8.15 Let T = ∑ X2 i , where Xi ∼ N(0,σ2) then T ∼ σ 2χ2(n) with

8.15 Let T = ∑ X2 i , where Xi ∼ N(0,σ2) then T ∼ σ 2χ2(n) with

ST 522-001: Statistical Theory II
Solution to Lab Exercises - 12
Prepared by Chen-Yen Lin
Apr. 20, 2011
8.15 Let T =
P
i
Xi2 , where Xi ∼ N (0, σ 2 ) then T ∼ σ 2 χ2 (n) with density function
t
n
t 2 −1 e− 2σ2
g(t|σ ) =
Γ(n/2)(2σ 2 )n/2
2
Suppose σ12 < σ22 , then the ratio
−
t
2
(2σ12 )n/2 e 2σ2
g(t|σ22 )
=
=
− t2
g(t|σ12 )
(2σ22 )n/2 e 2σ1
µ
σ12
σ22
¶2
t
e2
(
1
1
2 − σ2 )
σ1
2
is monotonic increasing in t. By definition, the family g(t|σ 2 ) has MLR property.
Subsequently, by Karlin-Rubin theorem, the UMP test of H0 : σ = σ0 versus H1 : σ =
σ1 where σ0 < σ1 has the form
½
P
1 if Pi Xi2 > c
φ(T ) =
2
0 if
i Xi ≤ c
To find the critical value of c, one can use χ2 (n) percentile
Ã
!
µP 2
¶
X
c
2
i Xi
α=P
Xi > c|σ0
= P
> 2 |σ0
σ02
σ0
i
c
= χ21−α (n)
σ02
c = σ02 χ21−α (n)
8.23 (a) The size of the test and power function are given by
µ
¶
µ ¶θ
Z 1
1
1
1
θ−1
α = sup P X > |θ = sup
θX dx = sup 1 −
=
2
2
2
θ∈Θ0
θ∈Θ0
1/2
µ
¶
µ ¶θ
1
1
β(θ) = P X > |θ = 1 −
2
2
1
1.0
0.8
0.6
0.0
0.2
0.4
Power
0
1
2
3
4
5
θ
Figure 1: Power function
(b) By Neymann-Pearson Lemma, we reject the null if
f (x|1)
<k ⇔ x>c
f (x|2)
where the cutoff c can be found as
α = P (X > c|θ = 1) = (1 − c) ⇒ c = 1 − α
(c) First show that {f (x|θ) : θ ∈ Θ} has MLR property. Let θ2 > θ1
f (x|θ2 )
θ2
= (x)θ2 −θ1
f (x|θ1 )
θ1
Since θ2 −θ1 > 0, the ratio is increasing in x, thus the density family has MLR property.
By Karlin-Rubin theorem, there exists a UMP test having the form X > c.
P
iid
8.25 (a) Let X1 , . . . , Xn ∼ N (θ, σ 2 ) and σ 2 is known, then T = ni=1 Xi is sufficient for θ
and T ∼ N (nθ, nσ 2 ). Let θ2 > θ1 , then
h
i
(t−nθ2 )2
2 −1/2
(2πnσ
)
exp
−
2nσ 2
g(t|θ2 )
h
i
=
(t−nθ1 )2
g(t|θ1 )
2
−1/2
(2πnσ )
exp − 2nσ2
¸
·
(t − nθ1 )2 − (t − nθ2 )2
= exp
2nσ 2
¸
·
2n(θ2 − θ1 )t − n2 (θ22 − θ12 )
= exp
2nσ 2
2
is monotone increasing in t. Thus, the normal family has MLR property.
P
iid
(b) Let X1 , . . . , Xn ∼ P oi(θ), then T = ni=1 Xi is sufficient for θ and T ∼ P oi(nθ).
Let θ2 > θ1 , then
g(t|θ2 )
(t!)−1 e−nθ2 (nθ2 )t
=
g(t|θ1 )
(t!)−1 e−nθ1 (nθ1 )t
µ ¶t
θ2
−n(θ2 −θ1 )
= e
θ1
is monotone increasing in t. Thus, the Poisson family has MLR property.
P
iid
(c) Let X1 , . . . , Xm ∼ Bin(n, θ) and n is known, then T = m
i=1 Xi is sufficient for θ
and T ∼ Bin(mn, θ). Let θ2 > θ1 , then
Ctnm θ2t (1 − θ2 )nm−t
g(t|θ2 )
=
g(t|θ1 )
Ctnm θ1t (1 − θ1 )nm−t
¶t µ
µ
¶nm
θ2 /(1 − θ2 )
1 − θ2
=
θ1 /(1 − θ1 )
1 − θ1
is monotone increasing in t. Thus, the Binomial family has MLR property.
2)
8.26 (a) Given that the family f (x|θ) has MLR property, WLOG, we assume ff (x|θ
is
(x|θ1 )
monotone increasing in x. Want to show that F (x|θ2 ) ≤ F (x|θ1 ). To start with,
denote G(x) = F (x|θ2 ) − F (x|θ1 ) and realize that limx→∞ G(x) = limx→−∞ G(x) = 0.
·
¸
d
f (x|θ2 )
G(x) = f (x|θ2 ) − f (x|θ1 ) = f (x|θ1 )
−1
dx
f (x|θ1 )
2)
Since ff (x|θ
is monotone increasing, it follows immediately that the derivative of G(x)
(x|θ1 )
only change sign once. More specifically, it can only change sign from negative to
positive. That is, the second order derivative is positive when evaluated at the point
where the first derivative is zero. So, the G(x) has a local minimum. As a result,
we can assert that G(x) is a valley-shape function which decreases initially and then
bounce back eventually. Moreover, G(x) = 0 at the boundary points. As a result,
G(x) is constantly smaller than or equal to zero and therefore F (x|θ2 ) ≤ F (x|θ1 ), ∀x,
X ∼ F (x|θ2 ) is stochastically larger.
(b) Consider two Cauchy distributions f (x|θ1 ) =Cauchy(θ1 ) and f (x|θ2 )=Cauchy(θ2 )
where θ2 > θ1 . the corresponding CDFs are given by
µ ¶
Z x
1 1
1
x
1
FX (x|θ1 ) =
³ ´2 dt = + arctan
2 π
θ1
t
−∞ πθ1 1 +
θ1
µ ¶
1 1
x
FX (x|θ2 ) =
+ arctan
2 π
θ2
3
Given that arctan is a monotone increasing function, it follows immediately that
F (x|θ2 ) < F (x|θ1 ), ∀x, i.e. Cauchy(θ2 ) is stochastically larger than Cauchy(θ1 ). However, as Exercise 8.29 suggests, Cauchy family does not have MLR property.
8.29 (a) Let θ2 > θ1 (location parameter), then the ratio of two Cauchy density has the
form
f (x|θ2 )
1 + (x − θ1 )2
=
f (x|θ1 )
1 + (x − θ2 )2
By solving the derivative of the ratio equaling to 0, we find there are two local extremums. Thus, location Cauchy family does not have MLR property.
f (x|0)
f (x|1)
<k
1.5
0.5
1.0
f(x|0)/f(x|1)
2.0
2.5
(b) By Neymann-Pearson Lemma, we reject the null if
−10
−5
0
5
10
x
Figure 2: Power function
By looking at the graph of the ratio, intuitively if k is chosen too large, say k > 1.5,
does not make sense because the null hypothesis is favorable in this case. So, if we
decide to set k = 1/2, that is equivalent to reject if 1 < x < 3. That is, φ(x) is a UMP
test. Moreover, type I error and type II error can be found as
1 1
1 1
+ tan−1 (3) − − tan−1 (1) = 0.147
2 π
2 π
1 1
1 1
1 − β = 1 − P (1 < X < 3|θ = 1) = 1 − − tan−1 (2) + + tan−1 (0) = 0.647
2 π
2 π
α = P (1 < X < 3|θ = 0) =
4
(c) This test cannot be generalized to arbitrary one-sided test of the form H0 : θ ≤ 0
versus H1 : θ > 0. As a counterexample, consider H0 : θ = −10 versus H1 : θ = 10, then
regardless what value of k we choose, we will never obtain a reject region 1 < x < 3.
8.30 (a) Let θ2 > θ1 (scale parameter), then the ratio of two Cauchy density has the form
f (x|θ2 )
θ1 π (1 + (x/θ1 )2 )
θ2 θ12 + x2
=
=
f (x|θ1 )
θ2 π (1 + (x/θ2 )2 )
θ1 θ22 + x2
Since the ratio is a parabola curve and has local minimum at 0, it follows Cauchy
family does not have MLR property.
(b) Given T = |X|, the conditional distribution of X|T has the form
½ 1
if x = t
2
fX|T =t (x|t) =
1
if x = −t
2
Since the conditional distribution of X|T does not depend on θ, T = |X| is sufficient
for θ. Moreover, the the density function for T = |X| is given by
f (t|θ) =
2
1
¡ ¢
πθ 1 + t 2
θ
Using similar argument as in (a), it can be shown that the family of T = |X| does not
have MLR property.
5