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**Unformatted text preview: **Suppose S = {X1 , . . . , Xn } is a simple random sample from a population with and nite variance 2 < . Show that the sample mean X is an unbiased estimator for , so is sample variance s2 for 2 . 1. Before moving to formal proofs, there are several properties regarding expectation and variance as we metioned in early chapters: E (aX + b) = aE (X ) + b. (1) This property is called linearity of expectation, which naturally implies n E (X1 + . . . + Xn ) = E (X1 ) + E (X2 + . . . + Xn ) = E (X1 ) + E (X2 ) + E (X3 + . . . + Xn ) = E (Xi ). i=1 The other properties are regarding variance, V ar(aX + b) = a2 V ar(X ), V ar(X1 + X2 ) = V ar(X1 ) + V ar(X2 ) + 2Cov (X1 , X2 ). (2) The property given in (2) implies: (a) If X1 , . . . , Xn are independent, then they are uncorrelated, i.e. Cov (Xi , Xj ), i = j . Therefore, n V ar{ n Xi } = i=1 n V ar(Xi ) + i=1 Cov (Xi , Xj ) = V ar(Xi ). i=1 i=j (b) If X1 , . . . , Xn are independent and identically distributed, n n V ar(X ) = V ar{n1 Xi } = n2 nV ar(X1 ) = Xi } = n2 V ar{ i=1 i=1 V ar(X1 ) . n 2. The following equality is helpful in many derivations, n n (Xi a)2 n{X a}2 , a. (Xi X )2 = (3) i=1 i=1 This is because n (Xi X )2 = i=1 Note that n {(Xi a) (X a)}2 = i=1 n i=1 (Xi [(Xi a)2 2(Xi a)(X a) + (X a)2 ]. i=1 a)(X a) = n(X a)(X a), the result in equation (3) follows. 3. There are two equivalent ways of expressiong variance, they are V ar(X1 ) = E {X E (X )}2 = E (X 2 ) {E (X )}2 . Now, n E ( X ) = n 1 E (Xi ) = n1 n = . i=1 E (s2 ) = E n i=1 (Xi X )2 n1 n = 1 E { (Xi X )2 } = n 1 i=1 n i=1 E (Xi )2 nE (X )2 . n1 Note that E (Xi )2 = V ar(Xi ) = 2 and E (X )2 = V ar(X ) = n1 V ar(Xi ) = n1 2 , we then have E (s2 ) = n 2 n n1 2 = 2 . n1 1...

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