# 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

what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive