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411_hw06_soln

# 411_hw06_soln - Stat 411 Homework 06 Solutions 1(a First...

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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 2 ) e - y 2 , 0 < 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 0 1 θ 2 e - y 2 dy 1 = y 2 θ 2 e - y 2 , 0 < 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 2 ) e - y 2 ( y 2 2 ) e - y 2 = 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 )] = θ 2 / 2 . To compare with Y 1 = X 1 , note first that E θ ( X 1 ) = θ so both Y 1 and g ( Y 2 ) are unbiased. However, V θ ( X 1 ) = θ 2 > θ 2 / 2 = V θ [ g ( Y 2 )] .

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