{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online