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STAT 710 2007 Third Exam Suggested Solution
1 Problem 1
(a) Suppose θ 2 > θ 1 > 0.
fθ 2 (x)
fθ 1 (x)
θ 2 x−(θ^2 +1)
θ 1 x−(θ^1 +1)^
θ 2
θ 1
x
)θ 2 −θ 1
Since
fθ 2 (x) fθ 1 (x) is increasing in^
1 x ,^ the family has monotone likelihood ratio in^ Y^ =^
1 X.
(b) From (a), a UMP test of size α is of the form
T∗ (X) =
1 , if 1 X
d; 0 , if 1 X ≤ d.
I.e.,
T∗ (X) =
1 , if X < c; 0 , if X ≥ c.
where c > 1 is a constant satisfying Eθ 0 [T∗ (X)] = α.
The cdf of the distribution is
Fθ (x) =
1 − x −θ , if x ≥ 1; 0 , if x < 1.
Hence Eθ 0 [T∗ (X)] = Pθ 0 [X < c] = Fθ 0 (c) = 1 − c −θ 0 .
Now 1 − c −θ 0 = α implies that
c = [1 − α]
− (^) θ^1 (^0).
(c)
βT∗ (θ) = Eθ [T∗ (X)] = Pθ [X < c]
= Pθ
[
X < [1 − α]
− (^) θ^1 0
]
[1 − α]
− (^) θ^1 0
)−θ
= 1 − [1 − α]
θ θ (^0).
(d)
log fθ (x) = log θ − (θ + 1) log x
d log fθ (x)
dx
θ
− log x
Hence ̂θ = 1 log X
Under H 0 , the MLE is ˜θ = min
θ 0 , θ̂
. Hence
λ (X) =
l
θ
l
˜θ
θX˜ −eθ
θX̂ −bθ^
1 , if θb θ 0 <^ 1; ( θb θ 0
exp
θb θ 0
, if θb θ 0 ≥^1.
Observe that h (t) = t − 1 exp
−t − 1
is decreasing in t for t ≥ 1. Therefore λ (X) is a decreasing function
of θb θ 0.
To see that
eθX−^ θe bθX− θb =
θb θ 0
exp
bθ θ 0
when θb θ 0 ≥^1 ,^ one just observe that^ X^
bθ = e, hence X = e
1 θ^ b (^).
Hence the likelihood ratio test rejects H 0 if
bθ θ 0 > c,^ the same as the UMP test given in (b). (e)
βT (θ) = Eθ [T (X)] = Pθ
[
c 1 <
X
< c 2
]
= Pθ
[
c 2
< X <
c 1
]
[
c 1
)−θ ]
−
[
c 2
)−θ ]
= c θ 2 −^ c
θ 1
Hence we have
c 2 − c 1 = α
c 2 2 −^ c
2 1 =^ α
We get c 1 = 1 −α 2 , c 2 = 1+α 2
2 Problem 2
(a) Let X• 1 = 1 n
∑n
i=
Xi 1 , X• 2 = 1 n
∑n
i=
Xi 2.
Let’s first derive MLE’s.
fθ (x) = (1 − θ 1 )
nX• 1 −n θ n 1 (1^ −^ θ^2 )
nX• 2 −n θ n 2 h^ (x) log fθ (x) =
nX• 1 − n
log (1 − θ 1 ) + n log θ 1 +
nX• 2 − n
log (1 − θ 2 ) + n log θ 2 + log h (x)
∂ log fθ (x)
∂θ 1
nX• 1 − n
1 − θ 1
n
θ 1
∂ log fθ (x)
∂θ 2
nX• 2 − n
1 − θ 2
n
θ 2
We have that the MLE θ̂ 1 = 1 X• 1
, θ̂ 2 = 1 X• 2
Under H 0 , θ 1 = θ 2 = θ,
fθ (x) = (1 − θ)
nX• 1 −n θ n (1 − θ)
nX• 2 −n θ n h (x) = (1 − θ)
nX• 1 +nX• 2 − 2 n θ 2 n h (x)
log fθ (x) =
nX• 1 + nX• 2 − 2 n
log (1 − θ) + 2n log θ + log h (x)
∂ log fθ (x)
∂θ
nX• 1 + nX• 2 − 2 n
1 − θ
2 n
θ
˜θ =
X• 1 + X• 2
I.e., under H 0 , ˜θ 1 = θ˜ 2 = 2 X• 1 +X• 2
Hence the likelihood ratio is
λn = λ (X) =
fe θ (x)
fb θ (x)
1 X• 1
)nX• 1 −n ( 1 X• 1
)n ( 1 − 1 X• 2
)nX• 2 −n ( 1 X• 2
)n
2 X• 1 +X• 2
)nX• 1 +nX• 2 − 2 n ( 2 X• 1 +X• 2
) 2 n
Rao’s score test is
Rn =
[
sn
˜θ
)]′ [
In
θ˜
)]− 1
sn
˜θ
n 2
[(
eθX• 1 − 1 eθ(θe− (^1) )
eθX• 2 − 1 θ^ e(eθ− (^1) )
) 2 ]
“ n 2 X• 1 +X• 2
” 2 “ 1 − (^) X^2
”
= n
θ˜X
˜θ
θ˜ − 1
2
θ˜X
˜θ
θ˜ − 1
2
θ˜
2
1 − θ˜
= n
˜θX
1 − θ˜
˜θX
1 − ˜θ
2 n
X• 1 −X• 2 X• 1 +X• 2
2 X• 1 +X• 2
Using Rao’s test statistic, we reject if H 0 if Rn > X 2 1 ,α.
