Page 56 - Elements of Distribution Theory

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42 Conditional Distributions and Expectation

This is an absolutely continuous distribution with density function
1
[exp(−x) + 2exp(−2x)].
2
Since, for y = 1, 2,

1 ∞ 1
Pr(Y = y) = y exp(−yx) dx = ,
2 0 2
it follows that Y has a discrete distribution with frequency function 1/2, y = 1, 2.

Independence
Consider a random vector (X, Y) with range X × Y.Wesay X and Y are independent if for
any A ⊂ X and B ⊂ Y, the events X ∈ A and Y ∈ B are independent events in the usual
sense of elementary probability theory; that is, if
Pr(X ∈ A, Y ∈ B) = Pr(X ∈ A)Pr(Y ∈ B).

Independence may easily be characterized in terms of either distribution functions or
expected values.

Theorem 2.1. Let (X, Y) denote a random vector with range X × Y and distribution func-
tion F. Let F X and F Y denote the marginal distribution functions of X and Y, respectively.
(i) X and Y are independent if and only if for all x, y
F(x, y) = F X (x)F Y (y).

(ii) X and Y are independent if and only if for all bounded functions g 1 : X → R and
g 2 : Y → R

E[g 1 (X)g 2 (Y)] = E[g 1 (X)]E[g 2 (Y)].

Proof. Suppose X and Y are independent. Let m denote the dimension of X and let n
denote the dimension of Y. Fix x = (x 1 ,..., x m ) and y = (y 1 ,..., y n ); let

A = (−∞, x 1 ] × ··· × (−∞, x m ]
and

B = (−∞, y 1 ] × ··· × (−∞, y n ]
so that

F(x, y) = Pr(X ∈ A, Y ∈ B), F X (x) = Pr(X ∈ A), and F Y (y) = Pr(Y ∈ B).

Then
F(x, y) = Pr(X ∈ A, Y ∈ B) = Pr(X ∈ A)Pr(Y ∈ B) = F X (x)F Y (y).

Now suppose F(x, y) = F X (x)F Y (y). Since F X (x)F Y (y)is the distribution function of
a random variable (X 1 , Y 1 ) such that X 1 and Y 1 are independent with marginal distribution
functions F X and F Y ,respectively,itfollowsthat(X, Y)hasthesamedistributionas(X 1 , Y 1 );
that is, X and Y are independent. This proves part (i).