Math 4650/5820 - Spring 2009

HW #1 Solutions

Problem #3The variables in parts d), f), h) are random variables.

Problem #12

a) Proof: Simple algebra shows that

s2 =1

n − 1{

n∑

i=1

X2

i − nX̄2}

Hence

E(s2) =1

n − 1{

n∑

i=1

EX2

i − nEX̄2}

We note that EX2

i = Var(Xi) + (EXi)2 = σ2 + µ2. By the same argument

coupled with simple random sampling without replacement, we have

EX̄2 = V ar(X̄) + (EX̄)2 =σ2

n+ µ2

Therefore

E(s2) =1

n − 1{n(σ2 + µ2) − n(σ2/n + µ2)} = σ2

b) Since the square root function is nonlinear, s is not an unbiased estimateof σ.

c) Since σ2

X̄= σ2/n, we conclude from part a) that

E(n−1s2) = n−1σ2 = σ2

X̄

d) Since σ2

T = N2σ2

X̄, it follows from part c) that n−1N2s2 is an unbiased

estimate of σ2

T .

e) Proof: In the dichotomous case, the population variance reduces top(1−p). Because of simple random sampling with replacement, we concludethat σ2

p̂ = p(1 − p)/n. The proof is complete by noting that

Ep̂(1 − p̂)

n − 1=

1

n − 1{Ep̂ − [V ar(p̂) + (Ep̂)2]}

=1

n − 1{p − [

p(1 − p)

n+ p2]} =

p(1 − p)

n

Problem #15

a) From Example A of Section 7.2, we know that N = 393 and σ = 589.7.Thus

P (|X̄ − µ| > 200) ≈ 2 − 2Φ(200

σX̄

),

where σX̄ = 589.7√

n

√

1 − n−1

392A plot of this probability versus sample size is

attached.

20 30 40 50 60 70 80 90 1000

0.02

0.04

0.06

0.08

0.1

0.12

Sample Size

Pro

babi

lity

b) By using the same CLT argument, we found that

n = 20, ∆1 = 211.5703, ∆2 = 86.7567

n = 40, ∆1 = 145.5368, ∆2 = 59.6789

n = 80, ∆1 = 96.9042, ∆2 = 39.7366

Problem #24

a) In order for the estimate to be unbiased, we must have

EX̄c =n

∑

i=1

ciEXi = µn

∑

i=1

ci = µ.

Thus the condition on the ci is∑n

i=1ci = 1.

b) Direct calculations show that

V arX̄c =n

∑

i=1

c2

i Var(Xi) =n

∑

i=1

c2

i σ2

To minimize this variance subject to the aforementioned condition, consider

L(c1, · · · , cn, λ) = σ2

n∑

i=1

c2

i + λ(n

∑

i=1

ci − 1)

For i = 1, · · · , n, we have

∂L

∂ci

= 2σ2ci + λ.

Setting these partial derivatives equal to zero, we have

ci = −λ

2σ2

Since∑n

i=1ci = 1, we obtain −λ = 2σ2/n. Plugging this expression in ci, we

conclude that ci = 1/n.