STAT 611
Solutions to Selected Problems from Assignment #9
8.5 c. Under H0 , T has the same distribution as
"
log
n
Y
#
Yi / min Yi ,
i
i=1
where Y1 , . . . , Yn are iid from density f (y) = y −2 I[1,∞) (y). Letting Y(1) < Y(2) < · · · < Y(n)
£Qn
¤
denote the ordered Yi ’s, T can be expressed as log i=2 (Y(i) /Y(1) ) . The probability
distribution of T can thus be found from the joint distribution of the ordered Yi ’s. The
Qn
joint density of Y(1) , . . . , Y(n) is n! i=1 f (ui ), where 1 < u1 < · · · < un . Define the random
variables Z1 , . . . , Zn by Z1 = Y(1) and Zi = Y(i) /Y(1) , i = 2, . . . , n. Using the standard
change of variables technique, the joint density of Z1 , . . . , Zn is
fZ1 ,...,Zn (z1 , . . . , zn ) =
−(n+1)
n!z1
n
Y
zi−2 ,
for z1 > 1, 1 < z2 < · · · < zn ,
i=2
and 0 otherwise. Integrate out z1 to get the joint density of Z2 , . . . , Zn . Upon doing so,
we see that Z2 , . . . , Zn have the same density as the order statistics in a random sample of
size n − 1 from the density f defined above. Together with previous conclusions, the last
statement implies that T has the same probability distribution as
S=
n−1
X
log W(i) =
i=1
n−1
X
log Wi ,
i=1
where W(1) , . . . , W(n−1) are the order statistics of a random sample W1 , . . . , Wn−1 from
density f . Finally, it is easy to verify (using independence of the Wi ’s) that S has a χ2
distribution with 2(n − 1) degrees of freedom.
8.33 a. It is easy to show that k = 1 − α1/n .
b. Clearly, if θ ≥ 1 − α1/n , then the power is equal to 1. So, let θ < 1 − α1/n , and consider
Power = Pθ (Y1 ≥ k ∪ Yn ≥ 1)
= Pθ (Y1 ≥ k) + Pθ (Yn ≥ 1) − Pθ (Y1 ≥ k ∩ Yn ≥ 1).
Now,
Pθ (Y1 ≥ k) = Pθ (all Xi ’s ≥ k)
= (θ + 1 − 1 + α1/n )n
= (θ + α1/n )n ,
1
and
Pθ (Yn ≥ 1) = Pθ (at least one Xi ≥ 1)
= 1 − Pθ (all Xi ’s < 1)
= 1 − (1 − θ)n .
Using the expression for the joint density of Y1 and Yn on p. 230 of Casella and Berger,
Z
θ+1
Z
θ+1
Pθ (Y1 ≥ k ∩ Yn ≥ 1) =
Z
k
1
1
θ+1
Z
=
k
n(n − 1)(yn − y1 )n−2 dyn dy1
1
Z
n(n − 1)(yn − y1 )n−2 I(y1 ,θ+1) (yn ) dyn dy1
θ+1
Z
θ+1
+
1
n(n − 1)(yn − y1 )n−2 dyn dy1
y1
= (θ + α1/n )n − α.
It follows that
½
Power(θ) =
1 + α − (1 − θ)n , θ < 1 − α1/n ,
1,
θ ≥ 1 − α1/n .
c. When θ ≥ 1, the power of the test in question is 1, and hence must be UMP for θ ≥ 1.
Now let 0 < θ1 < 1 and consider testing the hypotheses
H0 : θ = 0
vs. H1 : θ = θ1 .
Show that the rejection region of the Neyman-Pearson test of these hypotheses has the
form
Y1 ≥ k or Yn ≥ 1.
Since k is chosen to make the test have size α, k has nothing to do with θ1 . It follows from
Theorem 8 of the class notes that the test is UMP for all θ.
d. The power of the test is 1 for every θ ≥ 1 − α1/n . Since 1 − α1/n ≤ 1, it is true by
default that the power is always greater than 0.8 when θ > 1.
2