Page 153 - Probability and Statistical Inference

130 3. Multivariate Random Variables

Example 3.5.7 Suppose that the joint pdf of X , X is given by
1 2

for some θ > 0 where k(> 0) is a constant. Obviously, we can write f(x , x ) =
2
1
kh (x )h (x ) for all positive numbers x , x where h (x ) = exp{–θx } and h (x )
1
1
2
1
1
2
1
1
2
2
2
+
= exp{–1/θx }. Also the supports χ = ℜ , χ = ℜ are unrelated to each other.
+
1
2
2
Hence, by appealing to the Theorem 3.5.3 we can claim that X and X are
2
1
independent random variables. One should note two interesting facts here. The
two chosen functions h (x ) and h (x ) are not even probability densities to
2
2
1
1
begin with. The other point is that the choices of these two functions are not
–1
unique. One can easily replace these by kdh (x ) and kd h (x ) respectively
2
2
1
1
where d(> 0) is any arbitrary number. Note that we did not have to determine k
in order to reach our conclusion that X and X are independent random vari-
1 2
ables. !
Example 3.5.8 (Example 3.5.7 Continued) One can check that f (x ) = θ
1
1
–1
exp{–θx } and f (x ) = θ exp{–1/θ . x }, and then one may apply (3.5.1) in-
2
1
2
2
stead to arrive at the same conclusion. In view of the Theorem 3.5.3 it becomes
really simple to write down the marginal pdf’s in the case of independence. !
Example 3.5.9 Suppose that the joint pdf of X , X , X is given by
1 2 3
where c(> 0) is a constant. One has f(x , x , x ) = ch (x )h (x )h (x ) with
3
1
1
1
3
3
2
2
2
exp{–2x }, and h (x ) = exp
3
1
3
, for all values of x ∈ ℜ , (x , x ) Î ℜ . Also, the supports χ ∈ ℜ , χ = ℜ, χ 3
2
+
+
1
3
1
2
2
= ℜ are unrelated to each other. Hence, by appealing to the Theorem 3.5.3 we
can claim that X , X and X are independent random variables. One should
3
1
2
again note the interesting facts. The three chosen functions h (x ), h (x ) and
1
2
1
2
h (x ) are not even probability densities to begin with. The other point is that the
3 3
choices of these three functions are not unique. !
Example 3.5.10 (Example 3.5.9 Continued) One should check these out:
By looking at h (x ) one can immediately guess that X has the Gamma(α =
1
1
1
1/2, β = 1/2) distribution so that its normalizing constant . By
looking at h (x ) one can immediately guess that X has the Cauchy dis-
2
2
2
tribution so that its normalizing constant 1/π. Similarly, by looking at
h (x ) one can immediately guess that X has the N(0, 1/2π) distribution
3
3
3
so that its normalizing constant is unity. Hence, (1/π)(1)=