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The formula that we just derived
for the monotonic case

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has a nice intuitive explanation
that we will

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develop now.

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Suppose that g is a monotonic
function of x and that it's

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monotonically increasing.

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Let us fix a particular x and a
corresponding y so that the

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two of them are related as
follows-- y is equal to g of

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x, or we could argue in terms
of the inverse function so

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that x is equal to h of y.

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Recall that h is the inverse
function, that given a value

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of y, tells us which one is the
corresponding value of x.

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Now let us consider a small
interval in the

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vicinity of this x.

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Whenever x falls somewhere in
this range, then y is going to

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fall inside another
small interval.

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The event that x belongs here is
the same as the event that

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y belongs there.

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So these two events have
the same probability.

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And we can, therefore, write
that the probability that Y

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falls in this interval is the
same as the probability that X

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falls in the corresponding
little interval on the x-axis.

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This interval has a certain
length delta 1.

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This interval has a certain
length delta 2.

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Now remember our interpretation
of

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probabilities of small intervals
in terms of PDFs so

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this probability here is
approximately equal to the PDF

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of Y evaluated at the point
y times the length of the

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corresponding interval.

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Similarly, on the other side,
the probability that X falls

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on the interval is the
PDF of X times the

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length of that interval.

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So this gives us already a
relation between the PDF of Y

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and the PDF of X, but it
involves those two numbers

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delta 1 and delta 2.

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How are these two
numbers related?

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If x moves up by the amount of
delta 1, how much is y going

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to move up?

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It's going to move up by an
amount which is delta 1 times

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the slope of the function g
at that particular point.

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So that gives us one relation
that delta 2 is approximately

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equal to delta 1 times the
derivative of the function of

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g at that particular x.

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However, it's more useful to
work the other way, thinking

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in terms of the inverse
function.

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The inverse function maps y to
x, and it maps y plus delta to

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2 to x plus delta 1.

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When y advances by delta 2, x
is going to advance by an

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amount which is how much y
advanced times the slope, or

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the derivative, of
the function that

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maps y's into x's.

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And this function is the
inverse function.

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So this is the relation that
we're going to use.

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And so we replace delta 1 by
this expression that we have

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here in terms of delta 2.

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And now we cancel the delta
2 from both sides of this

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equality, and we obtain the
final formula that the PDF of

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Y evaluated at a certain point
is equal to the PDF of x

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evaluated at the corresponding
point, or we could write this

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as the PDF of X evaluated at the
value x that's associated

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to that y that's given by the
inverse function, times the

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derivative of the function
h, the inverse function.

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And this is just the same
formula as the one that we had

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derived earlier using CDFs.

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This derivation is
quite intuitive.

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It associates probabilities of
small intervals on the x-axis

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to probabilities of
corresponding small intervals

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on the y-axis.

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These two probabilities have to
be equal, and this implies

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a certain relation between
the two PDFs.