# 100AHW5 - STAT 100A HWV Solution Problem 1: For both...

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Unformatted text preview: STAT 100A HWV Solution Problem 1: For both discrete and continuous cases, prove (1) E[ a + bX ] = a + b E[ X ]. A: For discrete case, let p ( x ) be the probability mass function of X . Then E[ a + bX ] = ∑ x ( a + bx ) p ( x ) = a ∑ x p ( x ) + b ∑ x xp ( x ) = a + b E[ X ]. For continuous case, let f ( x ) be the probability density function of X . Then E[ a + bX ] = R ( a + bx ) f ( x ) dx = a R f ( x ) dx + b R xf ( x ) dx = a + b E[ X ]. (2) Var[ a + bX ] = b 2 Var[ X ]. A: Let μ X = E[ X ], and μ a + bX = E[ a + bX ]. Var[ a + bX ] = E[(( a + bX )- μ a + bX ) 2 ] = E[(( a + bX )- ( a + bμ X )) 2 ] = E[ b 2 ( X- μ X ) 2 ] = b 2 Var[ X ]. Problem 2: For both discrete and continuous cases, prove Var[ X ] = E[ X 2 ]- E[ X ] 2 . A: Var[ X ] = E[( X- μ ) 2 ] = E[ X 2- 2 μX + μ 2 ] = E[ X 2 ]- 2 μ E[ X ] + μ 2 = E[ X 2 ]- E[ X ] 2 . Problem 3: For U ∼ Uniform[0 , 1], calculate E[ U ], E[ U 2 ], and Var[ U ]....
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## This note was uploaded on 02/21/2010 for the course STATS 100A 262303210 taught by Professor Wu during the Spring '09 term at UCLA.

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100AHW5 - STAT 100A HWV Solution Problem 1: For both...

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