数理统计期末考试试题答案
江汉大学文理学院地址-范秀明
1. Let
X
1
,X
2
,
,X
n
be
a random sample from the
Gamma(
,
)
distribution <
br>f(x|
,
)
1
(
)
x
1
e
,
x0
,
0
,
0
.
x
(a) ( 8 %) Find the method of moment
estimates of
and
.
(b) ( 7
%) Find the MLE of
, assuming
is known.
(c) ( 7 %) Giving
0
,
find the Cramer-Rao lower bound of estimates of
.
(d) ( 8 %) Giving
0
,
find the UMVUE of
.
2. Suppose that
X
1
,X
2
,
,X
n
are
iid ~
B(2,p)
,
p(0,1)
. Let
(p)2p(1p)
.
(a) ( 5 %) Show that
T
X
i
is a sufficient
statistic for
p
.
i
1
n
1, if
X
1
1
(b) ( 5 %) Let
Y
. Show that
Y
is an
unbiased estimate of
(p)
.
0,
if X
1
1
(c) (10%) Find the UMVUE
W
of
(p)
.
3. Let
X
1
, X
2
,
,X
n
be a random sample from a
Poisson(
)
,
0
,
distribution.
Consider testing
H
0
:
1
vs
H
1
:
3
.
(a) (10%) Find
a UMP level
test,
0
1
.
(b) ( 7 %) For
n3
, the test rejects
H
0
, if
X
1
X
2
X
3
5
.
Find the power function
(
)
of the test.
(c) ( 8
%) For
n3
, the test rejects
H
0
, if
X
1
X
2
X
3
5
.
Evaluate the size and the power of the test.
4. (10%) Let
X
1
, X
2
,
,X
n
be iid
Poisson()
distribution, and let the prior
distribution
of
be a
Gamma(
,
)
distribution,
0
,
0
. Find the
posterior distribution of
.
5.
Let
X
1
, X
2
,
,X
n
be a random sample from an
exponential distribution with mean
,
0
.
(a) ( 5 %) Show
that
T
X
i
is a sufficient
statistic n for
.
i
1
n
(b) ( 5 %) Show that the
Poisson family has a monotone likelihood ratio,
MLR.
(c) ( 5 %) Find a UMP level
test of
H
0
:0
1
vs
H
1
:
1
by the
Karlin-Rubin Theorem shown below.
[Definition] A family of pdfs or pmfs
{g(t|
)|
}
has a
monotone likelihood ratio,
g(t|
2
)
MLR, if for every
2
1
,
is a monotone
function of
t
.
g(t|
1
)
[Karlin-Rubin Theorem]
Suppose that
T
is a sufficient statistic
for
and the pdfs or
pmfs
{g(t|
)|
}
has a non-
decreasing monotone likelihood ratio.
Consider
testing
H
0
:
0
vs
H
1
:
0
. A UMP
level
test rejects
H
0
if and
only if
Tt
0
, where
P
0
(Tt
0
)
.
數理統計期末考試試題答案
1. (a) Since
E(X)
0
(
)
1
(
1)
1
xed
x
and
x
<
br>(
)
E(X)
2<
br>
0
(
)
2<
br>m
1
1
x
1
(
<
br>
2)
2
edx
(
1)
2
,
x
(
)
Let
m
1
and
m
2
(
1)
2
~
m
2
2
m
1
1
2
m
1
m
2
m
1
,
~
.
2
m
1
m
2
m
1
1
n
2
1
n
(n
1)
222
Furthermore,
m
1
X
,
m
2
m
1
X
i
X
(X
i2
X)
2
S
,
n
i
1
n
i
1
n
~
(n1)S
2
nX
2
~
The MME of
.and
are
,
2
nX
(n
1)S
~
(b)
L(
|
,
~
x)
[
i
1
n
1
(
)
1
x
i
e]
x
i
1
[
(
)
]
n
i
1
n
1
(
x
i
)e
n
n
i
1
x
i
n
x)
nln
(
)
n
ln
(
1)
lnx
i
i
1
lnL(
|
,
~
i
1
x
1
n
1
n
1
n
x
~
ˆ
lnL(
|
,x)
x
0
x
Let
i
i
.
2
i
1
n
i
1
n
2
n
n
2nx
~<
br>lnL(
|
,x)
lnx
Furthermore,
i
223
23
i
1
n
2nxnx
2nx
n
ˆ
|
,
~
lnL(
x)
0
,
22332
ˆˆ
xx
ˆ
X
is the MLE of
. So,
2
2
n
2
n
n
2n
n
~
lnL(
|
,x)]
E
(
X)
(c)
E
[
223
i
232<
br>
i
1
2
CRLB =
E
[
(d) Since
E<
br>(
X
)
2
2
2
n
~
lnL(
|
<
br>,x)]
1
ˆ
X
is an
unbiased estimate of
, and
,
X1
2<
br>
2
ˆ
X
is the UMVUE of
.
Var()
CRLB,
n
2
n
1
(
)
(
)
Given
,
{f(x|
)}
is an
exponential family in
.
I
(0,
)
(x)x
1
e
x
[Or]
f(x|
,
)
1
1
I
(0,
)
(x)x
1
exp[x(
)]
T
X
i
is a sufficient
statistic for
.
i
1
n
ˆ
X
T
is an unbiased estimate of
and a
function of sufficient Since
n
ˆ
X
is the
UMVUE of
. statistics
T
, by Rao-
Blackwell Theorem,
nn
2
x
i
2
x
i
2. (a)
f(x
1
,x
2
,,x
n
|p)
<
br>f(x
i
|p)
[
I(x)p(1<
br>
p)]
{0,1,2}i
x
i
1i
1
i
2
2
p
x
i
p
i
1
2
[
I
{0,1,2}
(x
i
)()(1
p)]
[
I(x)]()(1
p)
2n
{0,1,2}i
<
br>xx
1
p1
p
i
1
i
i
1
i
n
2
p
T(
~
x)2n
~
)
(1p)
and
h(x)
Let
g
(T(x),p)
(
I
{0,1,2}
(
x
i
)
. By
x
1
p
i
1
i
nn
x
i
n
factorization theorem,
T
X
i
is a sufficient
statistic for
p
.
i
1
n
2
p
x2
x
2
2
[Or]
f(x|p)
I(x)p(1
p)
I(x)(1
p)exp[xln()]
x
{0,1,2}
x
{0,1,
2}
1
p
{f(x|p)}
is an exponential family
T
X
i
is a
sufficient statistic.
i
1
n
2
12
1
(b)
E(Y)
1
P(X
1
1)
0
P(X
1
1)
p(1p)2p(1p)<
br>, so
Y
is an
1
unbiased estimate of
(p)
.
(c) If
X
1
,X
2
,
,X
n
,
nN
, are iid ~
B(2,p)
, then
T
X
i
~
B
(2
n
,
p<
br>)
.
i
1
n
E(Y|T
t)
P(Y
1
&T
t)
P(X
1
1 &T
t)<
br>
P(T
t)P(T
t)
n
P(X<
br>1
1 &
X
i
t
1)
i
2
n
P(T
t)
2n
2
t
12n
t<
br>
1
2p(1
p)
p(1
p)
t
1
i
2
P(T
t)
2n
t2n
t
t
p(1
p)
2(2n
2)!t!(2n
t)!
t(2n
t)
,
t0,1,2,,2n
.
(t
1)!(2n
t
1)!(2n)!n(
2n
1)
T(n
T)
By Rao-
Blackwell Theorem,
W
E(Y|T)
is the UMVUE of
(
)e
.
2(2n
1)
3. (a) By Neyman-Pearson
Lemma, a UMP level
test rejects
H
0
if and only if
f(x
1
,x2
,,x
n
|
3)
kf(
x
1
,x
2
,,x
n
|
1)
.
P(X
1
1)P(
X
i
t
1)
n
n
1
x
i
3<
br>x
i
3
x
i
2n
1
ke
(
x
i
)ln3
2n
lnk
e]
k
[
e]
3
[
i
1
(x
i
)!
i
1
(x
i
)!
i
<
br>1
n
2
lnk
c
x
i
ln3
i
1
n
Since
n
i
1
X
i
~
Poisson
(
n
)
, a UMP
level
test rejects
H
0
if and
only if
n
n
i
n
Xi
c
, where
c
is the smallest
integer satisfying
i!
e
.
i
1
i
c
1
[Or]
T
X
i
is sufficient for
and
T~Poisson(n
)
.
i
1
n
By the corollary of
Neyman-Pearson Lemma, a UMP level
test
rejects
H
0
if and only if
g(t|
3)kg(t|
1)
.
n<
br>3
t
3
1
t
1
t2
e
k
e
3
ke
(
x
1
)ln3
2
ln
k
t!t!
i
1
(b)
(
)
P
(
X
1
X
2
X
3<
br>
5)
1
P
(
X
1
X
2
X
3
4)
(3
)
0
(3
)
1
(3
)
2
(3
)
3
(3
)
4
3<
br>
1
[
]
e
,
0
0!1!2!3!4!
3
0
3
1
3
2
3
3
3
4
3
]e
0.1847
(c) The size of this test
is
(1)
1
[
0!1!2!3!4!<
br>9
0
9
1
9
2
9
3
9
4<
br>
9
]e
0.9450
The power
of this test is
(3)
1
[
0!1!2!3!4!
4. Since
T
X
i
is sufficient for
and
T~Poisson(n
)
.
i
1
n
f
T|
(t|
)
(n
)
n
1
e
;
and
f
(
)
t!
(
)
t
t
1
e
f(t,
)
f
T
(t)
(n
)
n
1
e
t!
(
)
n
t
1
e
n
t
t!
(
)
n
t
t
1
e
1
(n
<
br>)
,
0
0
t!
(
)
t
1
1
(n
)
e
d
t!
(
)<
br>
1
(n
)
(
t
)(
t
)
n
1
1
(n
)
n
t
f(t,
)
t
1
e
t!
(
)
,
0
f
|t
(
|t)
t
f
T
(t)
n
t
(t
)(
)
t
(t
)()
n
1
n
1
t!
(
)
)
. The
posterior distribution of
is
Gamma(t
,
n
1
x
i
x
1
n
1
1
5. (a)
f(x
1
,x
2
,
,x
n
|
)
f(x
i
|
)
(eI
(0,
)
(x
i
)
)
e
I
(0,
)
(x
i<
br>)
n
i
1
i
<
br>1i
1
T(
~
x)
n
1
~
~
Let
g(T(x),
)
e
and
h
(x
)
I
(0,
)
(
x
i
)
. By factorization theorem,
n
i
<
br>1
t
1
e
<
br>
n
n
T
X
i
is a sufficient statistic for
.
i
1
n
1
11
[Or]
f(x|
)
eI
(0,
)
(x)
I
(0,
)
(x)exp[x(
)]
x
{f(x|
)}
is
an exponential family.
T
X
i
is a sufficient
statistic.
i
1
n
ˆ
X
T
is an unbiased estimate of
and a
function of sufficient Since
n
ˆ
X
is the
UMVUE of
. statistics
T
, by Rao-
Blackwell Theorem,
n
(n
)
t
n
e
(b)
T
X<
br>i
~
Poisson
(
n
)
g(t|
)
,
t0,1,2,
t!
i
1
(n
2
)
t
n
2
t
e
g(t|
2<
br>)
t!
2
e
n(
2
1
)
g(t|
1
)
(n
1
)
t
n
1
1
e
t!
g(t|
2
)
If
2
1
2
1
is an increasing
function of
t
,
1
g(t|
1
)
Hence
{g(t|
)|
0}
of
T
has MLR.
(c)
T
X
i
~
Gam
ma
(
n
,
)
g(t|
)
i
1
n
1
(
n)
n
t
n
1
e
,
t0
t
1
g(t|
2
)
g(t|
1
)
n
<
br>(n)
2
t
n
1
e
2
t
n
1
e
1
t
n
(
1
1
)t
1
n
(n)
1
1
2
<
br>1
e
,
t0
t
2
g(t|
2
)
11
If
2
1
(
)
21
0
is
increasing in
t
.
2
1
1
2
g(t|
1
)
Hence
{g(t|
)|
0}
of
T
has an MLR.
By Karlin-Rubin Theorem, the UMP
size
test rejecting
H
0
if
T
X
i
c
, where
c
i
1
n
satisfies that
P
{
X
i
c|
1}
;
i.e.,
i
1
n
c
1
n
1
x
(n)
xedx
.
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1.
Let
X
1
,X
2
,
,X
n
be
a random sample from the
Gamma(
,
)
distribution <
br>f(x|
,
)
1
(
)
x
1
e
,
x0
,
0
,
0
.
x
(a) ( 8 %) Find the method of moment
estimates of
and
.
(b) ( 7
%) Find the MLE of
, assuming
is known.
(c) ( 7 %) Giving
0
,
find the Cramer-Rao lower bound of estimates of
.
(d) ( 8 %) Giving
0
,
find the UMVUE of
.
2. Suppose that
X
1
,X
2
,
,X
n
are
iid ~
B(2,p)
,
p(0,1)
. Let
(p)2p(1p)
.
(a) ( 5 %) Show that
T
X
i
is a sufficient
statistic for
p
.
i
1
n
1, if
X
1
1
(b) ( 5 %) Let
Y
. Show that
Y
is an
unbiased estimate of
(p)
.
0,
if X
1
1
(c) (10%) Find the UMVUE
W
of
(p)
.
3. Let
X
1
, X
2
,
,X
n
be a random sample from a
Poisson(
)
,
0
,
distribution.
Consider testing
H
0
:
1
vs
H
1
:
3
.
(a) (10%) Find
a UMP level
test,
0
1
.
(b) ( 7 %) For
n3
, the test rejects
H
0
, if
X
1
X
2
X
3
5
.
Find the power function
(
)
of the test.
(c) ( 8
%) For
n3
, the test rejects
H
0
, if
X
1
X
2
X
3
5
.
Evaluate the size and the power of the test.
4. (10%) Let
X
1
, X
2
,
,X
n
be iid
Poisson()
distribution, and let the prior
distribution
of
be a
Gamma(
,
)
distribution,
0
,
0
. Find the
posterior distribution of
.
5.
Let
X
1
, X
2
,
,X
n
be a random sample from an
exponential distribution with mean
,
0
.
(a) ( 5 %) Show
that
T
X
i
is a sufficient
statistic n for
.
i
1
n
(b) ( 5 %) Show that the
Poisson family has a monotone likelihood ratio,
MLR.
(c) ( 5 %) Find a UMP level
test of
H
0
:0
1
vs
H
1
:
1
by the
Karlin-Rubin Theorem shown below.
[Definition] A family of pdfs or pmfs
{g(t|
)|
}
has a
monotone likelihood ratio,
g(t|
2
)
MLR, if for every
2
1
,
is a monotone
function of
t
.
g(t|
1
)
[Karlin-Rubin Theorem]
Suppose that
T
is a sufficient statistic
for
and the pdfs or
pmfs
{g(t|
)|
}
has a non-
decreasing monotone likelihood ratio.
Consider
testing
H
0
:
0
vs
H
1
:
0
. A UMP
level
test rejects
H
0
if and
only if
Tt
0
, where
P
0
(Tt
0
)
.
數理統計期末考試試題答案
1. (a) Since
E(X)
0
(
)
1
(
1)
1
xed
x
and
x
<
br>(
)
E(X)
2<
br>
0
(
)
2<
br>m
1
1
x
1
(
<
br>
2)
2
edx
(
1)
2
,
x
(
)
Let
m
1
and
m
2
(
1)
2
~
m
2
2
m
1
1
2
m
1
m
2
m
1
,
~
.
2
m
1
m
2
m
1
1
n
2
1
n
(n
1)
222
Furthermore,
m
1
X
,
m
2
m
1
X
i
X
(X
i2
X)
2
S
,
n
i
1
n
i
1
n
~
(n1)S
2
nX
2
~
The MME of
.and
are
,
2
nX
(n
1)S
~
(b)
L(
|
,
~
x)
[
i
1
n
1
(
)
1
x
i
e]
x
i
1
[
(
)
]
n
i
1
n
1
(
x
i
)e
n
n
i
1
x
i
n
x)
nln
(
)
n
ln
(
1)
lnx
i
i
1
lnL(
|
,
~
i
1
x
1
n
1
n
1
n
x
~
ˆ
lnL(
|
,x)
x
0
x
Let
i
i
.
2
i
1
n
i
1
n
2
n
n
2nx
~<
br>lnL(
|
,x)
lnx
Furthermore,
i
223
23
i
1
n
2nxnx
2nx
n
ˆ
|
,
~
lnL(
x)
0
,
22332
ˆˆ
xx
ˆ
X
is the MLE of
. So,
2
2
n
2
n
n
2n
n
~
lnL(
|
,x)]
E
(
X)
(c)
E
[
223
i
232<
br>
i
1
2
CRLB =
E
[
(d) Since
E<
br>(
X
)
2
2
2
n
~
lnL(
|
<
br>,x)]
1
ˆ
X
is an
unbiased estimate of
, and
,
X1
2<
br>
2
ˆ
X
is the UMVUE of
.
Var()
CRLB,
n
2
n
1
(
)
(
)
Given
,
{f(x|
)}
is an
exponential family in
.
I
(0,
)
(x)x
1
e
x
[Or]
f(x|
,
)
1
1
I
(0,
)
(x)x
1
exp[x(
)]
T
X
i
is a sufficient
statistic for
.
i
1
n
ˆ
X
T
is an unbiased estimate of
and a
function of sufficient Since
n
ˆ
X
is the
UMVUE of
. statistics
T
, by Rao-
Blackwell Theorem,
nn
2
x
i
2
x
i
2. (a)
f(x
1
,x
2
,,x
n
|p)
<
br>f(x
i
|p)
[
I(x)p(1<
br>
p)]
{0,1,2}i
x
i
1i
1
i
2
2
p
x
i
p
i
1
2
[
I
{0,1,2}
(x
i
)()(1
p)]
[
I(x)]()(1
p)
2n
{0,1,2}i
<
br>xx
1
p1
p
i
1
i
i
1
i
n
2
p
T(
~
x)2n
~
)
(1p)
and
h(x)
Let
g
(T(x),p)
(
I
{0,1,2}
(
x
i
)
. By
x
1
p
i
1
i
nn
x
i
n
factorization theorem,
T
X
i
is a sufficient
statistic for
p
.
i
1
n
2
p
x2
x
2
2
[Or]
f(x|p)
I(x)p(1
p)
I(x)(1
p)exp[xln()]
x
{0,1,2}
x
{0,1,
2}
1
p
{f(x|p)}
is an exponential family
T
X
i
is a
sufficient statistic.
i
1
n
2
12
1
(b)
E(Y)
1
P(X
1
1)
0
P(X
1
1)
p(1p)2p(1p)<
br>, so
Y
is an
1
unbiased estimate of
(p)
.
(c) If
X
1
,X
2
,
,X
n
,
nN
, are iid ~
B(2,p)
, then
T
X
i
~
B
(2
n
,
p<
br>)
.
i
1
n
E(Y|T
t)
P(Y
1
&T
t)
P(X
1
1 &T
t)<
br>
P(T
t)P(T
t)
n
P(X<
br>1
1 &
X
i
t
1)
i
2
n
P(T
t)
2n
2
t
12n
t<
br>
1
2p(1
p)
p(1
p)
t
1
i
2
P(T
t)
2n
t2n
t
t
p(1
p)
2(2n
2)!t!(2n
t)!
t(2n
t)
,
t0,1,2,,2n
.
(t
1)!(2n
t
1)!(2n)!n(
2n
1)
T(n
T)
By Rao-
Blackwell Theorem,
W
E(Y|T)
is the UMVUE of
(
)e
.
2(2n
1)
3. (a) By Neyman-Pearson
Lemma, a UMP level
test rejects
H
0
if and only if
f(x
1
,x2
,,x
n
|
3)
kf(
x
1
,x
2
,,x
n
|
1)
.
P(X
1
1)P(
X
i
t
1)
n
n
1
x
i
3<
br>x
i
3
x
i
2n
1
ke
(
x
i
)ln3
2n
lnk
e]
k
[
e]
3
[
i
1
(x
i
)!
i
1
(x
i
)!
i
<
br>1
n
2
lnk
c
x
i
ln3
i
1
n
Since
n
i
1
X
i
~
Poisson
(
n
)
, a UMP
level
test rejects
H
0
if and
only if
n
n
i
n
Xi
c
, where
c
is the smallest
integer satisfying
i!
e
.
i
1
i
c
1
[Or]
T
X
i
is sufficient for
and
T~Poisson(n
)
.
i
1
n
By the corollary of
Neyman-Pearson Lemma, a UMP level
test
rejects
H
0
if and only if
g(t|
3)kg(t|
1)
.
n<
br>3
t
3
1
t
1
t2
e
k
e
3
ke
(
x
1
)ln3
2
ln
k
t!t!
i
1
(b)
(
)
P
(
X
1
X
2
X
3<
br>
5)
1
P
(
X
1
X
2
X
3
4)
(3
)
0
(3
)
1
(3
)
2
(3
)
3
(3
)
4
3<
br>
1
[
]
e
,
0
0!1!2!3!4!
3
0
3
1
3
2
3
3
3
4
3
]e
0.1847
(c) The size of this test
is
(1)
1
[
0!1!2!3!4!<
br>9
0
9
1
9
2
9
3
9
4<
br>
9
]e
0.9450
The power
of this test is
(3)
1
[
0!1!2!3!4!
4. Since
T
X
i
is sufficient for
and
T~Poisson(n
)
.
i
1
n
f
T|
(t|
)
(n
)
n
1
e
;
and
f
(
)
t!
(
)
t
t
1
e
f(t,
)
f
T
(t)
(n
)
n
1
e
t!
(
)
n
t
1
e
n
t
t!
(
)
n
t
t
1
e
1
(n
<
br>)
,
0
0
t!
(
)
t
1
1
(n
)
e
d
t!
(
)<
br>
1
(n
)
(
t
)(
t
)
n
1
1
(n
)
n
t
f(t,
)
t
1
e
t!
(
)
,
0
f
|t
(
|t)
t
f
T
(t)
n
t
(t
)(
)
t
(t
)()
n
1
n
1
t!
(
)
)
. The
posterior distribution of
is
Gamma(t
,
n
1
x
i
x
1
n
1
1
5. (a)
f(x
1
,x
2
,
,x
n
|
)
f(x
i
|
)
(eI
(0,
)
(x
i
)
)
e
I
(0,
)
(x
i<
br>)
n
i
1
i
<
br>1i
1
T(
~
x)
n
1
~
~
Let
g(T(x),
)
e
and
h
(x
)
I
(0,
)
(
x
i
)
. By factorization theorem,
n
i
<
br>1
t
1
e
<
br>
n
n
T
X
i
is a sufficient statistic for
.
i
1
n
1
11
[Or]
f(x|
)
eI
(0,
)
(x)
I
(0,
)
(x)exp[x(
)]
x
{f(x|
)}
is
an exponential family.
T
X
i
is a sufficient
statistic.
i
1
n
ˆ
X
T
is an unbiased estimate of
and a
function of sufficient Since
n
ˆ
X
is the
UMVUE of
. statistics
T
, by Rao-
Blackwell Theorem,
n
(n
)
t
n
e
(b)
T
X<
br>i
~
Poisson
(
n
)
g(t|
)
,
t0,1,2,
t!
i
1
(n
2
)
t
n
2
t
e
g(t|
2<
br>)
t!
2
e
n(
2
1
)
g(t|
1
)
(n
1
)
t
n
1
1
e
t!
g(t|
2
)
If
2
1
2
1
is an increasing
function of
t
,
1
g(t|
1
)
Hence
{g(t|
)|
0}
of
T
has MLR.
(c)
T
X
i
~
Gam
ma
(
n
,
)
g(t|
)
i
1
n
1
(
n)
n
t
n
1
e
,
t0
t
1
g(t|
2
)
g(t|
1
)
n
<
br>(n)
2
t
n
1
e
2
t
n
1
e
1
t
n
(
1
1
)t
1
n
(n)
1
1
2
<
br>1
e
,
t0
t
2
g(t|
2
)
11
If
2
1
(
)
21
0
is
increasing in
t
.
2
1
1
2
g(t|
1
)
Hence
{g(t|
)|
0}
of
T
has an MLR.
By Karlin-Rubin Theorem, the UMP
size
test rejecting
H
0
if
T
X
i
c
, where
c
i
1
n
satisfies that
P
{
X
i
c|
1}
;
i.e.,
i
1
n
c
1
n
1
x
(n)
xedx
.
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