Order Statistics Problem: Wackerly/Mendenhall/Scheaffer, 5th Ed., Problem 6.58

Question

Problem Statement:

Suppose that the number of occurrences of a certain event in time interval $(0,t)$ has a Poisson distribution. If we know that $n$ such events have occurred in $(0,t),$ then the actual times, measured from $0,$ for the occurrences of the event in question form an ordered set of random variables, which we denote by $W_{(1)}\le W_{(2)}\le\cdots\le W_{(n)}.$ [$W_{(i)}$ actually is the waiting time from $0$ until the occurrence of the $i$th event.] It can be shown that the joint density function for $W_{(1)}, W_{(2)},\dots,W_{(n)}$ is given by $$f(w_1, w_2,\dots,w_n)= \begin{cases} \dfrac{n!}{t^n},&w_1\le w_2\le\cdots\le w_n\\ 0,&\text{elsewhere.} \end{cases} $$ [This is the density function for an ordered sample of size $n$ from a uniform distribution on the interval $(0,t).$] Suppose that telephone calls coming into a switchboard follow a Poisson distribution with a mean of ten calls per minute. A slow period of $2$ minutes' duration had only four calls.

1. Find the probability that all four calls came in during the first minute; that is, find $P(W_{(4)}\le 1).$
2. Find the expected waiting time, from the start of the $2$-minute period, until the fourth call.

My Work So Far:

What's extremely confusing to me in this problem is how to interpret all the numbers I'm given. So I'm told the underlying Poisson distribution has $\lambda=10.$ Where does that figure into solving this problem, if at all? Then we're examining a slow period of $2$ minutes: where does that figure into solving this problem? Should $n=4$ in the joint density function above? Or should $n=10?$

I think if I could please have a nudge in the right direction for the first part, I imagine I could easily perform the integral to get the second.

Many thanks for your time!

Answer 1

[EDIT] Only keeping this answer to make sense of the comments from whuber, which have the solution implicit in them.

Well, I've got an answer to Part 1, I hope:

We start with the underlying Poisson distribution: $$p(y)=\frac{10^y\,e^{-10}}{y!},\; y=0, 1, 2,\dots$$ We will need the cumulative distribution function $$F(y) =P(Y\le y) =\sum_{x=0}^y \frac{10^x\,e^{-10}}{x!} =\frac{\Gamma(y+1, 10)}{\Gamma(y+1)}.$$ The function in the numerator is the upper incomplete gamma function. This is a discrete distribution, so computing the order statistics is completely different from continuous distributions. Following the wikipedia article, we define \begin{align*} p_1(y)&:=P(Y<y) =F(y)-p(y) =\frac{\Gamma(y+1, 10)}{\Gamma(y+1)}-\frac{10^y\,e^{-10}}{y!} =\frac{\Gamma(y+1, 10)-10^y\,e^{-10}}{y!}\\ p_2(y)&:=P(Y=y) =p(y) =\frac{10^y\,e^{-10}}{y!}\\ p_3(y)&:=P(Y>y) =1-F(y) =1-\frac{\Gamma(y+1, 10)}{\Gamma(y+1)}. \end{align*} Now then, for the order statistics, we have that $$P(W_{(k)}\le y)=\sum_{j=0}^{n-k}\binom{n}{j}(p_3(y))^j (p_1(y)+p_2(y))^{n-j}$$ in general, so that \begin{align*} P(W_{(4)}\le 1) &=\sum_{j=0}^{10-4}\binom{10}{j}(p_3(1))^j (p_1(1)+p_2(1))^{10-j}\\ &=\sum_{j=0}^{6}\binom{10}{j}(p_3(1))^j (p_1(1)+p_2(1))^{10-j}. \end{align*} Note that \begin{align*} p_1(1)&=e^{-10}\\ p_2(1)&=10\,e^{-10}\\ p_3(1)&=1-11\,e^{-10}. \end{align*} Then we simplify \begin{align*} P(W_{(4)}\le 1) &=\sum_{j=0}^{6}\binom{10}{j}\left(1-11\,e^{-10}\right)^j \left(11\,e^{-10}\right)^{10-j}\\ &\approx 1.30308\times 10^{-11}. \end{align*}

[Correct Answer]

1. As whuber mentions in the comments, the probability of a call occurring in the first minute is $1/2$ with a uniform distribution. Hence the probability of all four calls occurring in the first minute is simply $1/16.$

2. We compute the expected value of $W_{(4)}.$ To do so, we need the density and distribution for the uniform distribution: \begin{align*} f(t)&= \begin{cases} \dfrac12, &t\in[0,2]\\ 0,&\text{elsewhere,} \end{cases}\\ F(t)&= \begin{cases} 0,&t<0\\ \dfrac{t}{2},&t\in[0,2]\\ 1,&t>2. \end{cases} \end{align*} Then, according to the work done in Section 6.6 of the book, the density function for the maximum order statistic is given by \begin{align*} g_{(4)}(t) &=4\,[F(t)]^3 f(t)\\ &= \begin{cases} 4(t/2)^3(1/2),&t\in[0,2]\\ 0,&\text{elsewhere} \end{cases}\\ &= \begin{cases} t^3/4,&t\in[0,2]\\ 0,&\text{elsewhere.} \end{cases} \end{align*} It follows that the expected value of $W_{(4)}$ is $$\int_0^2 \frac{t^4}{4}\,dt=\frac85.$$ This makes sense: we would expect the value to be greater than the midpoint, but certainly not greater than $2.$