06_19 - STAT 410 Examples for 06/19/2009 Summer 2009 Mixed...

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STAT 410 Examples for 06/19/2009 Summer 2009 Mixed Random Variables : 1. Consider a random variable X with c.d.f. F ( x ) = < + - < 2 1 2 1 4 2 2 1 0 2 x x x x x a) Find μ X = E ( X ). b) Find σ X 2 = Var ( X ). Discrete portion of the probability distribution of X: p ( 1 ) = 1 / 4 , p ( 2 ) = 1 / 2 . Continuous portion of the probability distribution of X: f ( x ) = < < - o.w. 0 2 1 2 1 x x . a) μ = E ( X ) = 1 1 / 4 + 2 1 / 2 + - 2 1 2 1 x x x d = 5 / 3 . b) E ( X 2 ) = 1 2 1 / 4 + 2 2 1 / 2 + - 2 1 2 2 1 x x x d = 71 / 24 . σ 2 = Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 13 / 72 .
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~ 1.9.19 Let X be a nonnegative continuous random variable with p.d.f. f ( x ) and c.d.f. F ( x ). Show that E ( X ) = ( 29 ( 29 - 0 1 x x d F . E ( X ) = ( 29 0 x x f x d = ( 29 0 0 x x f y d x d = ( 29 0 0 x y x f d x d ( 29 0 0 x y x f d x d = ( 29 0 y x x f d y d E ( X ) = ( 29 0 y x x f d y d = ( 29 0 X P y y d = ( 29 ( 29 - 0 1 y y d F .
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1.9.20 Let X be a random variable of the discrete type with pmf p ( x ) that is positive on the nonnegative integers and is equal to zero elsewhere. Show that E ( X ) = ( 29 [ ] - = 0 F 1 x x , where F ( x ) is the cdf of X. Idea : 1 – F ( x ) = P ( X > x ) = p ( x + 1 ) + p ( x + 2 ) + p ( x + 3 ) + p ( x + 4 ) + … 1 – F ( 0 ) p ( 1 ) + p ( 2 ) + p ( 3 ) + p ( 4 ) + p ( 5 ) + p ( 6 ) + p ( 7 ) + 1 – F ( 1 ) p ( 2 ) + p ( 3 ) + p ( 4 ) + p ( 5 ) + p ( 6 ) + p ( 7 ) + 1 – F ( 2 ) p ( 3 ) + p ( 4 ) + p ( 5 ) + p ( 6 ) + p ( 7 ) + 1 – F ( 3 ) p ( 4 ) + p ( 5 ) + p ( 6 ) + p ( 7 ) + 1 – F ( 4 ) p ( 5 ) + p ( 6 ) + p ( 7 ) + ( 29 [ ] - = 0 F 1 x x = 1 × p ( 1 ) + 2 × p ( 2 ) + 3 × p ( 3 ) + 4 × p ( 4 ) + = E ( X ). Proof : E ( X ) = ( 29 = 0 x x p x = ( 29 = 1 x x p x = ( 29 = = 1 1 1 x x y x p = ( 29 = = 1 1 x x y x p . ( 29
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06_19 - STAT 410 Examples for 06/19/2009 Summer 2009 Mixed...

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