20082ee131A_1_HW7SOL

20082ee131A_1_HW7SOL - 2 dx Z ∞ x π(1 x 2 dx Consider...

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EE 131A Homework #7 Solution Spring 2008 K. Yao 1. Since y = 2 - x 3 , then for 0 x < , then -∞ < y 2 . Furthermore, x = (2 - y ) 1 / 3 . Then dy dx = - 3 x 2 , dy dx = 3 x 2 . f Y ( y ) = f X ( x ) dy dx | x =(2 - y ) 1 / 3 = e - x 3 x 2 | x =(2 - y ) 1 / 3 = e - (2 - y ) 1 / 3 3(2 - y ) 2 / 3 , -∞ < y 2 . 2. a . P (reject) = P ( X 2 . 05) + P ( X 1 . 95) = 2 P ( X 1 . 95) = 2 P ( X - 2 0 . 08 1 . 95 - 2 0 . 08 ) = 2Φ( - 0 . 625) = 0 . 530 b . P (reject) = 0 . 2. From Part a, 2 P ( X 1 . 95) = 0 . 2, and 2 P ( X - 2 σ 1 . 95 - 2 σ ) = 0 . 2. Thus, Φ( - 0 . 05 ) = 0 . 1, or 0 . 05 = 1 . 28, and σ = 0 . 039. 3. E [ X ] = n k =1 kP [ X = k ] = n k =1 k n = 1 n n k =1 k = n ( n + 1) 2 n = n + 1 2 . σ 2 X = E [ X 2 ] - E [ X ] 2 = n k =1 k 2 n - n +1 2 2 = ( n +1)(2 n +1) 6 - ( n +1) 2 4 = n 2 - 1 12 . 4. a. E [ X ] = k =0 k α k k ! e - α = α k =1 α k - 1 ( k - 1)! e - α = α k =0 α k k ! e - α = α · 1 = α. E [ X 2 ] = k =0 k 2 α k k ! e - α = α k =1 k α k - 1 ( k - 1)! e - α = α k =0 ( k + 1) α k k ! e - α = α ( α + 1) . σ 2 X = E [ X 2 ] - E [ X ] 2 = α ( α + 1 - α ) = α. b. E[X] = α = λ t = arrival rate × t, which is the average of the rv X. 5.
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Unformatted text preview: 2 ) dx + Z ∞ x π (1 + x 2 ) dx. Consider the latter term: Z y x π (1 + x 2 ) dx = 1 2 π ln(1 + x 2 ) ± ± y = ln(1 + y 2 ) 2 π → ∞ , as y → ∞ . Thus, the integrals do not exist ⇒ E [ X ] does not exist. 6. Since G N ( z ) = exp( α ( z-1)) and G M ( z ) = exp( β ( z-1)) , then G N ( z ) G M ( z ) = exp(( α + β )( z-1)) . Thus, G N ( z ) G M ( z ) is the pgf of a Poisson rv with the parameter ( α + β ) ....
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