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# 09_20ans - STAT 400 Fall 2011 Examples for random variables...

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STAT 400 Examples for 09/20/2011 Fall 2011 random variables discrete continuous probability mass function p.m.f. p ( x ) = P ( X = x ) probability density function p.d.f. f ( x ) x 0 p ( x ) 1 ( ) x x p all = 1 x f ( x ) 0 ( ) - x x f d = 1 cumulative distribution function c.d.f. F ( x ) = P ( X x ) F ( x ) = ( ) x y y p F ( x ) = ( ) - x d y y f expected value E ( X ) = μ X discrete continuous If x x p x all ) ( < , E ( X ) = x x p x all ) ( If - x x f x d ) ( < , E ( X ) = - x x f x d ) ( discrete continuous If x x p x g all ) ( ) ( < , E ( g ( X ) ) = x x p x g all ) ( ) ( If - x x f x g d ) ( ) ( < , E ( g ( X ) ) = - x x f x g d ) ( ) (

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variance Var ( X ) = 2 X σ = E ( [ X - μ X ] 2 ) = E ( X 2 ) [ E ( X ) ] 2 discrete continuous Var ( X ) = ( ) - x x p x all 2 X ) ( μ = [ ] 2 all 2 ) X ( E ) ( x x p x - Var ( X ) = ( ) - - x x x d f ) ( 2 X μ = [ ] 2 2 ) X ( E ) ( x x x d f - - moment-generating function M X ( t ) = E ( e t X ) discrete continuous M X ( t ) = ( ) x x t x p e all M X ( t ) = ( ) - dx x f x t e Example 1: f X ( x ) = < < o.w. 0 1 0 2 x x P ( 0.3 < X < 0.7 ) = 7 . 0 3 . 0 2 x x d = 0.49 – 0.09 = 0.40. P ( X > 0.3 ) = 1 3 . 0 2 x x d = 1 – 0.09 = 0.91. E ( X ) = 1 0 2 x x x d = 1 0 2 2 x x d = 3 2 = μ . E ( X 2 ) = 1 0 2 2 x x x d = 1 0 3 2 x x d = 4 2 = 2 1 . Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 2 3 2 2 1 - = 18 1 = σ 2 .
F X ( x ) = P ( X x ) .

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