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3. Multivariate Random Variables 135

(ii) Let us denote and a = {2π(1–
. Utilizing (3.3.5), (3.6.4)-(3.6.5) and (3.6.11), the conditional pdf
of X | X = x is given by
1 2 2

The expression of f (x ) found in the last step in (3.6.12) resembles the
1
1|2
pdf of a normal random variable with mean and
variance
(iii) Its proof is left as the Exercise 3.6.2. ¢
Example 3.6.1 Suppose that (X , X ) is distributed as N (0,0,4,1,ρ) where
2
1
2
ρ = 1/2. From the Theorem 3.6.1 (i) we already know that E[X ] = 0 and
2
V[X ] = 1. Utilizing part (iii), we can also say that E[X | X = x ] = 1/4x and
2
1
1
1
2
V[X | X = x ] = 3/4. Now, one may apply the Theorem 3.3.1 to find indi-
2
1
1
rectly the expressions of E[X ] and V[X ]. We should have E[X ] = E{E[X |
2
2
2
2
X = x ]} = E[1/4 X ] = 0, whereas V[X ] = V{E[X | X = x ]} + E{V[X | X 1
1
2
2
1
1
1
1
2
= x ]} = V[1/4X ] + E[3/4] = 1/16V[X ] + 3/4 = 1/16(4) + 3/4 = 1. Note that
1
1
1
in this example, we did not fully exploit the form of the conditional distribu-
tion of X | X = x . We merely used the expressions of the conditional mean
2
1
1
and variance of X | X = x . !
2 1 1
In the Example 3.6.2, we exploit more fully the form
of the conditional distributions.
Example 3.6.2 Suppose that (X , X ) is distributed as N (0,0,1,1,ρ) where
2
2
1
ρ = 1/2. We wish to evaluate E{exp[1/2X X ]}. Using the Theorem 3.3.1 (i),
2
1
we can write
But, from the Theorem 3.6.1 (iii) we already know that the conditional distri-
bution of X | X = x is normal with mean 1/2x and variance 3/4. Now,
1
1
1
2
E(exp1/2x X ] | X = x ) can be viewed as the conditional mgf of X | X = x .
1
1
1
1
1
2
2
Thus, using the form of the mgf of a univariate normal random variable from
(2.3.16), we obtain