Page 65 - Elements of Distribution Theory

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2.3 Conditional Distributions 51

Theorem 2.3. Consider a random vector (X, Y) where either or both of X and Y may be
vectors.
(i) Suppose that (X, Y) has an absolutely continuous distribution with density p and
let p Y denote the marginal density function of Y. Then the conditional distribution
of X given Y = yis absolutely continuous with density p X|Y given by

p(x, y)
p X|Y (x|y) =
p Y (y)
provided that p Y (y) > 0;if p Y (y) = 0,p X|Y (x|y) may be taken to be any finite
value.
(ii) Suppose that (X, Y) has a discrete distribution with frequency function p and let p Y
denote the marginal frequency function of Y. Then the conditional distribution of X
given Y = yis discrete with frequency function p X|Y (x|y), given by
p(x, y)
p X|Y (x|y) =
p Y (y)
provided that p Y (y) > 0;if p Y (y) = 0,p X|Y (x|y) may be taken to have any finite
value.

Proof. Consider case (i) in which the distribution of (X, Y)is absolutely continuous. Then
Pr(X ∈ A|Y = y) must satisfy

Pr(X ∈ A, Y ∈ B) = Pr(X ∈ A|Y = y)p Y (y) dy.
B
For y satisfying p Y (y) > 0, let

p(x, y)

q(A, y) = dx.
A p Y (y)
Then

q(A, y)p Y (y) dy = p(x, y) dx dy = Pr(X ∈ A, Y ∈ B)
B B A
so that
p(x, y)

Pr(X ∈ A|Y = y) = dx.
A p Y (y)
Note that the value of q(A, y) for those y satisfying p Y (y) = 0is irrelevant. Clearly this
distribution is absolutely continuous with density p X|Y as given in the theorem.
The result for the discrete case follows along similar lines.

Example 2.11 (Bivariate distribution). Consider the distribution considered in Exam-
ple 2.1. The random vector (X, Y) has an absolutely continuous distribution with density
function

p(x, y) = 6(1 − x − y), x > 0, y > 0, x + y < 1
and the marginal density of Y is

2
p Y (y) = 3(1 − y) , 0 < y < 1.