# 15.1 Some random selection problems  (Page 2/6)

 Page 2 / 6

## Message routing

A junction point in a network has two incoming lines and two outgoing lines. The number of incoming messages N 1 on line one in one hour is Poisson (50); on line 2 the number is ${N}_{2}\sim$ Poisson (45). On incoming line 1 the messages have probability ${p}_{1a}=0.33$ of leaving on outgoing line a and $1-{p}_{1a}$ of leaving on line b. The messages coming in on line 2 have probability ${p}_{2a}=0.47$ of leaving on line a. Under the usual independence assumptions, what is the distribution of outgoing messages on line a?What are the probabilities of at least 30, 35, 40 outgoing messages on line a?

SOLUTION

By the Poisson decomposition, ${N}_{a}\sim$ Poisson $\left(50·0.33+45·0.47=37.65\right)$ .

ma = 50*0.33 + 45*0.47 ma = 37.6500Pa = cpoisson(ma,30:5:40) Pa = 0.9119 0.6890 0.3722

VERIFICATION of the Poisson decomposition

1. ${N}_{k}=\sum _{i=1}^{N}{I}_{{E}_{ki}}$ .
This is composite demand with ${Y}_{k}={I}_{{E}_{ki}}$ , so that ${g}_{{Y}_{k}}\left(s\right)={q}_{k}+s{p}_{k}=1+{p}_{k}\left(s-1\right)$ . Therefore,
${g}_{{N}_{k}}\left(s\right)={g}_{N}\left[{g}_{{Y}_{k}}\left(s\right)\right]={e}^{\mu \left(1+{p}_{k}\left(s-1\right)-1\right)}={e}^{\mu {p}_{k}\left(s-1\right)}$
which is the generating function for ${N}_{k}\sim$ Poisson $\left(\mu {p}_{k}\right)$ .
2. For any ${n}_{1},\phantom{\rule{0.166667em}{0ex}}{n}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{n}_{m}$ , let $n={n}_{1}+{n}_{2}+\cdots +{n}_{m}$ , and consider
$A=\left\{{N}_{1}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2}={n}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{N}_{m}={n}_{m}\right\}=\left\{N=n\right\}\cap \left\{{N}_{1n}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2n}={n}_{2},\cdots ,\phantom{\rule{0.277778em}{0ex}}{N}_{mn}={n}_{m}\right\}$
Since N is independent of the class of ${I}_{{E}_{ki}}$ , the class
$\left\{\left\{N=n\right\},\phantom{\rule{0.166667em}{0ex}}\left\{{N}_{1n}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2n}={n}_{2},\cdots ,\phantom{\rule{0.277778em}{0ex}}{N}_{mn}={n}_{m}\right\}\right\}$
is independent. By the product rule and the multinomial distribution
$P\left(A\right)={e}^{-\mu }\frac{{\mu }^{n}}{n!}·n!\prod _{k=1}^{m}\frac{{p}_{k}^{{n}_{k}}}{\left({n}_{k}\right)!}=\prod _{k=1}^{m}{e}^{-\mu {p}_{k}}\frac{{p}_{k}^{{n}_{k}}}{{n}_{k}!}=\prod _{k=1}^{m}P\left({N}_{k}={n}_{k}\right)$
The second product uses the fact that
${e}^{\mu }={e}^{\mu \left({p}_{1}+{p}_{2}+\cdots +{p}_{m}\right)}=\prod _{k=1}^{m}{e}^{\mu {p}_{k}}$
Thus, the product rule holds for the class $\left\{{N}_{k}:1\le k\le m\right\}$ , so that it is independent.

## Extreme values

Consider an iid class $\left\{{Y}_{i}:1\le i\right\}$ of nonnegative random variables. For any positive integer n we let

${V}_{n}=min\left\{{Y}_{1},\phantom{\rule{0.166667em}{0ex}}{Y}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{Y}_{n}\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{W}_{n}=max\left\{{Y}_{1},\phantom{\rule{0.166667em}{0ex}}{Y}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{Y}_{n}\right\}$

Then

$P\left({V}_{n}>t\right)={P}^{n}\left(Y>t\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({W}_{n}\le t\right)={P}^{n}\left(Y\le t\right)$

Now consider a random number N of the Y i . The minimum and maximum random variables are

${V}_{N}=\sum _{n=0}^{\infty }{I}_{\left\{N=n\right\}}{V}_{n}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{W}_{N}=\sum _{n=0}^{\infty }{I}_{\left\{N=n\right\}}{W}_{n}$

$\square$

Computational formulas

If we set ${V}_{0}={W}_{0}=0$ , then

1. ${F}_{V}\left(t\right)=P\left(V\le t\right)=1+P\left(N=0\right)-{g}_{N}\left[P\left(Y>t\right)\right]$
2. ${F}_{W}\left(t\right)={g}_{N}\left[P\left(Y\le t\right)\right]$

These results are easily established as follows. $\left\{{V}_{N}>t\right\}=\underset{n=0}{\overset{\infty }{\bigvee }}\left\{N=n\right\}\left\{{V}_{n}>t\right\}$ . By additivity and independence of $\left\{N,\phantom{\rule{0.166667em}{0ex}}{V}_{n}\right\}$ for each n

$P\left({V}_{N}>t\right)=\sum _{n=0}^{\infty }P\left(N=n\right)P\left({V}_{n}>t\right)=\sum _{n=1}^{\infty }P\left(N=n\right){P}^{n}\left(Y>t\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{since}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({V}_{0}>t\right)=0$

If we add into the last sum the term $P\left(N=0\right){P}^{0}\left(Y>t\right)=P\left(N=0\right)$ then subtract it, we have

$P\left({V}_{N}>t\right)=\sum _{n=0}^{\infty }P\left(N=n\right){P}^{n}\left(Y>t\right)-P\left(N=0\right)={g}_{N}\left[P\left(Y>t\right)\right]-P\left(N=0\right)$

A similar argument holds for proposition (b). In this case, we do not have the extra term for $\left\{N=0\right\}$ , since $P\left({W}_{0}\le t\right)=1$ .

Special case . In some cases, $N=0$ does not correspond to an admissible outcome (see [link] , below, on lowest bidder and [link] ). In that case

${F}_{V}\left(t\right)=\sum _{n=1}^{\infty }P\left({V}_{n}\le t\right)P\left(N=n\right)=\sum _{n=1}^{\infty }\left[1-{P}^{n}\left(Y>t\right)\right]P\left(N=n\right)=\sum _{n=1}^{\infty }P\left(N=n\right)-\sum _{n=1}^{\infty }{P}^{n}\left(Y>t\right)P\left(N=n\right)$

Add $P\left(N=0\right)={P}^{0}\left(Y>t\right)P\left(N=0\right)$ to each of the sums to get

${F}_{V}\left(t\right)=1-\sum _{n=0}^{\infty }{P}^{n}\left(Y>t\right)P\left(N=n\right)=1-{g}_{N}\left[P\left(Y>t\right)\right]$

$\square$

## Maximum service time

The number N of jobs coming into a service center in a week is a random quantity having a Poisson (20) distribution. Suppose the service times (in hours) for individualunits are iid, with common distribution exponential (1/3). What is the probability the maximum service time for the units is no greater than 6, 9, 12, 15, 18 hours?SOLUTION

## Solution

$P\left({W}_{N}\le t\right)={g}_{N}\left[P\left(Y\le t\right)\right]={e}^{20\left[{F}_{Y}\left(t\right)-1\right]}=exp\left(-20{e}^{-t/3}\right)$
t = 6:3:18; PW = exp(-20*exp(-t/3));disp([t;PW]')6.0000 0.0668 9.0000 0.369412.0000 0.6933 15.0000 0.873918.0000 0.9516

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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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