5.5 Functions of random variables Let X be a random variable on (,F , P) and suppose that g : R → R. Then Y = g(X) is a mapping from  into R, defined by Y (ω) = g[X(ω)] for ω ∈ . Actually, Y is not generally a random variable since it need not satisfy condition (5.1). It turns out, however, that (5.1)

is valid for Y whenever g is sufficiently well behaved (such as g is a continuous function, or a monotone function, or . . . ), and so we neglect this difficulty, assuming henceforth that all quantities of the form Y = g(X) are random variables. The main question is now the following: if we knowthe distribution of X, then how do we find the distribution of Y = g(X)?

If X is discrete with mass function pX , then (2.25) provides the answer, and we consider next the case when X is continuous with density function fX . We begin with an example.

Example 5.49 If X is continuous with density function fX , and g(x) = ax + b when a > 0, then Y = g(X) = aX + b has distribution function given by P(Y ≤ y) = P(aX + b ≤ y)

= P

????

X ≤ a−1(y − b)

= FX

????

a−1(y − b)

.

By differentiation with respect to y, fY (y) = a−1 fX

????

a−1(y − b)

for y ∈ R. △

The next theorem generalizes the result of this example.

Theorem 5.50 If X is a continuous random variable with density function fX , and g is a strictly increasing and differentiable function from R into R, then Y = g(X) has density function fY (y) = fX

????

g−1(y)

d dy [g−1(y)] for y ∈ R, (5.51)

where g−1 is the inverse function of g.

72 Distribution functions and density functions Proof First, we find the distribution function of Y :

P(Y ≤ y) = P

????

g(X) ≤ y

= P

????

X ≤ g−1(y)

since g is increasing.

We differentiate this with respect to y, noting that the right-hand side is a function of a function, to obtain (5.51). 2 If, in Theorem 5.50, g were strictly decreasing, then the same argument gives that Y =

g(X) has density function fY (y) = − fX

????

g−1(y)

d