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Unformatted text preview: STAT5101 Lecture Notes 9 Discrete distributions Bernoulli Binomial Geometric Poisson Continuous distributions Normal Bivariate Normal Gamma Exponential Chisquare Briefly mention Negative Binomial Lognormal Beta 1 Bernoulli Distribution and Bernoulli trials We say X follows a Bernoulli distribution with parameter p if X take only 0 and 1 and the probabilities are P ( X = 1) = p and P ( X = 0) = 1 p . We write X Ber ( p ). We have studied the basic properties of Ber ( p ). For example, EX = p , V ar ( X ) = p (1 p ) and its MGF is ( t ) = 1 p + pe t . n Bernoulli trials: n independent Ber ( p ) ran dom variables form n Bernoulli trails. For each Bernoulli trial, if X = 1 then we say we have a success, otherwise we have a failure. So p = P ( X = 1) is also called the probability of success. 2 Binomial Distribution Definition: We say X follows a Binomial dis tribution with parameters n and p if P ( X = k ) = n k p k (1 p ) n k , k = 0 , 1 ,...,n We write X Bin ( n,p ) . We have studied the basic properties of Bin ( n,p ). For example, EX = np , V ar ( X ) = np (1 p ) and the MGF is ( t ) = (1 p + pe t ) n . The model Suppose we have n Bernoulli trails with probability of success p . Let Y be the number of success in the n Bernoulli trails. Then Y Bin ( n,p ) . Why? let X i be the ith Ber ( p ) rv in the n Bernoulli trails. Then we observe Y = X 1 + X 2 + + X n 3 Geometric Distribution Definition: We say X follows a Geometric dis tribution with parameter of success p if P ( X = k ) = (1 p ) k p, k = 0 , 1 , 2 ,..., We write X Geo ( p ) . For convenience we also let q = 1 p . P ( X = k ) = q k p . The model Suppose we keep doing the Bernoulli trails until we have a success. Let X be the number of trials before the success . Then X Geo ( p ) . MGF of Geo ( p ) ( t ) = E [ e tX ] = X k =0 e tk q k p = p X k =0 ( e t q ) k ( t ) = p 1 e t q for t < log(1 /q ) 4 Mean and Variance of Geo ( p ) If X Geo ( p ) then EX = q p V ar ( X ) = q p 2 . Explanation: (1) ( t ) = p (1 e t q ) 2 qe t EX = (1) (0) = p (1 q ) 2 q = q p (2) ( t ) = p (1 e t q ) 2 qe t + 2 p (1 e t q ) 3 qe t qe t E ( X 2 ) = (2) (0) = p (1 q ) 2 q + 2 p (1 q ) 3 q 2 E ( X 2 ) = q p + 2 q 2 p 2 V ar ( X ) = E ( X 2 ) ( EX ) 2 = q p 2 5 Poisson Distribution Definition: We say X follows a Poisson dis tribution with parameter if P ( X = k ) = e k k ! , k = 0 , 1 , 2 ,..., We write X Poi ( ) . MGF of Poi ( ) : Let X Poi ( ) then its MGF is ( t ) = e ( e t 1) . ( t ) = E ( e t X ) = X k =0 e tk e k k ! = e X k =0 ( e t ) k k !...
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 Fall '02
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 Statistics, Bernoulli, Binomial

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