100AHW5S - STAT 100A HWV Solution Problem 1 For a discrete random variable X prove(1 E[aX b = aE[X b A Let p(x be the probability mass function of

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STAT 100A HWV Solution Problem 1: For a discrete random variable X , prove (1) E[ aX + b ] = a E[ X ] + b . A: Let p ( x ) be the probability mass function of X . Then E[ aX + b ] = x ( ax + b ) p ( x ) = a x xp ( x ) + b x p ( x ) = a E[ X ] + b . (2) Var[ aX + b ] = a 2 Var[ X ]. A: Let μ X = E[ X ], and μ aX + b = E[ aX + b ]. Var[ aX + b ] = E[(( aX + b ) - μ aX + b ) 2 ] = E[(( aX + b ) - ( X + b )) 2 ] = E[ a 2 ( X - μ X ) 2 ] = a 2 Var[ X ]. Problem 2: For two discrete random variables X and Y , if X and Y are independent, prove (1) E[ X + Y ] = E[ X ] + E[ Y ]. A: Let p ( x ) be the probability mass function of X . Let q ( y ) be the probability mass function of Y . The joint probability mass function is p ( x,y ) = p ( x ) q ( y ). E[ X + Y ] = X x X y ( x + y ) p ( x ) q ( y ) = X x X y xp ( x ) q ( y ) + X x X y yp ( x ) q ( y ) = X x xp ( x ) X y q ( y ) + X x p ( x ) X y yq ( y ) = E[ X ] + E[ Y ] .
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This note was uploaded on 12/16/2008 for the course STATS 100A taught by Professor Wu during the Fall '07 term at UCLA.

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100AHW5S - STAT 100A HWV Solution Problem 1 For a discrete random variable X prove(1 E[aX b = aE[X b A Let p(x be the probability mass function of

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