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
α MM = 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)
1
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
α MM = 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(θ ) + bias2(θ ) (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 σ 2 = n 1n s
2:
V ar(σ 2) =
(
n 1
n
)2 2σ4
n 1 =
2(n 1)σ4
n2
(12)
Thus,
MSE(σ 2) =
2(n 1)σ4
n2
+
(
n 1
n
1
)2
σ4 =
2n 1
n2
σ4 (13)
Compare these two MSE, we get that σ 2 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 .
2
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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