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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 = i = 1 N Y i (composite ``demand'' with N - 1 geometric p )

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

g N ( s ) = p s / ( 1 - q s ) , so that

M T ( s ) = g N [ M Y ( s ) ] = p M Y ( s ) 1 - q M Y ( s )

There are two useful special cases:

  1. Y i exponential ( λ ) , so that M Y ( s ) = λ λ - s .
    M T ( s ) = p λ / ( λ - s ) 1 - q λ / ( λ - s ) = p λ p λ - s
    which implies T exponential ( p λ ) .
  2. Y i - 1 geometric ( p 0 ) , so that g Y ( s ) = p 0 s 1 - q 0 s
    g T ( s ) = p p 0 s / ( 1 - q 0 s ) 1 - p p 0 s / ( 1 - q 0 s ) = p p 0 s 1 - ( 1 - p p 0 ) s
    so that T - 1 geometric ( p p 0 ) .

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 exponential ( 0 . 2 · 3 = 0 . 6 ) , so that P ( T 4 ) = 1 - e - 0 . 6 · 4 = 0 . 9093 .

P ( W t ) = g N [ P ( Y t ) ] = 0 . 2 ( 1 - e - 3 t ) 1 - 0 . 8 ( 1 - e - 3 t ) = 1 - e - 3 t 1 + 4 e - 3 t

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.

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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 geometric ( p ) ,

g N ( s ) = p s 1 - q s = p s k = 0 ( q s ) k = p s k = 0 n ( q s ) k + k = n + 1 ( q s ) k = p s k = 0 n ( q s ) k + ( q s ) n + 1 k = 0 ( q s ) k
= p s k = 0 n ( q s ) k + ( q s ) n + 1 g N ( s ) = g n ( s ) + ( q s ) n + 1 g N ( s )

Note that g n ( s ) 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 ( s ) = g n [ g Y ( s ) ] + ( q s ) n + 1 g N [ g Y ( s ) ]

The evaluation involves convolution of coefficients which effectively sets s = 1 . Since g N ( 1 ) = g Y ( 1 ) = 1 ,

( q s ) n + 1 g N [ g Y ( s ) ] for s = 1 reduces to q n + 1 = P ( N > n )

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 [ g Y ( s ) ] , where

g n ( s ) = p s + p q s 2 + p q 2 s 3 + + 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 { 1 , 2 , , 10 } . Determine the distribution for

T = 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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Questions & Answers

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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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