09_03 - 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 … … … ±& ∞-= F 1 x x = 1 × p 1

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STAT 410 Examples for 09/03/2008 Fall 2008 Mixed Random Variables : 1. Consider a random variable X with c.d.f. F ( x ) = ± ² ³ < + - < 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 ) = ± ² ³ < < - 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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1.9.20 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 ) = ( ) [ ] - = 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
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Unformatted text preview: ( 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 ) + … … … ± ( ) [ ] & ∞-= F 1 x x = 1 × p ( 1 ) + 2 × p ( 2 ) + 3 × p ( 3 ) + 4 × p ( 4 ) + … = E ( X ). Proof : E ( X ) = ( ) & ∞ = ⋅ x x p x = ( ) & ∞ = ⋅ 1 x x p x = ( ) & & ∞ ² ² ³ ´ µ µ ¶ · = = 1 1 1 x x y x p = ( ) & & ∞ ² ² ³ ´ µ µ ¶ · = = 1 1 x x y x p . → ( ) & & ∞ ² ² ³ ´ µ µ ¶ · = = 1 1 x x y x p = ( ) & & ∞ ² ² ³ ´ µ µ ¶ · ∞ = = 1 y y x x p ± E ( X ) = ( ) & & ∞ ² ² ³ ´ µ µ ¶ · ∞ = = 1 y y x x p = ( ) [ ] & ∞ ≥ = 1 X P y y = ( ) [ ] & ∞--= 1 1 F 1 y y = ( ) [ ] & ∞-= F 1 x x ....
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This note was uploaded on 10/22/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.

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09_03 - 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 … … … ±& ∞-= F 1 x x = 1 × p 1

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