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In this important segment, we
will develop a method for

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finding the PDF of a general
function of a continuous

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random variable, a function g of
X, which, in general, could

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be nonlinear.

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The method is very general
and involves two steps.

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The first step is to find the
CDF of Y. And then the second

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step is to take the derivative
of the CDF and

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then find the PDF.

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Most of the work lies here in
finding the CDF of Y. And how

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do we do that?

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Well, since Y is a function of
the random variable X, we

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replace Y by g of X. And now
we're dealing with a

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probability problem that
involves a random variable, X,

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with a known PDF.

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And we somehow calculate
this probability.

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So let us illustrate
this procedure

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through some examples.

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In our first example, we let X
be a random variable which is

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uniform on the range
from 0 to 2.

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And so the height of
the PDF is 1/2.

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And we wish to find the PDF of
the random variable Y which is

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defined as X cubed.

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So since X goes all the way
up to 2, Y goes all

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the way up to 8.

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The first step is to find the
CDF of Y. And since Y is a

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specific function of X, we
replace that functional form.

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And we write it this way.

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So we want to calculate the
probability that x cubed is

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less than or equal to
a certain number y.

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Let us take cubic roots of both
sides of this inequality.

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This is the same as the
probability that X is less

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than or equal to y to the 1/3.

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Now, we only care about values
of y that are between 0 and 8.

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So this calculation is going to
be for those values of y.

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For other values of y, we know
that the PDF is equal to 0.

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And there's no work that
needs to be done there.

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OK.

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Now, y is less than or equal to
8, so the cubic root of y

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is less than or equal to 2.

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So y to the 1/3 is going
to be a number

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somewhere in this range.

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Let's say this number.

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We want the probability
that X is less than or

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equal to that value.

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So that probability is equal to
this area under the PDF of

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X. And since it is uniform,
this area is easy to find.

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It's the height, which is
1/2 times the base,

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which is y to the 1/3.

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So we continue this calculation,
and we get 1/2

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times y to the 1/3.

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So this is the formula for the
CDF of Y for values of little

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y between 0 and 8.

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This completes step one.

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The second step is
simple calculus.

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We just need to take the
derivative of the CDF.

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And the derivative is 1/2 times
1/3, this exponent, y to

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the power of minus 2/3.

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Or in a cleaner form,
1/6 times 1 over y

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to the power 2/3.

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So the form of this PDF is
not a constant anymore.

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Y is not a uniform
random variable.

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The PDF becomes larger and
larger as y approaches 0.

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And in fact, in this example,
it even blows up when y

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becomes closer and
closer to 0.

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So this is the shape
of the PDF of Y.

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Our second example
is as follows.

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You go to the gym, you jump on
the treadmill, and you set the

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speed on the treadmill to some
random value which we call X.

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And that random value is
somewhere between 5 and 10

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kilometers per hour.

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And the way that you set it is
chosen at random and uniformly

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over this interval.

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So X is uniformly distributed
on the interval

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between 5 and 10.

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You want to run a total
of 10 kilometers.

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How long is it going
to take you?

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Let the time it takes you be
denoted by Y. And the time

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it's going to take you is the
distance you want to travel,

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which is 10 divided
by the speed with

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which you will be going.

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So the random variable y is
defined in terms of x through

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this particular expression.

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We want to find the PDF of y.

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First let us look at the range
of the random variable Y.

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Since x takes values between
5 and 10, Y takes values

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between 1 and 2.

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Therefore, the PDF of
Y is going to be 0

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outside that range.

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And let us now focus on values
of Y that belong to this

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interesting range.

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So 1 less than y less
than or equal to 2.

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And now we start with our
two-step program.

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We want to find the CDF of Y,
namely, the probability that

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capital Y takes a value less
than or equal to a certain

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little y in this range.

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We recall the definition of
capital Y. So now we're

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dealing with a probability
problem that involves the

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random variable capital
X, whose

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distribution is given to us.

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Now, we rewrite this
event as follows.

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We move X to the other side.

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This is the probability that X
is larger than or equal after

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we move the little y also
to the left-hand side.

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X being larger than or equal
to 10 over little y.

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Now, y is between 1 and 2.

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10/y is going to be a number
between 5 and 10.

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So 10/y is going to be somewhere
in this range.

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We're interested in the
probability that X is larger

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than or equal to that number.

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And this probability is going
to be the area of this

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rectangle here.

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And the area of that rectangle
is equal to the height of the

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rectangle--

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now, the height of this
rectangle is going to be 1/5.

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This is the choice that makes
the total area under this

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curve be equal to 1--

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times the base.

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And the length of the
base is this number

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10 minus that number.

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It's 10 minus 10/y.

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So this is the form of the CDF
of Y for y's in this range.

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To find the PDF of Y, we just
take the derivative.

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And we get 1/5 times the
derivative of this term, which

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is minus 10, divided
by y squared.

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But when we take the derivative
of 1/y, that gives

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us another minus sign.

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The two minus signs
cancel, and we

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obtain 2 over y squared.

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And if you wish to plot
this, it starts at 2.

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And then as y increases, the
PDF actually decreases.

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And this is the form of the PDF
of the random variable y.

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This is the form which is true
when y lies between 1 and 2.

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And of course, the PDF is
going to be 0 for other

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choices of little y.

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So what we have seen here is a
pretty systematic approach

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towards finding the PDF of the
random variable Y. Again, the

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first step is to look at the
CDF, write the CDF in terms of

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the random variable X, whose
distribution is known, and

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then solve a probability problem
that involves this

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particular random variable.

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And then in the last step, we
just need to differentiate the

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CDF in order to obtain the PDF.