HW7_Solutions

HW7_Solutions - PSTAT 120A Spring 2010 HW 7 Solutions 1 X...

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Unformatted text preview: PSTAT 120A Spring 2010 HW 7 Solutions 1. X ∼ exp( λ ) E [ X ] = ˆ ∞-∞ xf ( x ) dx, = λ ˆ ∞ xe- λx dx. To evaluate this integral, we use integration by parts ˆ b a u ( x ) dv ( x ) = u ( x ) v ( x ) | x = b x = a- ˆ b a v ( x ) du ( x ) . For our purposes we set u ( x ) = x and dv ( x ) = e- λx dx . Then we get E [ X ] = λ- xλ- 1 e- λx | x = ∞ x =0 + λ- 1 ˆ ∞ e- λx dx . To rigorously evaluate the boundary term at x = ∞ , we note xe- λx = x e λx . Both the numerator and the denominator tend towards ∞ as x goes to ∞ . L’Hospital’s rule (see math review sheet) tells us that we can evaluate the limit in this case by lim x →∞ x e λx = lim x →∞ d dx ( x ) d dx ( e λx ) , = lim x →∞ 1 λe λx , = 0 . Then we continue with the calculation of E [ X ], E [ X ] = λ λ- 1 ˆ ∞ e- λx dx , =- λ- 1 e- λx | x = ∞ x =0 , = λ- 1 . To get the variance use the formula Var( X ) = E [ X 2 ]- ( E [ X ]) 2 . (1) We need E [ X 2 ]. E [ X 2 ] = ˆ ∞-∞ x 2 f ( x ) dx, = λ ˆ ∞ x 2 e- λx dx. We again rely on integration by parts, setting u ( x ) = x 2 and dv ( x ) = e- λx dx . E [ X 2 ] = λ- x 2 λ- 1 e- λx | x = ∞ x =0 +2 λ- 1 ˆ ∞ xe- λx dx ....
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This note was uploaded on 06/05/2010 for the course STAT PStat 120a taught by Professor Rohinikumar during the Spring '10 term at UCSB.

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HW7_Solutions - PSTAT 120A Spring 2010 HW 7 Solutions 1 X...

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