Page 62 - Elements of Distribution Theory

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

48 Conditional Distributions and Expectation

The conclusion of this lemma can be stated as follows: for any A ⊂ X, q 1 (A, y) =
q 2 (A, y) for almost all y (F Y ). As in the statement of the lemma, this means that the set
of y for which q 1 (A, y) = q 2 (A, y) does not hold has probability 0 under the distribution
given by F Y . Alternatively, we may write that q 1 (A, ·) = q 2 (A, ·) almost everywhere (F Y ),
or, more simply, q 1 (A, ·) = q 2 (A, ·) a.e. (F Y ).

Proof of Lemma 2.2. Fix a set A ⊂ X.For n = 1, 2,..., define
Y n ={y ∈ Y: |q 1 (A, y) − q 2 (A, y)|≥ 1/n}.

Note that Y 1 ⊂ Y 2 ··· and
∞

Y n = Y 0 .
n=1
For each n = 1, 2,..., let

B n ={y ∈ Y: q 1 (A, y) − q 2 (A, y) ≥ 1/n}
and

C n ={y ∈ Y: q 2 (A, y) − q 1 (A, y) ≥ 1/n}
so that Y n = B n ∪ C n , n = 1, 2,....
Fix n. Since both q 1 and q 2 satisfy (2.3),
1 1

0 = q 1 (A, y) dF Y (y) − q 2 (A, y) dF Y (y) ≥ dF Y (y) = Pr(Y ∈ B n )
n n
B n B n B n
so that Pr(Y ∈ B n ) = 0. Similarly, Pr(Y ∈ C n ) = 0. Hence, for each n = 1, 2,...,
Pr(Y ∈ Y n ) = 0.
By condition (P4) on probability distributions, given in Chapter 1, together with the fact
that Y 0 = Y 1 ∪ Y 2 ∪· · · ,
Pr(Y ∈ Y 0 ) = lim Pr(Y ∈ Y n ) = 0,
n→∞
proving the result.

Here we will refer to the conditional probability, with the understanding that there may
be another version of the conditional probability that is equal to the first for y in a set of
probability 1.
It is important to note that, when Y has an absolutely continuous distribution, conditional
probabilities of the form Pr(X ∈ A|Y = y 0 ) for a specific value y 0 ∈ Y are not well defined,
except as a function q(A, y) satisfying (2.3) evaluated at y = y 0 , and this fact can sometimes
cause difficulties.
For instance, suppose we wish to determine Pr(X ∈ A|(X, Y) ∈ B), for some sets A
and B, A ⊂ X and B ⊂ X × Y, where Pr((X, Y) ∈ B) = 0. Suppose further that the event
(X, Y) ∈ B can be described in terms of two different random variables W and Z, each of
which is a function of (X, Y); that is, suppose there exist functions W and Z and values w 0
and z 0 in the ranges of W and Z, respectively, such that

{(x, y) ∈ X × Y: W(x, y) = w 0 }={(x, y) ∈ X × Y: Z(x, y) = z 0 }= B.