Page 70 - Elements of Distribution Theory

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

56 Conditional Distributions and Expectation

Proof. Part (i) follows immediately from Equation (2.6) with B taken to be Y.
Let f (y) = E[g(X)|Y = y]; then, for almost all y (F Y ),

f (y) = g(x) dF X|Y (x|y).
X
Hence, for any B ⊂ Y,

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

,
= I {y∈B} g(x)h(y) dF X,Y (x, y)
X×Y
= E[I {Y∈B} g(X)h(Y)]
so that (2.6) is satisfied; part (ii) follows.
By (2.6), for any B ⊂ Y,
E{E[g(X)|Y, T (Y)]I {Y∈BT (Y)∈T } }= E[g(X)I {Y∈B, T (Y)∈T } ] = E[g(X)I {Y∈B} ];

part (iii) now follows from (2.6).
Let ¯ g(Y) = E[g(X)|Y]. Then, by (2.6),

E{E[¯ g(Y)|h(Y)]I {h(Y)∈B 0 } }= E[¯ g(Y)I {h(Y)∈A} ]
for any subset A of the range of h(y). Let B ⊂ Y denote a set satisfying
with probability 1.
I {h(Y)∈A} = I {Y∈B}
Then, by (2.6),
E[¯ g(Y)I {h(Y)∈A} ] = E{E[g(X)|Y]I {Y∈B} }= E[g(X)I {Y∈B} ] = E[g(X)I {h(Y)∈A} ].

That is, for all subsets A of the range of h(·),
E{E[¯ g(Y)|h(Y)]I {h(Y)∈A} }= E[g(X)I {h(Y)∈A} ].

It now follows from (2.6) that
E[¯ g(Y)|h(Y)] = E[g(X)|h(Y)],

proving part (iv).
Note that E[g(X)|h(Y)] is a function of Y such that

E{|E[g(X)|h(Y)]|} ≤ E{E[|g(X)||h(Y)]}= E[|g(X)|] < ∞;
part (v) now follows from part (iii).

Example 2.16. Let Y = (Y 1 , Y 2 ) where Y 1 , Y 2 are independent, real-valued random
variables, each with a uniform distribution on (0, 1). Let X denote a real-valued random
variable such that the conditional distribution of X given Y = y has an absolutely continuous
distribution with density
1
p X|Y (x|y) = exp{−x/(y 1 + y 2 )}, x > 0
y 1 + y 2
where y = (y 1 , y 2 ) and y 1 > 0, y 2 > 0.