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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 wemetioned in early chapters:E (aX + b) = aE (X ) + b.(1)This property is called linearity of expectation, which naturally impliesnE (X1 + . . . + Xn ) = E (X1 ) + E (X2 + . . . + Xn ) = E (X1 ) + E (X2 ) + E (X3 + . . . + Xn ) =E (Xi ).i=1The 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,nV ar{nXi } =i=1nV ar(Xi ) +i=1Cov (Xi , Xj ) =V ar(Xi ).i=1i=j(b) If X1 , . . . , Xn are independent and identically distributed,nnV ar(X ) = V ar{n1Xi } = n2 nV ar(X1 ) =Xi } = n2 V ar{i=1i=1V ar(X1 ).n2. The following equality is helpful in many derivations,nn(Xi a)2 n{X a}2 , a.(Xi X )2 =(3)i=1i=1This is becausen(Xi X )2 =i=1Note thatn{(Xi a) (X a)}2 =i=1ni=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 areV ar(X1 ) = E {X E (X )}2 = E (X 2 ) {E (X )}2 .Now,nE ( X ) = n 1E (Xi ) = n1 n = .i=1E (s2 ) = Eni=1 (Xi X )2n1n=1E { (Xi X )2 } =n 1 i=1ni=1E (Xi )2 nE (X )2.n1Note that E (Xi )2 = V ar(Xi ) = 2 and E (X )2 = V ar(X ) = n1 V ar(Xi ) = n1 2 , we then haveE (s2 ) =n 2 n n1 2= 2 .n11...

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