Unformatted text preview: Suppose S = {X1 , . . . , Xn } is a simple random sample from a population with µ and ﬁnite variance
¯
σ 2 < ∞. Show that the sample mean X is an unbiased estimator for µ, so is sample variance s2 for σ 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{n−1 Xi } = n−2 · nV ar(X1 ) = Xi } = n−2 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 ) = n−1 · nµ = µ.
i=1 E (s2 ) = E n
i=1 (Xi ¯
− X )2
n−1 n = 1
¯
E { (Xi − X )2 } =
n − 1 i=1 n
i=1 ¯
E (Xi − µ)2 − nE (X − µ)2
.
n−1 ¯
¯
Note that E (Xi − µ)2 = V ar(Xi ) = σ 2 and E (X − µ)2 = V ar(X ) = n−1 V ar(Xi ) = n−1 σ 2 , we then have
E (s2 ) = nσ 2 − n · n−1 σ 2
= σ2 .
n−1 1 ...
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This note was uploaded on 02/01/2012 for the course STAT 330B taught by Professor Zhou during the Spring '11 term at Iowa State.
 Spring '11
 Zhou
 Variance

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