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Unformatted text preview: Stat 116 Homework 5 Solutions May 8, 2008 1. P ( Z > t ) = P ( X > t,Y > t ) = P ( X > t ) P ( Y > t ) = e ( Î» + Î¼ ) t Also, P ( I X<Y = 1) = P ( X < Y ) = Z âˆž Z âˆž x Î»Î¼e Î»x Î¼y dydx = Î» Î» + Î¼ P ( I X<Y = 0) = Î¼ Î» + Î¼ Therefore, for âˆ€ t > 0, P ( I X<Y = 1 ,Z > t ) = P ( t < X < Y ) = Z âˆž t Z âˆž x Î»Î¼e Î»x Î¼y dxdy = Î» Î» + Î¼ e Î» + Î¼ t = P ( X < Y ) P ( Z > t ) = P ( I X<Y = 1) P ( Z > t ) Similarly, we could derive P ( I X<Y = 0 ,Z > t ) = P ( I X<Y = 0) P ( Z > t ) , âˆ€ t > 0. And Hence P ( I X<Y = i,Z < t ) = P ( I X<Y = i ) P ( Z < t ) for all i = 0 , 1 ,t > 0, and Z is independent of I X<Y . 2. Y is distributed as negative binomial: P ( Y = l ) = l 1 5 p 6 (1 p ) l 6 Given Y = l , X = k  Y = l is hypergeometric, since for the first l1 trials, one get 5 success in all and among those success, k success come from the first 5 trials: for 5 â‰¥ k â‰¥ 11 l,l â‰¥ 6 P ( X = k  Y = l ) = 5 k l 6 5 k l 1 5 1 P ( X = k,Y = l ) = P ( X = k  Y = l ) P ( Y = l ) = 5 k l 6 5 k l 1 5 l...
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 Spring '07
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 Normal Distribution, Probability, Î», AirTrain Newark, 1 5 L, 5 K, 6 5 k

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