Page 43 - Elements of Distribution Theory

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1.8 Expectation 29

Example 1.29 (Cauchy distribution). Let X denote a real-valued random variable with an
absolutely continuous distribution with density function
1
p(x) = , −∞ < x < ∞;
2
π(1 + x )
this is a standard Cauchy distribution.IfE(X)exists, it must be equal to
x x x

∞ ∞ ∞
dx = dx − dx.
2
2
2
−∞ π(1 + x ) 0 π(1 + x ) 0 π(1 + x )
Since
x
∞

dx =∞,
2
π(1 + x )
0
it follows that E(X) does not exist.
Now suppose that X is a random vector X = (X 1 ,..., X d ), where X j , j = 1,..., d,is
real-valued. Then E(X)is simply the vector (E(X 1 ),..., E(X d )).
Example 1.30 (Uniform distribution on the unit cube). Let X = (X 1 , X 2 , X 3 ) denote a
3
three-dimensional random vector with the uniform distribution on (0, 1) ; see Examples 1.7
and 1.25. Then X has an absolutely continuous distribution with density function
3
p(x) = 1, x ∈ (0, 1) .
It follows that
1
1
E(X 1 ) = xdx = .
0 2
Similarly, E(X 2 ) = E(X 3 ) = 1/2. It follows that E(X) = (1/2, 1/2, 1/2).

Expectation of a function of a random variable
Let X denote a random variable, possibly vector-valued, and let g denote a real-valued
function defined on X, the range of X. Let Y = g(X) and let H denote the distribution
function of Y. Then

∞
E(Y) = yd H(y)
−∞
provided that the integral exists. An important result is that we may also compute E(Y)by

E(Y) = g(x) dF(x) (1.3)
X
so that the probability distribution of Y is not needed to compute its expected value.
When X has a discrete distribution, with frequency function p, proof of this result is
straightforward. Let f denote the frequency function of Y. Then

E(Y) = yf (y).
y
Note that

f (y) = Pr(Y = y) = Pr(X ∈{x: g(x) = y}) = p(x).
x:g(x)=y