300 Some Elementary Statistical InferencesAssume then thatXhas a Γ(α,1) distribution, whereαis not an integer. As-
sume first thatα>1. LetYhave a Γ([α], 1 /b) distribution, whereb<1ischosen
later and, as usual, [α] means the greatest integer less than or equal toα.Toes-
tablish rule (4.8.9), consider the ratio, withh(x)andt(x) proportional to the pdfs
ofxandy, respectively, given by
h(x)
t(x)
=b−[α]xα−[α]e−(1−b)x, (4.8.11)where we have ignored some of the normalizing constants. We next determine the
constantb.
As Exercise 4.8.14 shows, the derivative of expression (4.8.11) is
d
dxb−[α]xα−[α]e−(1−b)x=b−[α]e−(1−b)x[(α−[α])−x(1−b)]xα−[α]−^1 , (4.8.12)which has a maximum critical value atx=(α−[α])/(1−b). Hence, using the
maximum ofh(x)/t(x),
h(x)
t(x)≤b−[α][
α−[α]
(1−b)e]α−[α]. (4.8.13)
Now, we need to find our choice ofb. Differentiating the right side of this inequality
with respect tob, we get, as Exercise 4.8.15 shows,
d
dbb−[α](1−b)[α]−α=−b−[α](1−b)[α]−α[
[α]−αb
b(1−b)]
, (4.8.14)which has a critical value atb=[α]/α <1. As shown in that exercise, this value
ofbprovides a minimum of the right side of expression (4.8.13). Thus, if we take
b=[α]/α <1, then equality (4.8.13) holds and it is the tightest inequality possible
and, hence, provides the highest acceptance rate. The final value ofMis the right
side of expression (4.8.13) evaluated atb=[α]/α <1.
What if 0<α<1? Then the above argument does not work. In this case
writeX=YU^1 /αwhereY has a Γ(α+1,1)-distribution,Uhas a uniform (0,1)-
distribution, andY andUare independent. Then, as the derivation in Exercise
4.8.16 shows,Xhas a Γ(α,1)-distribution and we are finished.
For further discussion, see Kennedy and Gentle (1980) and Robert and Casella
(1999).
EXERCISES4.8.1.Prove the converse of Theorem MCT. That is, letXbe a random variable
with a continuous cdfF(x). Assume thatF(x) is strictly increasing on the space
ofX. Consider the random variableZ =F(X). Show thatZhas a uniform
distribution on the interval (0,1).