The formula that we just derived for the monotonic case
has a nice intuitive explanation that we will
develop now.
Suppose that g is a monotonic function of x and that it's
monotonically increasing.
Let us fix a particular x and a corresponding y so that the
two of them are related as follows-- y is equal to g of
x, or we could argue in terms of the inverse function so
that x is equal to h of y.
Recall that h is the inverse function, that given a value
of y, tells us which one is the corresponding value of x.
Now let us consider a small interval in the
vicinity of this x.
Whenever x falls somewhere in this range, then y is going to
fall inside another small interval.
The event that x belongs here is the same as the event that
y belongs there.
So these two events have the same probability.
And we can, therefore, write that the probability that Y
falls in this interval is the same as the probability that X
falls in the corresponding little interval on the x-axis.
This interval has a certain length delta 1.
This interval has a certain length delta 2.
Now remember our interpretation of
probabilities of small intervals in terms of PDFs so
this probability here is approximately equal to the PDF
of Y evaluated at the point y times the length of the
corresponding interval.
Similarly, on the other side, the probability that X falls
on the interval is the PDF of X times the
length of that interval.
So this gives us already a relation between the PDF of Y
and the PDF of X, but it involves those two numbers
delta 1 and delta 2.
How are these two numbers related?
If x moves up by the amount of delta 1, how much is y going
to move up?
It's going to move up by an amount which is delta 1 times
the slope of the function g at that particular point.
So that gives us one relation that delta 2 is approximately
equal to delta 1 times the derivative of the function of
g at that particular x.
However, it's more useful to work the other way, thinking
in terms of the inverse function.
The inverse function maps y to x, and it maps y plus delta to
2 to x plus delta 1.
When y advances by delta 2, x is going to advance by an
amount which is how much y advanced times the slope, or
the derivative, of the function that
maps y's into x's.
And this function is the inverse function.
So this is the relation that we're going to use.
And so we replace delta 1 by this expression that we have
here in terms of delta 2.
And now we cancel the delta 2 from both sides of this
equality, and we obtain the final formula that the PDF of
Y evaluated at a certain point is equal to the PDF of x
evaluated at the corresponding point, or we could write this
as the PDF of X evaluated at the value x that's associated
to that y that's given by the inverse function, times the
derivative of the function h, the inverse function.
And this is just the same formula as the one that we had
derived earlier using CDFs.
This derivation is quite intuitive.
It associates probabilities of small intervals on the x-axis
to probabilities of corresponding small intervals
on the y-axis.
These two probabilities have to be equal, and this implies
a certain relation between the two PDFs.