(1−p)n+1(n+1)p \mathbb {E}\left[\frac{1}{1+X}\right] = \frac{1-(1-p)^{n+1}}{(n+1) p}

Find the limit of this expression as n→∞n \rightarrow \infty and p→0p \rightarrow 0, the limit being taken in such a way that np→λn p \rightarrow \lambda where 0<λ<∞0 < \lambda < \infty. Comment.

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Problem 5.2.5

A coin is tossed repeatedly, and heads turns up with probability pp on each toss. Let hnh_{n} be the probability of an even number of heads in the first nn tosses, with the convention that 0 is an even number. Find a difference equation for the hnh_{n} and deduce that they have generating function 12{(1+2ps−s)−1+(1−s)−1}\frac{1}{2}\left\{ (1+2 p s-s)^{-1}+(1-s)^{-1}\right\}.

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Problem 5.2.6

An unfair coin is flipped repeatedly, where P(H)=p=1−q\mathbb {P}\left(\mathrm{H}\right) = p = 1-q. Let XX be the number of flips until HTH first appears, and YY the number of flips until either HTH or THT appears. Show that E[sX]=(p2qs3)/(1−s+pqs2−pq2s3)\mathbb {E}\left[s^{X}\right] = \left(p^{2} q s^{3}\right) /\left(1-s+p q s^{2}-p q^{2} s^{3}\right) and find E[sY]\mathbb {E}\left[s^{Y}\right].

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Problem 5.2.7

The pile of (by now dog-eared) letters is dropped again and enveloped at random, yielding XnX_{n} matches. Show that P(Xn=j)=(j+1)P(Xn+1=j+1)\mathbb {P}\left(X_{n} = j\right) = (j+1) \mathbb {P}\left(X_{n+1} = j+1\right). Deduce that the derivatives of the Gn(s)=E[sXn]G_{n}(s) = \mathbb {E}\left[s^{X_{n}}\right] satisfy Gn+1′=GnG_{n+1}^{\prime } = G_{n}, and hence derive the conclusion of Example (3.4.3), namely:

P(Xn=r)=1r!(12!−13!+⋯+(−1)n−r(n−r)!) \mathbb {P}\left(X_{n} = r\right) = \frac{1}{r!}\left(\frac{1}{2!}-\frac{1}{3!}+\cdots +\frac{(-1)^{n-r}}{(n-r)!}\right)
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Problem 5.2.8

Let XX have a Poisson distribution with parameter Λ\Lambda, where Λ\Lambda is exponential with parameter μ\mu. Show that XX has a geometric distribution.

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Problem 5.2.9

Recall from Exercise (3.3.2) that each packet of an overpriced commodity contains a worthless plastic object. There are four types of object, and each packet is equally likely to contain any of the four. Let TT be the number of packets you open until you first have the complete set. Find E[sT]\mathbb {E}\left[s^{T}\right] and P(T=k)\mathbb {P}\left(T = k\right).

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Problem 5.2.10

A library permits a reader to hold at most mm books at any one time. Your holding before a visit is BB books, having the binomial bin⁡(m,p)\operatorname {bin}(m, p) distribution. On each visit, a held book is retained with probability rr, and returned otherwise; given the number RR of retained books, you borrow a further NN books with the bin⁡(m−R,α)\operatorname {bin}(m-R, \alpha ) distribution. Assuming the usual independence, find the distribution of the number A=R+NA = R+N of books held after a visit. Show that, after a large number of visits, the number of books held has approximately the bin⁡(m,s)\operatorname {bin}(m, s) distribution where s=α/[1−r(1−α)]s = \alpha /[1-r(1-\alpha )]. You may assume that 0<α,p,r<10 < \alpha , p, r < 1.

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Problem 5.2.11

Matching yet again, Exercise (5.2.7) revisited. Each envelope is addressed correctly with probability tt, and incorrectly otherwise (assume the usual independence). It is decided that an incorrectly addressed envelope cannot count as a match even when containing the correct letter. Show that the number YnY_{n} of correctly addressed matches satisfies

P(Yn=y)=tyy!∑k=0n−y(−1)ktkk!→e−ttyy! as n→∞, for y=0,1,2,… \mathbb {P}\left(Y_{n} = y\right) = \frac{t^{y}}{y!} \sum _{k = 0}^{n-y} \frac{(-1)^{k} t^{k}}{k!} \rightarrow \frac{e^{-t} t^{y}}{y!} \quad \text{ as } n \rightarrow \infty , \text{ for } y = 0,1,2, \ldots

Note the Poisson limit distribution.

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Problem 5.2.12

A random variable is called simple if it may take only finitely many values. Show that a simple random variable XX may be simulated using a finite sequence of Bernoulli random variables, in the sense that there exist independent Bernoulli random variables {Bj:j=1,2,…,m}\left\{ B_{j}: j = 1,2, \ldots , m\right\} with respective parameters bjb_{j}, and real numbers {rj}\left\{ r_{j}\right\}, such that Y=∑jrjBjY = \sum_{j} r_{j} B_{j} has the same distribution as XX.

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Problem 5.2.13
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(a)

For a generating function A(z)=∑i=0∞aiziA(z) = \sum_{i = 0}^{\infty } a_{i} z^{i}, show that a0+a1+⋯+ama_{0}+a_{1}+\cdots +a_{m} is the coefficient of zmz^{m} in B(z):=A(z)(1−zm+1)/(1−z)B(z): = A(z)\left(1-z^{m+1}\right) /(1-z)

(b)

With the definitions and notation of Example (5.2.11), show that, for rr an even integer,

Si−(i+1i)Si+1+⋯−(i+r+1i)Si+r+1≤P(X=i)≤Si−(i+1i)Si+1+⋯+(i+ri)Si+r \begin{aligned} S_{i}-\binom {i+1}{i} S_{i+1}+\cdots - & \binom {i+r+1}{i} S_{i+r+1} \\ & \leq \mathbb {P}\left(X = i\right) \leq S_{i}-\binom {i+1}{i} S_{i+1}+\cdots +\binom {i+r}{i} S_{i+r} \end{aligned}