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Unformatted text preview: Homework Solutions Assignment Number 4 Due February 12, 2007 Section 5.4. Problem 19. (a) (i) E ( ˆ θ 1 ) = E ( Y 1 ) = Z ∞ y 1 θ e y/θ dy = θ (ii) E ( ˆ θ 2 ) = E ( Y ) = E 1 n n X i =0 Y i ! = 1 n n X i =0 E ( Y i ) = 1 n · nθ = θ (iii) First, we will compute the pdf for Y min , and use that to compute the pdf for n · Y min : f Y min ( y ) = nf Y ( y )(1 F Y ( y )) n 1 = n 1 θ e y/θ [1 (1 e y/θ )] n 1 = n 1 θ e ny/θ = nf Y ( ny ) . Now, to compute the pdf for n · Y min , we consider that P ( a < n · Y min < b ) = P ( a/n < Y min < b/n ) = R b/n a/n f Y min ( y ) dy = R b a 1 n f Y min ( y/n ) dy , so the pdf f n · Y min ( y ) is f n · Y min ( y ) = 1 n f Y min ( y/n ) = 1 n · nf Y ( y/n · n ) = f Y ( y ) . This tells us that E ( ˆ θ 3 ) = E ( n · Y min ) = E ( Y ) = θ . (b) (i) Var( Y 1 ) = E ( Y 2 1 ) E ( Y 1 ) 2 = Z ∞ y 2 1 θ e y/θ dy θ 2 = 2 θ 2 θ 2 = θ 2 (ii) Var( Y ) = Var( 1 n n X i =1 Y i ) = 1 n 2 n X i =1 Var( Y i ) = 1 n 2 ·...
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 Spring '07
 Xu,Lily
 Statistics, Derivative, Normal Distribution, 1g, Yi, 1 W

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