# 15.1 Some random selection problems  (Page 4/6)

 Page 4 / 6

## Bernoulli trials with random execution times or costs

Consider a Bernoulli sequence with probability p of success on any component trial. Let N be the number of the trial on which the first success occurs. Let Y i be the time (or cost) to execute the i th trial. Then the total time (or cost) from the beginning to the completion of the first success is

$T=\sum _{i=1}^{N}{Y}_{i}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{(composite}\phantom{\rule{4.pt}{0ex}}\text{demand''}\phantom{\rule{4.pt}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}N-1\sim \phantom{\rule{4.pt}{0ex}}\text{geometric}\phantom{\rule{4.pt}{0ex}}p\right)$

We suppose the Y i form an iid class, independent of N . Now $N-1\sim$ geometric $\left(p\right)$ implies

${g}_{N}\left(s\right)=ps/\left(1-qs\right)$ , so that

${M}_{T}\left(s\right)={g}_{N}\left[{M}_{Y}\left(s\right)\right]=\frac{p{M}_{Y}\left(s\right)}{1-q{M}_{Y}\left(s\right)}$

There are two useful special cases:

1. ${Y}_{i}\sim$ exponential $\left(\lambda \right)$ , so that ${M}_{Y}\left(s\right)=\frac{\lambda }{\lambda -s}$ .
${M}_{T}\left(s\right)=\frac{p\lambda /\left(\lambda -s\right)}{1-q\lambda /\left(\lambda -s\right)}=\frac{p\lambda }{p\lambda -s}$
which implies $T\sim$ exponential $\left(p\lambda \right)$ .
2. ${Y}_{i}-1\sim$ geometric $\left({p}_{0}\right)$ , so that ${g}_{Y}\left(s\right)=\frac{{p}_{0}s}{1-{q}_{0}s}$
${g}_{T}\left(s\right)=\frac{p{p}_{0}s/\left(1-{q}_{0}s\right)}{1-p{p}_{0}s/\left(1-{q}_{0}s\right)}=\frac{p{p}_{0}s}{1-\left(1-p{p}_{0}\right)s}$
so that $T-1\sim$ geometric $\left(p{p}_{0}\right)$ .

## Job interviews

Suppose a prospective employer is interviewing candidates for a job from a pool in which twenty percent are qualified. Interview times (in hours) Y i are presumed to form an iid class, each exponential (3). Thus, the average interview time is 1/3 hour (twentyminutes). We take the probability for success on any interview to be $p=0.2$ . What is the probability a satisfactory candidate will be found in four hours or less?What is the probability the maximum interview time will be no greater than 0.5, 0.75, 1, 1.25, 1.5 hours?

SOLUTION

$T\sim$ exponential $\left(0.2·3=0.6\right)$ , so that $P\left(T\le 4\right)=1-{e}^{-0.6·4}=0.9093$ .

$P\left(W\le t\right)={g}_{N}\left[P\left(Y\le t\right)\right]=\frac{0.2\left(1-{e}^{-3t}\right)}{1-0.8\left(1-{e}^{-3t}\right)}=\frac{1-{e}^{-3t}}{1+4{e}^{-3t}}$

MATLAB computations give

t = 0.5:0.25:1.5; PWt = (1 - exp(-3*t))./(1 + 4*exp(-3*t));disp([t;PWt]')0.5000 0.4105 0.7500 0.62931.0000 0.7924 1.2500 0.89251.5000 0.9468

The average interview time is 1/3 hour; with probability 0.63 the maximum is 3/4 hour or less; with probability 0.79 the maximum is one hour or less; etc.

In the general case, solving for the distribution of T requires transform theory, and may be handled best by a program such as Maple or Mathematica.

For the case of simple Y i , we may use approximation procedures based on properties of the geometric series. Since $N-1\sim$ geometric $\left(p\right)$ ,

${g}_{N}\left(s\right)=\frac{ps}{1-qs}=ps\sum _{k=0}^{\infty }{\left(qs\right)}^{k}=ps\left[\sum _{k=0}^{n},{\left(qs\right)}^{k},+,\sum _{k=n+1}^{\infty },{\left(qs\right)}^{k}\right]=ps\left[\sum _{k=0}^{n},{\left(qs\right)}^{k},+,{\left(qs\right)}^{n+1},\sum _{k=0}^{\infty },{\left(qs\right)}^{k}\right]$
$=ps\left[\sum _{k=0}^{n},{\left(qs\right)}^{k}\right]+{\left(qs\right)}^{n+1}{g}_{N}\left(s\right)={g}_{n}\left(s\right)+{\left(qs\right)}^{n+1}{g}_{N}\left(s\right)$

Note that ${g}_{n}\left(s\right)$ has the form of the generating function for a simple approximation N n which matches values and probabilities with N up to $k=n$ . Now

${g}_{T}\left(s\right)={g}_{n}\left[{g}_{Y}\left(s\right)\right]+{\left(qs\right)}^{n+1}{g}_{N}\left[{g}_{Y}\left(s\right)\right]$

The evaluation involves convolution of coefficients which effectively sets $s=1$ . Since ${g}_{N}\left(1\right)={g}_{Y}\left(1\right)=1$ ,

${\left(qs\right)}^{n+1}{g}_{N}\left[{g}_{Y}\left(s\right)\right]\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}s=1\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{reduces}\phantom{\rule{4.pt}{0ex}}\text{to}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{q}^{n+1}=P\left(N>n\right)$

which is negligible if n is large enough. Suitable n may be determined in each case. With such an n , if the Y i are nonnegative, integer-valued, we may use the gend procedure on ${g}_{n}\left[{g}_{Y}\left(s\right)\right]$ , where

${g}_{n}\left(s\right)=ps+pq{s}^{2}+p{q}^{2}{s}^{3}+\cdots +p{q}^{n}{s}^{n+1}$

For the integer-valued case, as in the general case of simple Y i , we could use mgd. However, gend is usually faster and more efficient for the integer-valued case. Unless q is small, the number of terms needed to approximate g n is likely to be too great.

## Approximating the generating function

Let $p=0.3$ and Y be uniformly distributed on $\left\{1,\phantom{\rule{0.166667em}{0ex}}2,\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}10\right\}$ . Determine the distribution for

$T=\sum _{k=1}^{N}{Y}_{k}$

SOLUTION

p = 0.3; q = 1 - p;a = [30 35 40]; % Check for suitable nb = q.^a b = 1.0e-04 * % Use n = 400.2254 0.0379 0.0064 n = 40;k = 1:n; gY = 0.1*[0 ones(1,10)]; gN = p*[0 q.^(k-1)]; % Probabilities, 0<= k<= 40 gendDo not forget zero coefficients for missing powers Enter gen fn COEFFICIENTS for gN gNEnter gen fn COEFFICIENTS for gY gY Values are in row matrix D; probabilities are in PD.To view the distribution, call for gD. sum(PD) % Check sum of probabilitiesans = 1.0000 FD = cumsum(PD); % Distribution function for Dplot(0:100,FD(1:101)) % See [link] P50 = (D<=50)*PD' P50 = 0.9497P30 = (D<=30)*PD' P30 = 0.8263

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