# 08_29 - STAT 410 Examples for 08/29/2011 Fall 2011 Mixed...

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STAT 410 Examples for 08/29/2011 Fall 2011 Mixed Random Variables : 1. Consider a random variable X with c.d.f. F ( x ) = < + - < 2 1 2 1 4 3 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 / 2 , p ( 2 ) = 1 / 4 . 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 / 2 + 2 1 / 4 + - 2 1 2 1 x x x d = 17 / 12 . b) E ( X 2 ) = 1 2 1 / 2 + 2 2 1 / 4 + - 2 1 2 2 1 x x x d = 53 / 24 . σ 2 = Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 29 / 144 .

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2. Let X be a random variable of the mixed type having the distribution function F ( x ) = < + < < . 2 1 , 2 1 4 1 , 1 0 4 , 0 0 2 x x x x x x a) Carefully sketch the graph of F ( x ). b) Find the mean and the variance of X. c) Find P ( X = 1 ), P ( X = 1 / 2 ), P ( 1 / 4 < X < 1 ), P ( 1 / 4 < X 1 ), P ( 1 / 2 X < 2 ), P ( 1 / 2 X 2 ). a) Discrete portion of the probability distribution of X: p ( 1 ) = 1 / 4 , p ( 2 ) = 1 / 4 .
Continuous portion of the probability distribution of X: f ( x ) = F ' ( x ) = < < < < o.w. 0 2 1 4 1 1 0 2 x x x . b) μ = E ( X ) = 1 1 / 4 + 2 1 / 4 + 1 0 2 x x x d + 2 1 4 1 x x d = 31 / 24 . E ( X 2 ) = 1 2 1 / 4 + 2 2 1 / 4 + 1 0 2 2 x x x d + 2 1 2 4 1 x x d = 47 / 24 . σ 2 = Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 47 / 24 ( 31 / 24 ) 2 = 167 / 576 . c) P ( X = 1 ) = F ( 1 ) – F ( 1 – ) = 1 / 2 1 / 4 = 1 / 4 . P ( X = 1 / 2 ) = F ( 1 / 2 ) – F ( 1 / 2 ) = 1 / 16 1 / 16 = 0 . P

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## This note was uploaded on 10/31/2011 for the course MATH 464 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.

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08_29 - STAT 410 Examples for 08/29/2011 Fall 2011 Mixed...

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