STA 130B HW3
1. Chapter 8 Problem 17
Proof. (a): Denote l as the log-likelihood, then
∂l
∂x
=
α− 1
x
− α− 1
1− x
=
(α− 1)(1− 2x)
x(1− x)
(1)
Thus, if:
• α > 1, the density will increase in the interval [0,1/2], and then decrease in [1/2,1]
• α = 1, the density will be a constant.
• α < 1, the density will decrease in the interval [0,1/2], and then increase in [1/2,1].
(b):From the property above, Var(X) = 14(2α+1) . Thus,
EX2 = V ar(X) + (EX)2
=
1
4(2α+ 1)
+
1
4
⇒αˆMM = n
8
∑n
i=1 x
2
i − 2n
− 1
2
(2)
(c): The mle equation: ∂L∂α = 0
Γ′(2α)
Γ(2α)
− Γ
′(α)
Γ(α)
+
1
2n
n∑
i=1
log [Xi (1−Xi)] = 0 (3)
(d):The asymptotic variance is 1nI(α) , where I(α) = − ∂
2l
∂α2(
2n
[
Γ′′(α)Γ(α)− Γ′(α)2
Γ(α)2
− 2Γ
′′(2α)Γ(2α)− 2Γ′(2α)2
Γ(2α)2
])−1
(4)
Remark 1. The answer is not unique for question (e)
f(x1, · · · , xn|α) = C(α)exp((α− 1) log(
n∏
i=1
xi(1− xi))) (5)
By factorization theorem, a sufficient statistic would be T (x1, · · · , xn) =
∏n
i=1 xi(1− xi)
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2. Chapter 8 Problem 18:
(a):From the property above, Var(X) = 29(3α+1) . Thus,
EX2 = V ar(X) + (EX)2
=
1
4(2α+ 1)
+
1
4
⇒αˆMM = 1
3
(
2n
9
∑n
i=1 x
2
i − n
− 1
) (6)
(c): The mle equation: ∂L∂α = 0
3Γ′(3α)
Γ(3α)
− Γ
′(α)
Γ(α)
− 2Γ
′(2α)
Γ(2α)
+
1
n
n∑
i=1
log
[
Xi (1−Xi)2
]
= 0 (7)
(d):The asymptotic variance is 1nI(α) , where I(α) = − ∂
2l
∂α2(
n
[
Γ′′(α)Γ(α)− Γ′(α)2
Γ(α)2
+
4Γ′′(2α)Γ(2α)− 4Γ′(2α)2
Γ(2α)2
− 9Γ
′′(3α)Γ(3α)− Γ′(3α)2
Γ2(3α)
])−1
(8)
(e):
f(x1, · · · , xn|α) = C(α)h(x)exp(α log(
n∏
i=1
xi(1− xi)2)) (9)
By factorization theorem, a sufficient statistic would be T (x1, · · · , xn) =
∏n
i=1 xi(1− xi)2
3. Chapter 8 Problem 57
(a): s2 is unbiased, which has been proved before.
Use the formula:
MSE(θ) = V ar(θˆ) + bias2(θˆ) (10)
(b) For s2, it is unbiased, thus the second part is 0:
V ar(s2) =
(
σ2
n− 1
)2
V ar(
(n− 1)s2
σ2
) =
σ4
(n− 1)2 2(n− 1) =
2σ4
n− 1 (11)
Thus the MSE of s2 is 2σ
4
n−1 .
For σˆ2 = n−1n s
2:
V ar(σˆ2) =
(
n− 1
n
)2 2σ4
n− 1 =
2(n− 1)σ4
n2
(12)
Thus,
MSE(σˆ2) =
2(n− 1)σ4
n2
+
(
n− 1
n
− 1
)2
σ4 =
2n− 1
n2
σ4 (13)
Compare these two MSE, we get that σˆ2 has smaller MSE. (c): From (b), replace n by 1/ρ, we
need to calculate the minumum value of 2n−1((n− 1)ρ)2 + ((n− 1)ρ− 1)2. Let t = (n− 1)ρ and
take derivative we end up with 2t− 2 + 4tn−1 , then t = n−1n+1 , then ρ = 1n+1 .
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4. Chapter 9 Problem 1
Denote X the number of heads. (a): The significance level of the test:
α = P (X = 0) + P (X = 10) = (1/2)9 = 0.002 (14)
(b): Denote H1 as the head probability is 0.1, then the power of the test:
1− β = P (X = 0|H1) + P (X = 10|H1) = 0.100.910 + 0.1100.90 = 0.349 (15)
5. Chapter 9 Problem 2
Simple hypothesis: all the parameters related to the distribution are stated. (b): the probability
of any side is p = 1/6 unless you say the die may have different faces than 6. (c) and (d), σ is
not known. Thus:
(a) and (b) are simple; (c) and (d) are composite.
6. Chapter 9 Problem 4
(a): Λ = P (H0|x)P (H1|x)
• x = x1,Λ = 0.20.1 = 2
• x = x2,Λ = 0.30.4 = 3/4
• x = x3,Λ = 0.30.1 = 3
• x = x4,Λ = 0.20.4 = 0.5
Thus the order of xi according to Λ is x3, x1, x2, x4
(b): The likelihood ratio test reject H0 when Λ < k, P (X = x4|H0) = 0.2 = α and
P (x = {x4, x2}|H0) = 0.2 + 0.3 = 0.5 > α. Thus a LR test at level α = .2 would be:
reject the null hypothesis when Λ < k, where k ∈ (0.5, 0.75].
Similarly, a LR test at level α = .5 would be :
Reject the null hypothesis when Λ < k, where k ∈ (0.75, 2]
(c).P (Hi|x) ∝ P (Hi)P (x|Hi), we have P (H0) = P (HA), thus larger value of P (x|Hi) would be
an evidence in favor of Hi. Based on the distribution, x1, x3 favor H0.
(d).Suppose the prior is: P (H0) = pi, P (HA) = 1 − pi. Then we would reject H0 if
piP (X = xi|H0) < (1− pi)P (X = xi|HA).
• To reject when x = x1:0.2pi < 0.1(1− pi),⇒ pi < 1/3
• To reject when x = x2:0.3pi < 0.4(1− pi),⇒ pi < 4/7
• To reject when x = x3:0.3pi < 0.1(1− pi),⇒ pi < 1/4
• To reject when x = x4:0.2pi < 0.4(1− pi),⇒ pi < 2/3
Thus, to have a α = 0.2 test, we can at most reject the null when x = x4 becasue 2/3 is the
largest number here, this requires pi ≥ 4/7.To have a α = 0.5 test, we could at most reject
x = x4 and x = x2, this requires pi ≥ 1/3
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