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052184472Xc02 CUNY148/Severini May 24, 2005 2:29

2.4 Conditional Expectation 57

It is straightforward to show that

E(X|(Y 1 , Y 2 )) = Y 1 + Y 2
and, since each of Y 1 , Y 2 has expected value 1/2, it follows from part (i) of Theorem 2.5
that E(X) = 1.
Now consider E(X|Y 1 ). Since E(X|(Y 1 , Y 2 )) = Y 1 + Y 2 ,it follows from part (iv) of
Theorem 2.5 that
E(X|Y 1 ) = E(Y 1 + Y 2 |Y 1 ) = Y 1 + E(Y 2 ) = Y 1 + 1/2.

The following characterization of conditional expected values is often useful.

Theorem 2.6. Let (X, Y) denote a random vector with range X × Y and let g denote a
real-valued function on X.
(i) Suppose that E[|g(X)|] < ∞ and let Z denote a real-valued function of Y such that
E(|Z|) < ∞.If
Z = E[g(X)|Y] with probability 1

then
E[Zh(Y)] = E[g(X)h(Y)] (2.7)

for all functions h : Y → R such that
E[|h(Y)|] < ∞ and E[|g(X)h(Y)|] < ∞.

(ii) If Z is a function of Y such that (2.7) holds for all bounded functions h : Y → R
then

Z = E[g(X)|Y] with probability 1.

Proof. Suppose Z = E[g(X)|Y] with probability 1. Let h be a real-valued function on Y
such that E[|h(Y)|] < ∞ and E[|g(X)h(Y)|] < ∞. Then, since
{Z − E[g(X)|Y]}h(Y) = 0 with probability 1,

E{(Z − E[g(X)|Y])h(Y)}= 0.

It follows from Theorem 2.5 that

E{E[g(X)|Y]h(Y)}= E[g(X)h(Y)] < ∞,

so that
E[Zh(Y)] = E{E[g(X)|Y]h(Y)}= E[g(X)h(Y)],

proving the first part of the theorem.
Now suppose that (2.7) holds for all bounded functions h :Y → R. Let B ⊂ Y. Since
h(y) = I {y∈B} is a bounded, real-valued function on Y,it follows that (2.6) holds for any B.
Part (ii) follows.