Page 68 - Elements of Distribution Theory

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54 Conditional Distributions and Expectation

It is sometimes convenient to define E[g(X)|Y = y] directly, without reference to
F X|Y (·|y). First consider the case in which g(x) = I {x∈A} for some set A ⊂ X. Then
E[g(X)|Y = y] = Pr[X ∈ A|Y = y]so that E[g(X)|Y = y] satisfies the equation

E[g(X)I {Y∈B} ] = E[g(X)|Y = y] dF Y (y)
B
for all B ⊂ Y.
This definition can be extended to an arbitrary function g. Suppose that g: X → R
satisfies E[|g(X)|] < ∞.We define E[g(X)|Y = y]tobeany function of y satisfying

E[g(X)I {Y∈B} ] = E[g(X)|Y = y] dF Y (y) (2.5)
B
for all sets B ⊂ Y. The issues regarding existence and uniqueness are essentially the same as
they are for conditional probabilities. If E[|g(X)|] < ∞, then the Radon-Nikodym Theorem
guarantees existence of the conditional expected value. Conditional expected values are not
unique, but any two versions of E[g(X)|Y = y] differ only for y in a set of probability 0.
Let F X|Y (·|y) denote the conditional distribution function of X given Y = y and consider

h(y) = g(x) dF X|Y (x|y).
X
Then

h(y) dF Y (y) = g(x) dF X|Y (x|y) dF Y (y) = I {y∈B} g(x) dF X,Y (x, y)
B B X X×Y
= E[g(X)I {Y∈B} ].

Hence, one choice for E[g(X)|Y = y]isgiven

E[g(X)|Y = y] = g(x) dF X|Y (x|y);

that is, the two approaches to defining E[g(X)|Y = y] considered here are in agreement.
Generally speaking, the expression based on F X|Y (·|y)is more convenient for computing
conditional expected values for a given distribution, while the definition based on (2.5) is
more convenient for establishing general properties of conditional expected values.

Example 2.14 (Bivariate distribution). Let (X, Y) denote a two-dimensional random vec-
tor with the distribution described in Example 2.1. This distribution is absolutely continuous
with density function
p(x, y) = 6(1 − x − y), x > 0, y > 0, x + y < 1;

itwasshowninExample2.11thattheconditionaldistributionof X given Y = y isabsolutely
continuous with density function

1 − x − y
p X|Y (x|y) = 2 , 0 < x < 1 − y
(1 − y) 2
where 0 < y < 1. It follows that

2 1−y 1
E[X|Y = y] = x(1 − x − y) dx = (1 − y), 0 < y < 1.
(1 − y) 2 0 3