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
$(50\xb70.33+45\xb70.47=37.65)$ .
ma = 50*0.33 + 45*0.47
ma = 37.6500Pa = cpoisson(ma,30:5:40)
Pa = 0.9119 0.6890 0.3722
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VERIFICATION of the Poisson decomposition

$N}_{k}=\sum _{i=1}^{N}{I}_{{E}_{ki}$ .
This is composite demand with
${Y}_{k}={I}_{{E}_{ki}}$ , so that
${g}_{{Y}_{k}}\left(s\right)={q}_{k}+s{p}_{k}=1+{p}_{k}(s1)$ .
Therefore,
$${g}_{{N}_{k}}\left(s\right)={g}_{N}\left[{g}_{{Y}_{k}}\left(s\right)\right]={e}^{\mu (1+{p}_{k}(s1)1)}={e}^{\mu {p}_{k}(s1)}$$
which is the generating function for
${N}_{k}\sim $ Poisson
$\left(\mu {p}_{k}\right)$ .
 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
$A=\{{N}_{1}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2}={n}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{N}_{m}={n}_{m}\}=\{N=n\}\cap \{{N}_{1n}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2n}={n}_{2},\cdots ,\phantom{\rule{0.277778em}{0ex}}{N}_{mn}={n}_{m}\}$
Since
N is independent of the class of
${I}_{{E}_{ki}}$ , the class
$$\{\{N=n\},\phantom{\rule{0.166667em}{0ex}}\{{N}_{1n}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2n}={n}_{2},\cdots ,\phantom{\rule{0.277778em}{0ex}}{N}_{mn}={n}_{m}\}\}$$
is independent. By the product rule and the multinomial distribution
$$P\left(A\right)={e}^{\mu}\frac{{\mu}^{n}}{n!}\xb7n!\prod _{k=1}^{m}\frac{{p}_{k}^{{n}_{k}}}{\left({n}_{k}\right)!}=\prod _{k=1}^{m}{e}^{\mu {p}_{k}}\frac{{p}_{k}^{{n}_{k}}}{{n}_{k}!}=\prod _{k=1}^{m}P({N}_{k}={n}_{k})$$
The second product uses the fact that
$${e}^{\mu}={e}^{\mu ({p}_{1}+{p}_{2}+\cdots +{p}_{m})}=\prod _{k=1}^{m}{e}^{\mu {p}_{k}}$$
Thus, the product rule holds for the class
$\{{N}_{k}:1\le k\le m\}$ , so that
it is independent.
Extreme values
Consider an iid class
$\{{Y}_{i}:1\le i\}$ of nonnegative random variables. For any
positive integer
n we let
$${V}_{n}=min\{{Y}_{1},\phantom{\rule{0.166667em}{0ex}}{Y}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{Y}_{n}\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{W}_{n}=max\{{Y}_{1},\phantom{\rule{0.166667em}{0ex}}{Y}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{Y}_{n}\}$$
Then
$$P({V}_{n}>t)={P}^{n}(Y>t)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P({W}_{n}\le t)={P}^{n}(Y\le t)$$
Now consider a random number
N of the
Y
_{i} . The minimum and maximum random variables are
$${V}_{N}=\sum _{n=0}^{\infty}{I}_{\{N=n\}}{V}_{n}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{W}_{N}=\sum _{n=0}^{\infty}{I}_{\{N=n\}}{W}_{n}$$
—
$\square $
Computational formulas
If we set
${V}_{0}={W}_{0}=0$ , then

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

${F}_{W}\left(t\right)={g}_{N}\left[P(Y\le t)\right]$
These results are easily established as follows.
$\{{V}_{N}>t\}=\underset{n=0}{\overset{\infty}{\bigvee}}\{N=n\}\{{V}_{n}>t\}$ . By additivity and independence of
$\{N,\phantom{\rule{0.166667em}{0ex}}{V}_{n}\}$ for each
n
$$P({V}_{N}>t)=\sum _{n=0}^{\infty}P(N=n)P({V}_{n}>t)=\sum _{n=1}^{\infty}P(N=n){P}^{n}(Y>t),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{since}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P({V}_{0}>t)=0$$
If we add into the last sum the term
$P(N=0){P}^{0}(Y>t)=P(N=0)$ then subtract it, we have
$$P({V}_{N}>t)=\sum _{n=0}^{\infty}P(N=n){P}^{n}(Y>t)P(N=0)={g}_{N}\left[P(Y>t)\right]P(N=0)$$
A similar argument holds for proposition (b). In this case, we do not have
the extra term for
$\{N=0\}$ , since
$P({W}_{0}\le t)=1$ .
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
${F}_{V}\left(t\right)=\sum _{n=1}^{\infty}P({V}_{n}\le t)P(N=n)=\sum _{n=1}^{\infty}[1{P}^{n}(Y>t)]P(N=n)}=\sum _{n=1}^{\infty}P(N=n)\sum _{n=1}^{\infty}{P}^{n}(Y>t)P(N=n)$
Add
$P(N=0)={P}^{0}(Y>t)P(N=0)$ to each of the sums to get
$${F}_{V}\left(t\right)=1\sum _{n=0}^{\infty}{P}^{n}(Y>t)P(N=n)=1{g}_{N}\left[P(Y>t)\right]$$
—
$\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
$$P({W}_{N}\le t)={g}_{N}\left[P(Y\le t)\right]={e}^{20[{F}_{Y}\left(t\right)1]}=exp(20{e}^{t/3})$$
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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