Page 63 - Elements of Distribution Theory

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

2.3 Conditional Distributions 49

Let q W (A, w) = Pr(X ∈ A|W = w) and q Z (A, z) = Pr(X ∈ A|Z = z). Then Pr(X ∈
A|(X, Y) ∈ B)isgiven by both q W (A, w 0 ) and q Z (A, z 0 ); however, there is no guaran-
tee that q W (A, w 0 ) = q Z (A, z 0 )so that these two approaches could yield different results.
This possibility is illustrated in the following example.

Example 2.10. Let X and Y denote independent random variables such that
1
Pr(X = 1) = Pr(X = c) = ,
2
for some constant c > 1, and Y has an absolutely continuous distribution with density
1
p(y) = , −1 < y < 1.
2
Let Z = XY. Note that the events Z = 0 and Y = 0 are identical; that is, Z = 0if and
only if Y = 0. However, it will be shown that
Pr(X = 1|Z = 0) = Pr(X = 1|Y = 0).

Using (2.3), for z ∈ R,
1 1
Pr(Z ≤ z) = Pr(Z ≤ z|X = 1) + Pr(Z ≤ z|X = c).
2 2
Since the events X = 1 and X = c both have nonzero probability, we know from elementary
probability theory that, for x = 1, c,

Pr(Z ≤ z|X = x) = 2Pr(XY ≤ z ∩ X = x) = 2Pr(Y ≤ z/x ∩ X = x)
= 2Pr(Y ≤ z/x)Pr(X = x) = Pr(Y ≤ z/x).
It follows that, for z > −c,
1 z/c 1 z
Pr(Z ≤ z) = dy + dy
4 −1 4 −1
so that Z has an absolutely continuous distribution with density

1 1
p Z (z) = I {|z|<c} + I {|z|<1} .
4c 4
Define

0 if |z|≥ 1
h(z) = . (2.4)
c/(c + 1) if |z| < 1
It will be shown that h(z) = Pr(X = 1|Z = z). To do this, first note that, for B ⊂ R,

1
Pr(X = 1 ∩ Z ∈ B) = Pr(X = 1 ∩ Y ∈ B) = dz.
4 B∩(−1,1)
Using (2.4),
c 1

h(z) dF Z (z) = I {|z|<1} p Z (z) dz = dz
B B c + 1 4 B∩(−1,1)
so that (2.3) is satisfied and, hence, Pr(X = 1|Z = z) = h(z)as claimed.