ST3239: Survey Methodology - The Department of Statistics and ...

Theorem 2.2.2
E(ȳ) = µ,
V ar(ȳ) = σ2
n
( ) N − n
.
N − 1
Proof.
Note ȳ = 1 n (y 1 + ... + y n ). So
E(ȳ) = 1 n (Ey 1 + ... + Ey n ) = 1 (nµ) = µ.
n
Now
V ar(ȳ) = 1 n Cov( ∑ n n∑
y 2 i , y j ) = 1 ∑ n n∑
Cov(y
i=1 j=1
n 2
i , y j )
i=1 j=1
⎛
⎞
= 1 ⎝ ∑ Cov(y
n 2 i , y j ) + ∑ Cov(y i , y j ) ⎠
i≠j
i=j
⎛
⎞
= 1 ⎝ ∑ (− σ2 n
n 2 i≠j
N − 1 ) + ∑
V ar(y i ) ⎠
i=1
= 1 (
)
n(n − 1)(−
σ2
n 2 N − 1 ) + nσ2
(
(n − 1)(− 1
= σ2
n
= σ2
n
( ) N − n
N − 1
N − 1 ) + 1 )
Remark: From Theorem 2.2.2, we see that ȳ is an unbiased estimator for µ. Also as n gets large
(but n ≤ N), V ar(ȳ) tends to 0. This implies that ȳ will be a more accurate estimator for µ as n gets
larger (but less than N). In particular, when n = N, we have a census and V ar(ȳ) = 0.
Remark: In our previous statistics course, we usually sample {y 1 , y 2 , · · · , y n } from the population
with replacement. Therefore, {y 1 , y 2 , · · · , y n } are independent and identically distributed (i.i.d.).
And recall we have results like
E iid (ȳ) = µ,
V ar iid (ȳ) = σ2
n .
Notice that V ar iid (ȳ) is different from V ar(ȳ) in Theorem 2.2.2. In fact, for n > 1,
V ar(ȳ) = σ2
n
( ) N − n
< σ2
N − 1 n = V ar iid(ȳ).
Thus, for the same sample size n, sampling without replacement produces a less variable estimator of
µ. Why?
7