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**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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