STAT 461/561
Kelin Pan
Assignment Two Solution
Question 1. Chapter 8. Ex.8.25 (p.406).
(a) and (b) has been discussed in class. Now for (c) Given BIN(n, θ) family, n is known.
The exponential family family can be found as
P (x|θ) =
n
Y
n!
θxi (1 − θ)n−xi = h(x)exp[T (x)η(θ) − B(θ)]
(n
−
x
)!x
!
i
i
i=1
Pn
where h(x) = exp[
n!
i=1 (n−xi )!xi ! ],
θ
T (x) = x̄, η(θ) = n(logθ − log(1 − θ)) = nlog 1−θ
, and
B(θ) = nlog(1 − θ). η is strictly increasing function of θ in the region (0, 1). Therefore
by the definition given in question 4, the statistic T is in MLR family.
Question 2. Chapter 8. Ex.8.28 (p.406).
The pdf and cdf of logistic distribution are
f (x|θ) =
ex−θ
,
(1 + ex−θ )2
F (x|θ) =
ex−θ
.
1 + ex−θ
1. For θ2 > θ1 , the likelihood ratio
λ(x, θ2 , θ1 ) =
f (x|θ2 )
1 + ex−θ1 2
= eθ1 −θ2 (
) .
f (x|θ1 )
1 + ex−θ2
The derivative of the quantity in the circle brackets is
d
ex−θ1 − ex−θ2
(.) =
.
dx
(1 + ex−θ2 )2
Because θ2 > θ1 , ex−θ1 > ex−θ2 , hence, the ratio is increasing in x. The family has
MLR.
2. Test hypothesis: H0 : θ = 0 vs. H1 : θ = 1.
The MP test to reject H0 is when
f (x|1)
f (x|0)
> k. By (1), this ratio is increasing in x.
Thus, this inequality is equivalent to rejecting H0 if x > k 0 . Using cdf of logistic
distribution, we have
0
α = supθ∈Θ0
ek
1
= P (X > k |θ = 0) = β(0) = 1 − F (k |0) = 1 −
.
0 =
k
1+e
1 + ek0
0
0
and
0
β = F (k 0 |1) =
ek −1
,
1 + ek0 −1
For a specified α = .2, k 0 = log( 1−α
α ) = 1.386 and β =
1
e.386
1+e.386
= .595.
3. From (1) and (2), the Karlin-Rubin theorem is satisfied. So the test is UMP of size
α.
Question 3. Chapter 8. Ex.8.31 (p.406).
Given X ∼ iid Poisson(λ)
1. Test hypothesis: H0 : λ ≤ λ0 vs. H1 : λ > λ0 .
By Karlin-Rubin theorem, the UMP test is to reject H0 if a sufficient statistic
T > k can be found.
Let T =
P
i Xi
∼ exp(nλ), which is Poisson distribution with single parameter
exponential family. We proved such a distribution has MLR (see Ex.8.25).
Karlin-Rubin theorem is satisfied if we chose a constant k to satisfy
P (T > k|λ0 ) = α.
2. Test hypothesis: H0 : λ ≤ 1 vs. H1 : λ > 1.
By CLT, T ∼ N (nλ, nλ), therefore
√
√
P (T > k|λ = 1) ≈ P (Z > (k − nλ)/ nλ|λ = 1) = P (Z > (k − n)/ n) = .05,
√
√
P (T > k|λ = 2) ≈ P (Z > (k − nλ)/ nλ|λ = 2) = P (Z > (k − 2n)/ 2n) = .9.
By normal table, we have
k−n
√
n
= 1.645 and
k−2n
√
2n
= −1.28, yielding n = 12 and
k = 17.70.
Question 4. Chapter 8. Ex.8.34 (p.407).
• Definition of MLR (Based on Bickel and Doksum book (Ref.1). I understand that
it may result in a confusion to have another definition for MLR. However without
this definition, it is hard to solve this question. As we discussed in class most
single-parameter exponential family pdf (with strictly increasing η(θ)) satisfies this
definition.)
Def.: The family of models {Pθ : θ ∈ Θ} with Θ ⊂ R (a Euclid space) is said to be
a monotone likelihood ratio (MLR) family if for θ1 > θ2 , the distribution Pθ1 and
Pθ2 are distinct and the ratio P (x, θ1 )/P (x, θ2 ) is an increasing function of T (x).
• (a) has been proved in class.
2
• For part (b), suppose we have continuous pdf. For θ1 > θ2 , we must show
F (t|θ1 ) ≤ F (t|θ2 ) or equivalently S(t|θ1 ) > S(t|θ2 ). Notice
d
f (t|θ1 )
[F (t|θ1 ) − F (t|θ2 )] = f (t|θ1 ) − f (t|θ2 ) = f (t|θ2 )(
− 1).
dt
f (t|θ2 )
Because pdf has MLR, the ratio on the right-hand side is increasing (according to
our definition of MLR). The derivative can only change the sign from negative to
positive, which means that any interior extremum is a minimum. Thus the
function in the square bracket is maximized by its value at ±∞, which is zero. It
means [.] ≤ 0, then we have F (t|θ1 ) ≤ F (t|θ2 ).
Question 5. Chapter 8. Ex.8.37 (p.407).
Given Xi ∼ N (θ, σ 2 )
Test hypothesis: H0 : θ ≤ θ0 vs. H1 : θ > θ0
1. If σ 2 is given, show that the test that reject H0 when X̄ > θ0 + zα σ 2 /n is a test
p
of size α. Show that the test can be derived as a LRT.
q
X̄ − θ0
P (X̄ > θ0 + zα σ 2 /n) = P ( p 2
> zα |θ0 ) = P (Z > zα |θ0 ) = α,
σ /n
for one sided-test, where Z ∼ N (0, 1). The normal test is equivalent to LRT when
C = exp[−zα σ 2 /2] (as we derived in lass for two-sided test).
2. From (1) we have X̄ > C. X̄ is sufficient and its pdf has MLR (it has been proved
in Exercise 8.25(a)). Then by Karlin-Rubin theorem, the test is UMP.
3. If σ 2 is unknown. In this case,
√
P (X̄ > θ0 + tn−1,α S/ n|θ0 ) = P (Tn−1 > tn−1,α ) = α
X̄−θ0
where Tn−1 = √
is a student’s t random variable with n − 1 degree of freedom.
2
S /n
If we define
1X
(xi − x̄)2 ,
n
1X
σ̂02 =
(xi − θ0 )2
n
then for X̄ > θ0 the LRT statistic can be expressed
σ̂ 2 =
(2πσ̂02 )−n/2 exp[− i (xi − θ0 )2 /2σ̂02 ]
λ(x) =
.
P
(2πσ̂ 2 )−n/2 exp[− i (xi − x̄)2 /2σ̂ 2 ]
P
3
(1)
Notice that in eq.(1), the numerator and denominator inside of square brackets of
exponential are same. Thus, we have
(
σ̂ 2 n/2
) .
σ̂0
For the case X̄ < θ0 , λ is simply equal to 1. Writing σ̂ 2 =
σ̂02 = (x̄ − θ0 )2 +
n−1 2
n S ,
n−1 2
n S
and
it is clear that the LRT is equivalent to the t-test because
λ < C when
(x̄ −
n−1 2
n S
2
θ0 )2 + n−1
n S
=
(n − 1)/n
< C, and x̄ ≥ θ0 ,
(x̄ − θ0 )2 /S 2 + (n − 1)/n
√
which is same as rejecting when (x̄ − θ0 )/(S/ n) is large.
Question 6. Chapter 8. Ex.8.49 (p.411).
1. p-value is
10
X
1
1
P r(X ≥ 7|θ = ) =
P r(X = i|θ = ) = .171875.
2
2
i=7
2. p-value is
P (X ≥ 3|λ = 1) = 1 − P (X < 3|λ = 1) = 1 − [
e−1 12 e−1 11 e−1 10
+
+
] = .0803.
2!
1!
0!
3. p-value is
X
P(
Xi ≥ 9|3λ = 3) = 1−P (T < 9|3λ = 3) = 1−e−3 [
i
38 37 36
30
+ + +...+ ] = .0038.
8! 7! 6!
0!
Question 7.
1. p-value is
P (H0 |X) = P (N ≤ 2|X = 1) = P (N = 1|X = 1)+P (N = 2|X = 1) = e−λ2 +λ2 e−λ2
= e−1.8 (1 + 1.8) = .463.
If we set α = .05, obviously we accept H0 . If we consider α = .5, we will reject H0 .
4