Let’s consider a continuous r.v., Y, which
is a differentiable, increasing function
of a second continuous r.v., X:
Y = g(X).
Because g is differentiable and increasing ,
g' and g-1 are guaranteed to exist.
Because g maps all
to
It fol lows that for small Δx:
Dividing by Δy, we get an approximate
expression for
in terms of
:
This ex pression is exact in the limit Δx→
0:
From calculus we know that:
Consequently
Substituting g-1(y) for x:
yields an expression for
in terms of
g', g-1, and
.
Linear Example
Consider a continuous r.v., Y, which is
a linear function of a continuous r.v., X.
Specifically, Y =aX +b. It follows that
g(x) = ax+b
g'(x) = a
g-1(y) = (y−b)/a.
Substituting the above into
yields
Let 
,then
Unfortunately, there is a problem. When
a = −1, the function g is not increasing
(it is decreasing). Consequently,
It follows that for decreasing functions,
However, we can derive a function which
is correct in both cases by replacing g'(.)
with |g'(.)| in the expression relating
(.)
and (.):
Quadratic Example
Consider a continuous r.v., Y, which is
a quadratic function of a continuous r.v.,
X. Specifically, Y = X2. It follows that
Substituting the above into
yields
Let
,then
Unfortunately, there is a problem:
Can any one see the mistake?
The mistake is that two different values
of Y satisfy Y = X2:
We decided to use the positive square
root arbitrarily and ignored the negative
square root . Hence the factor of
two error . In general, if a function does
not have a unique inverse, we must sum
over all possible inverse values :
Let 
and
then
The above p.d.f. defines a distribution
called the chi square distribution.
Kinetic Energy
Recall from elementary physics, that the
kinetic energy, K, of a moving particle
is given by
where m is mass and V is velocity. Let
V be a normally distributed random variable
with mean, μ, and variance, 
:
We would like to compute
, the
p.d.f for the continuous random variable,
K.
We start by computing, g',
, and 
:
Substituting 
and the above into
yields
