By Mark A. Pinsky, Samuel Karlin

ISBN-10: 0123814162

ISBN-13: 9780123814166

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**Extra info for An Introduction to Stochastic Modeling, Fourth Edition**

**Sample text**

N. A person draws a chip, returns it, draws another, returns it, and so on, until a chip is drawn that has been drawn before. Let X be the number of drawings. Find the probability distribution for X. It is easier to compute Pr{X > k} first. Then, Pr{X > 1} = 1, since at least two draws are always required. The event {X > 2} occurs when distinct numbers appear on the first two draws, whence Pr{X > 2} = (n/n)[(n − 1)/n]. Continuing in this manner, we Introduction 35 obtain Pr{X > k} = 1 1 − 1 n 1− 2 k−1 ··· 1 − , n n for k = 1, .

39) Introduction 29 Note that the parameter is the reciprocal of the mean and not the mean itself. The exponential distribution is fundamental in the theory of continuous-time Markov chains (see Chapter 5), due in major part to its memoryless property, as now explained. Think of T as a lifetime and, given that the unit has survived up to time t, ask for the conditional distribution of the remaining life T − t. Equivalently, for x > 0 determine the conditional probability Pr{T − t > x|T > t}. 39)] = e−λx .

Letting Y denote the result of tossing a single shaved die, we postulate that the probability mass function is given by pY (3) = pY (4) = 1 + 2ε ≡ p+ , 6 pY (1) = pY (2) = pY (5) = pY (6) = 1 − ε ≡ p− , 6 where ε > 0 is a small quantity depending on the amount by which the die has been biased. If both dice are shaved in the same manner, the mass function for their sum can be determined in a straightforward manner from the following joint table: Die #1 Die #2 1 p− 2 p− 3 p+ 4 p+ 5 p− 6 p− 1 p− p2− p2− p+ p− p+ p− p2− p2− 2 p− p2− p2− p+ p− p+ p− p2− p2− p2+ p2+ p+ p− p+ p− 3 p+ p+ p− p+ p− 4 p+ p+ p− p+ p− p2+ p2+ p+ p− p+ p− 5 p− p2− p2− p+ p− p+ p− p2− p2− 6 p− p2− p2− p+ p− p+ p− p2− p2− It is easily seen that the probability mass function for the sum of the dice is p(2) = p2− = p(12), p(3) = 2p2− = p(11), p(4) = p− ( p− + 2p+ ) = p(10), p(5) = 4p+ p− = p(9), p(6) = p2− + ( p+ + p− )2 = p(8), p(7) = 4p2− + 2p2+ .

### An Introduction to Stochastic Modeling, Fourth Edition by Mark A. Pinsky, Samuel Karlin

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