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HW2 - 8.2 ^ a Show that E(3 = ^ ^ ^ ^ ^ E(3 = E(a1(1 a)2 =...

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Unformatted text preview: 8.2 ^ a Show that E(3 ) = ^ ^ ^ ^ ^ E(3 ) = E(a1 + (1 - a)2 ) = aE(1 ) + (1 - a)E(2 ) = a + (1 - a) = ^ ^ b Find a = arg min V ar()3 under 1 and 2 are independent. ^ ^ ^ ^ ^ Let L = var(3 ) = V ar(a1 + (1 - a)2 ) = a2 V ar(1 ) + (1 - a)2 var(2 ) = a2 2 + (1 - a)2 2 1 2 To find minimizer of L, take a derivative a L 2 2 = 2a1 - 2(1 - a)2 = 0 2 2 2 2 1 +2 Therefore a = 2 L a2 Take the second derivative to check whether a is minimizer or maximizer 2 2 = 21 + 22 0 8.8 Since Y (, + 1), fy = 1 (+1)- =1 a E(Y ) = E(Y ) = Therefore +1 yf (y)dy Bais=E(Y ) - = 1 2 =+ 1 2 b Define Z = Y - 1 , then E(Z) = 2 Therefore Z is the unbiased estimator of , which is a function of Y 1 c MSE(Y ) = V ar(Y )+Bias(Y ) = 12n + 1 4 ) = 1 Var(Y)= 1 +1 y 2 dy + E(Y )2 = V ar(Y n n 1 n 1 3 ( + 1)3 - 1 3 + + 3 1 2 = 1 12n 8.9 a E nY 1 - n Y n =E Y - Y2 n = np - V ar(Y )+E(Y )2 n = (n - 1)p(1 - p) = np(1 - p) = V (Y ) Therefore the suggested estimator is a biased estimator of V (Y ). b Let Z = n Y n-1 n 1- Y n , then E(Z) = np(1 - p) = V (Y ) Therefore Z is an unbiased estimator of V (Y ) 1 Problems 3) Find 1 20th percentile of Normal using standard normal table. -2.3263479 -2.0537489 -1.8807936 -1.7506861 -1.6448536 -1.5547736 -1.4757910 -1.4050716 -1.3407550 -1.2815516 -1.2265281 -1.1749868 -1.1263911 -1.0803193 -1.0364334 -0.9944579 -0.9541653 -0.9153651 -0.8778963 -0.8416212 Then draw plot. Y axis is the percentile of normal above, and X axis is the given percentile. 2 ...
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HW2 - 8.2 ^ a Show that E(3 = ^ ^ ^ ^ ^ E(3 = E(a1(1 a)2 =...

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