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Unformatted text preview: Stat 311 Introduction to Mathematical Statistics Zhengjun Zhang Department of Statistics, University of Wisconsin, Madison, WI 53706, USA October 2729, 2009 Example Let X 1 , X 2 , ..., X n be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k . Let Y denote the minimum of the X i s. Find the distribution of Y . Geometric Distribution The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ... } , the probability distribution of the number Y = X 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... } . Which of these one calls the geometric distribution is a matter of convention and convenience. We use the first one. Example T be the number of trials up to and including the first success. Then P ( T = 1) = p , P ( T = 2) = qp , P ( T = 3) = q 2 p , and in general, P ( T = n ) = q n 1 p . Memoryless property It is often assumed that the length of time required to service a customer also has a geometric distribution but with a different value for p . P ( T > r + s  T > r ) = P ( T > r + s ) P ( T > r ) = q r + s q r = q s . Thus, the probability that the customers service takes s more time units is independent of the length of time r that the customer has already been served. Because of this interpretation, this property is called the memoryless property, and is also obeyed by the exponential distribution. Example X is geometric( p ), compute P (5 X 9). Negative Binomial Distribution Suppose we are given a coin which has probability p of coming up heads when it is tossed. We fix a positive integer k , and toss the coin until the k th head appears. We let X represent the number of tosses. When k = 1, X is geometrically distributed. For a general k , we say that X has a negative binomial distribution. Formula If X = x , then it must be true that there were exactly k 1 heads thrown in the first x 1 tosses, and a head must have been thrown on the x th toss. There are x 1 k 1 sequences of length x with these properties, and each of them is assigned the same probability, namely p k 1 q x k . Therefore, if we define u ( x,k,p ) = P ( X = x ) , then u ( x,k,p ) = x 1 k 1 p k q x k ....
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 Fall '08
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