Simulation by Sheldon M. Ross

By Sheldon M. Ross

  • ''I have constantly beloved Ross’ books, as he's at the same time mathematically rigorous and intensely drawn to functions. the most important power I see is the infrequent mix of mathematical rigor and representation of ways the mathematical methodologies are utilized in perform. Books with functional standpoint are not often this rigourous and mathematically specific. I additionally just like the number of workouts, that are particularly not easy and significant excellence from students.''
    --Prof. Krzysztof Ostaszewski, Illinois nation University.

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J i=0 pi Since, for 0 < a < b < 1, p{a ≤ U < b} = b − a, we have that p{X = x j } = p j−1 i=0 pi ≤ U < j i=0 pi = p j and so X has desired distribution. Simulation. 00004-8 © 2013 Elsevier Inc. All rights reserved. 47 48 4 Generating Discrete Random Variables Remarks 1. The preceding can be written algorithmically as Generate a random number U If U < p0 set X = x0 and stop If U < p0 + p1 set X = x1 and stop If U < p0 + p1 + p2 set X = x2 and stop .. 2. If the xi , i ≥ 0, are ordered so that x0 < x1 < x2 < · · · and if we let F k denote the distribution function of X , then F(xk ) = i=0 pi and so X will equal x j if F(x j−1 ) ≤ U < F(x j ) In other words, after generating a random number U we determine the value of X by finding the interval [F(x j−1 ), F(x j ) in which U lies [or, equivalently, by finding the inverse of F(U )].

1. We now prove that the rejection method works. Theorem X such that The acceptance–rejection algorithm generates a random variable P{X = j} = p j , j = 0, . . In addition, the number of iterations of the algorithm needed to obtain X is a geometric random variable with mean c. Proof To begin, let us determine the probability that a single iteration produces the accepted value j. First note that P{Y = j, it is accepted} = P{Y = j}P{accept|Y = j} pj = qj cq j pj = c Summing over j yields the probability that a generated random variable is accepted: P{accepted} = j 1 pj = c c 58 4 Generating Discrete Random Variables As each iteration independently results in an accepted value with probability 1/c, we see that the number of iterations needed is geometric with mean c.

6. The continuous random variable X has a probability density function given by f (x) = cx, 0 < x < 1 Find P{X > 21 }. 7. If X and Y have a joint probability density function specified by f (x, y) = 2e−(x+2y) , 0 < x < ∞, 0 < y < ∞ Find P{X < Y }. 8. Find the expected value of the random variable specified in Exercise 5. 9. Find E [X ] for the random variable of Exercise 6. 10. There are 10 different types of coupons and each time one obtains a coupon it is equally likely to be any of the 10 types.

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