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Unformatted text preview: Stat 411 – Homework 06 Solutions 1. (a) First need to find the joint PDF of Y 1 = X 1 and Y 2 = X 1 + X 2 . The Jacobian of the (inverse) transformation is 1, so the joint PDF of ( Y 1 ,Y 2 ) is given by f Y 1 ,Y 2 ( y 1 ,y 2 ) = f X 1 ,X 2 ( y 1 ,y 2 y 1 ) = (1 /θ 2 ) e y 2 /θ , < y 1 < y 2 < ∞ . The marginal PDF for Y 2 is obtained by integrating f Y 1 ,Y 2 ( y 1 ,y 2 ) over all y 1 . In this case, f Y 2 ( y 2 ) = Z y 2 1 θ 2 e y 2 /θ dy 1 = y 2 θ 2 e y 2 /θ , < y 2 < ∞ . (An alternative derivation of this formula sees that the MGF of Y 2 is that of a gamma random variable with suitable parameters.) Now the conditional PDF of Y 1 given Y 2 = y 2 is the ratio f Y 1  Y 2 ( y 1  y 2 ) = f Y 1 ,Y 2 ( y 1 ,y 2 ) f Y 2 ( y 2 ) = (1 /θ 2 ) e y 2 /θ ( y 2 /θ 2 ) e y 2 /θ = 1 y 2 ; that is, given Y 2 = y 2 , Y 1 ∼ Unif (0 ,y 2 ). (b) Since, given Y 2 = y 2 , Y 1 ∼ Unif (0 ,y 2 ), it follows immediately that g ( y 2 ) = E θ ( Y 1  Y 2 = y 2 ) = E { Unif (0 ,y 2 ) } = y 2 / 2. This expression does not depend on θ , as it shouldn’t since Y 2 is a sufficient statistic for θ . (c) Since g ( Y 2 ) = Y 2 / 2, it is clear that E θ [ g ( Y 2 )] = E θ ( Y 2 ) / 2 and V θ [ g ( Y 2 )] = V θ ( Y 2 ) / 4. Moreover, since Y 2 = X 1 + X 2 , we know that E θ ( Y 2 ) = 2 E θ ( X 1 ) and V θ ( Y 2 ) = 2 V θ ( X 1 ). Further, we know that E θ ( X 1 ) = θ and V θ ( X 1 ) = θ 2 because X 1 is an exponential random variable. Therefore, E θ [ g ( Y 2 )] = θ and V θ [ g ( Y 2 )] =...
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This note was uploaded on 03/12/2012 for the course STAT 411 taught by Professor Staff during the Spring '08 term at Ill. Chicago.
 Spring '08
 STAFF

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