Page 219 - Probability and Statistical Inference

196 4. Functions of Random Variables and Sampling Distribution

express x = y y and x = y (1 – y ). It is easy to verify that
1 1 2 2 1 2

so that det(J) = –y y – y (1 – y ) = –y . Now, writing the constant c instead of
1
1
1 2
2
the expression {β α +α 2 Γ(α )Γ(α )} , the joint pdf of X and X can be written
–1
2
1
2
1
1
as
for 0 < x , x < ∞. Hence using (4.4.4) we can rewrite the joint pdf of Y and
1
1
2
Y as
2
for 0 < y < ∞, 0 < y < 1. The terms involving y and y in (4.4.5) factorize
2
1
2
1
and also either variable’s domain does not involve the other variable. Refer
back to the Theorem 3.5.3 as needed. It follows that Y and Y are indepen-
1
2
dent random variables, Y is distributed as Gamma(α + α ,β) and Y is dis-
2
1
1
2
tributed as Beta(α , α ) since we can rewrite c as {β α +α 2 Γ(α + α )} {b(α ,
–1
1
1
1
2
2
1
α )} where b(α , α ) stands for the beta function, that is
–1
2 1 2
One may refer to (1.6.19) and (1.6.25)-(1.6.26) to review the gamma and
beta functions. !
Example 4.4.6 (Example 4.4.5 Continued) Suppose that X , ..., X are
1
n
independent random variables where X is Gamma(α , β), i = 1, ..., n. In view
i
i
of the Example 4.4.5, we can conclude that X + X + X = (X +X )+X is then
3
1
1
2
2
3
Gamma(α +α +α , β) whereas X /(X +X +X ) is Beta(α , α +α ) and these are
2
3
2
3
3
1
1
1
2
1
independent random variables. Thus, by the mathematical induction one can claim
that is Gamma is Beta
whereas these are also distributed independently. !
In the next two examples one has X and X dependent.
1
2
Also, the transformed variables Y and Y are dependent.
1 2
Example 4.4.7 Suppose that X and X have their joint pdf given by
1 2