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lect21 - MOMENT GENERATING FUNCTIONS JOINT MGF Outline...

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MOMENT GENERATING FUNCTIONS JOINT MGF Outline MOMENT GENERATING FUNCTIONS The Uniqueness Theorem and the Multiplication Rule JOINT MGF 1 / 15 Xinghua Zheng Lect 21: Moment Generating Functions
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MOMENT GENERATING FUNCTIONS JOINT MGF The Uniqueness Theorem and the Multiplication Rule MOMENT GENERATING FUNCTIONS Let X be a random variable. For any number t , set M ( t ) := E ( e tX ) = X x e tx f ( x ) , if X is discrete with pmf f , Z -∞ e tx f ( x ) dx , if X is continuous with density f , Formal interchanges of “ E ” and “ d / dt ” gives M 0 ( t ) = d dt E ( e tX ) = E d dt e tX = E ( Xe tX ) , M 00 ( t ) = d dt E ( Xe tX ) = E d dt Xe tX = E ( X 2 e tX ) , M ( k ) ( t ) = d dt M ( k - 1 ) ( t ) = E d dt X k - 1 e tX = E ( X k e tX ) , for each positive integer k . In particular, E ( X ) = , E ( X 2 ) = , and E ( X k ) = for each k . Because M ( t ) generates the moments of X in the above manner, it is called the moment generating function (MGF) of X . 2 / 15 Xinghua Zheng Lect 21: Moment Generating Functions
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MOMENT GENERATING FUNCTIONS JOINT MGF The Uniqueness Theorem and the Multiplication Rule SOME DISCRETE EXAMPLES Suppose X Poisson ( λ ) . Then M ( t ) = E ( e tX ) = k = 0 e tk · e - λ λ k k ! = = exp ( λ ( e t - 1 )) . Taking derivatives of M ( t ) gives M 0 ( t ) = exp ( λ ( e t - 1 )) λ e t ; M 00 ( t ) = exp ( λ ( e t - 1 )) λ 2 e 2 t + exp ( λ ( e t - 1 )) λ e t hence E ( X ) = M 0 ( 0 ) = ; E ( X 2 ) = M 00 ( 0 ) = ; Var ( X ) = . Suppose X Binomial ( n , p ) . Then M ( t ) = E ( e tX ) = n k = 0 e tk · ( n k ) p k q n - k = = ( pe t + q ) n . Taking derivatives of M ( t ) gives M 0 ( t ) = n ( pe t + q ) n - 1 pe t ; M 00 ( t ) = n ( n - 1 )( pe t + q ) n - 2 p 2 e 2 t + n ( pe t + q ) n - 1 pe t hence E ( X ) = M 0 ( 0 ) = ; E ( X 2 ) = M 00 ( 0 ) = ; Var ( X ) = . 3 / 15 Xinghua Zheng Lect 21: Moment Generating Functions
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MOMENT GENERATING FUNCTIONS JOINT MGF The Uniqueness Theorem and the Multiplication Rule GAMMA RANDOM VARIABLES Suppose X Gamma ( r , λ ) . Then for t < λ , M ( t ) = E ( e tX ) = Z 0 e tx · λ r e - λ x x r - 1 Γ( r ) dx = λ r ( λ - t ) r Z 0 ( λ - t ) r e - ( λ - t ) x x r - 1 Γ( r ) dx = λ λ - t r Taking derivatives of M ( t ) gives M 0 ( t ) = r λ r ( λ - t ) - r - 1 ; M 00 ( t ) = r ( r + 1 ) λ r ( λ - t ) - r - 2 hence E ( X ) = M 0 ( 0 ) = ; E ( X 2 ) = M 00 ( 0 ) = ; Var ( X ) = . Some special cases: Exponential with parameter λ = Gamma ( , ) , Chisquare with n degrees of freedom = Gamma ( , ) .
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