STAT 410
Examples for 07/22/2009
Spring 2009
1.
Let
X
1
, X
2
, … , X
n
be a random sample of size
n
from a Poisson distribution
with mean
λ
.
That is,
f
(
k
;
λ
) = P
(
X
1
=
k
) =
!
λ
λ
k
e
k

⋅
,
k
= 0, 1, 2, 3, … .
a)
Use Factorization Theorem to find
Y =
u
(
X
1
,
X
2
,
…
,
X
n
),
a sufficient statistic
for
λ
.
f
(
x
1
;
λ
)
f
(
x
2
;
λ
)
…
f
(
x
n
;
λ
)
=
∏
=

⋅
n
i
i
x
x
e
i
1
!
λ
λ
=
∏
∑
=

⋅
⋅
=
n
i
i
n
x
x
e
n
i
i
1
!
λ
1
1
λ
.
By Factorization Theorem,
Y =
∑
=
n
i
i
1
X
is a sufficient statistic for
λ
.
[
⇒
X
is also a sufficient statistic for
λ
.
]
b)
Show that
P
(
X
1
=
x
1
, X
2
=
x
2
, … , X
n
=
x
n

Y =
y
)
does not depend on
λ
.
Since
Y =
∑
=
n
i
i
1
X
has a Poisson distribution with mean
n
λ
,
if
∑
=
n
i
i
x
1
=
y
,
P
(
X
1
=
x
1
, X
2
=
x
2
, … , X
n
=
x
n

Y =
y
)
=
=
( 29
!
λ
!
λ
!
λ
!
λ
2
1
λ
λ
...
λ
λ
2
1
y
n
x
x
x
n
y
n
x
x
x
e
e
e
e
n




⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
y
n
n
x
x
x
y
1
...
!
!
!
!
2
1
⋅
does not depend on
λ
.
[
P
(
X
1
=
x
1
, X
1
=
x
1
, … , X
n
=
x
n

Y =
y
)
= 0
if
∑
=
n
i
i
x
1
≠
y
.
]
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Let
X
1
, X
2
, … , X
n
be a random sample of size
n
from an Exponential
distribution with probability with mean
θ
.
( 29
∑

=
=
⋅
=
∏
n
i
i
n
n
i
i
x
x
f
1
1
1
1
θ
exp
θ
θ
;
.
By Factorization Theorem,
∑
=
n
i
i
1
X
is a sufficient statistic for
θ
.
OR
f
(
x
;
θ
)
=
θ
1
θ
x
e

=
exp
[
–
θ
1
⋅
x
–
ln
θ
].
K
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 Summer '08
 AlexeiStepanov
 Poisson Distribution, Probability theory, θ, Sufficient statistic, joint sufficient statistics

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