Page 67 - Elements of Distribution Theory

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

2.4 Conditional Expectation 53

Since
x
Pr(X ≤ x|Y = y) = y exp(−yx) dx,
0
it follows that, for y = 1, 2, the conditional distribution of X given Y = y is absolutely
continuous with density function y exp(−yx), x > 0.
It is worth noting that this same result may be obtained by the following informal method.
Since, for A ⊂ Y and y = 1, 2,
1
Pr(X ∈ A, Y = y) = y exp(−yx) dx,
2 A
the function
1
y exp(−yx)
2
plays the role of a density function for (X, Y) with the understanding that we must integrate
with respect to x and sum with respect to y. Since the marginal distribution of Y is discrete
with frequency function 1/2, y = 1, 2, the conditional density function of X given Y = y
is y exp(−yx), x > 0.
Now consider the conditional distribution of Y given X = x. Recall that the marginal
distribution of X is absolutely continuous with density function

1
[exp(−x) + 2exp(−2x)], x > 0.
2
Using the informal method described above, the conditional distribution of Y given X = x,
x > 0, has frequency function

y exp(−yx)
, y = 1, 2.
exp(−x) + 2exp(−2x)
It is easy to verify that this result is correct by noting that
1
Pr(Y = y, X ∈ A) = y exp(−yx) dx
2 A
y exp(−yx)

= p X (x) dx, y = 1, 2.
A exp(−x) + 2exp(−2x)
Hence, the conditional distribution of Y given X = x is discrete with
1
Pr(Y = 1|X = x) = 1 − Pr(Y = 2|X = x) =
1 + 2exp(−x)
for x > 0.

2.4 Conditional Expectation
Let (X, Y) denote a random vector with range X × Y and let F X|Y (·|y) denote the distribu-
tion function corresponding to the conditional distribution of X given Y = y.For a function
g: X → R such that E[|g(X)|] < ∞,E[g(X)|Y = y] may be defined by

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