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Section 3-7/Geometric and Negative Binomial Distributions     89


                     Example 3-23    Lack of Memory Property  In Example 3-20, the probability that a bit is transmitted in error is
                                     equal to 0.1. Suppose that 50 bits have been transmitted. The mean number of bits until the next
                                =
                     error is 1 0 1 10—the same result as the mean number of bits until the Ð  rst error.
                              .
                                         Negative Binomial Distribution
                                         A generalization of a geometric distribution in which the random variable is the number of
                                         Bernoulli trials required to obtain r successes results in the negative binomial distribution.


                     Example 3-24    Digital Channel  As in Example 3-20, suppose that the probability that a bit transmitted through
                                     a digital transmission channel is received in error is 0.1. Assume that the transmissions are inde-
                     pendent events, and let the random variable X denote the number of bits transmitted until the fourth error.
                        Then X has a negative binomial distribution with r = 4. Probabilities involving X can be found as follows. For example,
                       (

                     P X = ) is the probability that exactly three errors occur in the irst 9 trials and then trial 10 results in the fourth error.
                           10

                     The probability that exactly three errors occur in the irst 9 trials is determined from the binomial distribution to be
                                                              ⎛ 9⎞   3  0 9) 6
                                                                  0 1) ( .
                                                              ⎜ ⎟  ( .
                                                               3⎠
                                                              ⎝

                     Because the trials are independent, the probability that exactly three errors occur in the irst 9 trials and trial 10 results
                     in the fourth error is the product of the probabilities of these two events, namely,
                                                     ⎛ 9⎞   3  0 9) (   ⎛ 9⎞   4  0 9) 6
                                                                 6
                                                         0 1) ( .
                                                                            0 1) ( .
                                                                   .
                                                     ⎜ ⎟  ( .     0 1) =  ⎜ ⎟  ( .
                                                                         3⎠
                                                                        ⎝
                                                      3⎠
                                                     ⎝
                                                                                                (
                                         In general, probabilities for X can be determined as follows. Here P X =  x) implies that r −1
                                         successes occur in the i rst x −1 trials and the rth success occurs on trial x. The probability
                                         that r −1 successes occur in the i rst x −1 trials is obtained from the binomial distribution to be
                                                                       ⎛ x − ⎞ 1  r−1  x r −
                                                                       ⎜ ⎝ r − ⎠ 1 ⎟  p  1 (  −  p)
                                         for r ≤  x. The probability that trial x is a success is p. Because the trials are independent, these
                                         probabilities are multiplied so that
                                                                           ⎛  x − ⎞ 1
                                                                                          −
                                                                  P X =  x) = ⎜ ⎝  r − ⎠ 1 ⎟  p r−1 ( −  p)  x r  p
                                                                   (
                                                                                     1
                                         This leads to the following result.
                         Negative Binomial
                             Distribution    In a series of Bernoulli trials (independent trials with constant probability p of a suc-
                                             cess), the random variable X that equals the number of trials until r successes occur is a
                                             negative binomial random variable with parameters 0 < p <  1 and r = 1 2 3, , ,…, and
                                                               ⎛  x − ⎞ 1  x r
                                                                           −
                                                                                      ,
                                                         f x ( ) = ⎜ ⎝  r − ⎠ 1 ⎟  (1 −  p)  p r  x =  r r +1, r + 2,…  (3-11)
                                         Because at least r trials are required to obtain r successes, the range of X is from r to ∞. In the
                                         special case that r = 1, a negative binomial random variable is a geometric random variable.
                                         Selected negative binomial distributions are illustrated in Fig. 3-10.
                                            The lack of memory property of a geometric random variable implies the following. Let
                                         X  denote the total number of trials required to obtain r  successes. Let X 1  denote the num-
                                         ber of trials required to obtain the i rst success, let X 2  denote the number of extra trials
                                         required to obtain the second success, let X 3  denote the number of extra trials to obtain the
                                         third success, and so forth. Then the total number of trials required to obtain r successes is
                                         X =  X +  X + … +  X r . Because of the lack of memory property, each of the random variables
                                                  2
                                              1
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