final-prac-prob-sol

# final-prac-prob-sol - EE 131A Probabilities Instructor...

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EE 131A Practice Problem Solution Set II Probabilities Instructor: Vwani Roychowdhury 1. Problem 3.153 (a) Pr[pass the test] = (1 - p ) + p (1 - α ) Pr[fail the test] = Pr[ k items] = [(1 - p ) + p (1 - α )] k - 1 (b) The answer is p (1 - α ) 1 - . (c) The answer is p (1 - α ) 1 - (1 - p ) b - . 2. Problem 3.156 X = X 1 + · · · + X n , X i is exponential with λ = 1 Pr[ X > 2] 0 . 9where X is an Erlang r.v. k =0 X n - 1 1 . 2 k k ! e - 1 . 2 0 . 9 Hence, n 5 3. Problem 3.59 (a) Pr[ Y y ] = Pr[ e X y ] = Pr[ X ln Y ] = F X (ln y ) , y > 0 f Y ( y ) = d dy F Y ( y ) = 1 y f X (ln y ) (b) For Gaussian X and y > 0 f Y ( y ) = exp[ - (ln y - m ) 2 / 2 σ 2 ] 2 π 1

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4. Problem 3.58 (a) Problem y < 0 : Pr[ Y y ] = Pr[ X + a y ] = F X ( y - a ) y = 0 : Pr[ Y 0] = Pr[ X a ] = F X ( a ) y < 0 : Pr[ Y y ] = Pr[ X - a y ] = F X ( y + a ) y < 0 : f Y ( y ) = f x ( y - a ) y = 0 : f Y ( y ) = ( F X ( a ) - F X ( - a - )) δ ( y ) y > 0 : f Y ( y ) = f X ( y + a ) (b) If f X ( x ) = β exp[ - β | x | ] 2 , then F Y ( y ) = 0 . 5 e β ( y - a ) , y < 0 F Y ( y ) = 1 - 0 . 5 e β ( y + a ) , y 0 y < 0 : f Y ( y ) = β 2 e - ( a - y ) y = 0 : f Y ( y ) = (1 - e βa ) δ ( y ) y > 0 : f Y ( y ) = 0 . 5 βe - β ( y + a ) 5. Problem 3.89 1 2 π Z + -∞ e -| w | e - jwx dw = 1 2 Z 0 -∞ e w e - jwx dw + 1 2 Z 0 e w e - jwx dw = 1 2 π [ 1 1 - jx + 1 1 + jx ] = 1 π (1 + x 2 ) 6. Problem 3.89
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