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We now move to the case of
continuous random variables.

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We will start with a special
case where we want to find the

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PDF of a linear function of a
continuous random variable.

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We will start by considering a
simple example, and study it

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using an intuitive argument.

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And afterwards, we will justify
our conclusions

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mathematically.

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So we start with a random
variable X that has a PDF over

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the form shown in this figure
so that it is a piecewise

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constant PDF.

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We then consider a random
variable z, which is defined

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to be 2 times X. The random
variable x takes values

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between minus 1 and 1.

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So z takes values between
minus 2 and 2.

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Now, values of X between minus
1 and 0 correspond to values

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of Z between minus 2 and 0.

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The different values of X in
this range are, in some sense,

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equally likely, because we
have a constant PDF.

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And that argues that the
corresponding values of Z

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should also be, in some
sense, equally likely.

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So the PDF should be constant
over this range.

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By a similar argument, the PDF
of Z should also be constant

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over the range from 0 to 2.

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And the PDF must, of course,
be 0 outside this range,

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because these are values of
Z that are impossible.

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Let us now try to figure out
the parameters of this PDF.

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The probability that X
is positive is the

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area of this rectangle.

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And the area of this
rectangle is 2/3.

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So the area of this rectangle
should also be 2/3.

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And that means that the height
of this rectangle should be

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equal to 1/3.

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Similarly, the probability that
X is negative is the area

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of this rectangle, and the
area of this rectangle is

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equal to 1/3.

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When X is negative, Z is also
negative, so the probability

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of a negative value should
be equal to 1/3.

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And for the area of this
rectangle to be 1/3, it means

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that the height of this
rectangle should be 1/6.

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So what happened here?

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We started with a PDF of X and
essentially stretched it out

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by a factor of 2 while keeping
the same shape.

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However, we also scaled
it down by a

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corresponding amount.

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So 2/3 became 1/3, and
1/3 became 1/6.

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The reason for this scaling down
is because we need the

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total probability, the total
area under this PDF, to be

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equal to 1.

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If we now add a number, let's
say 3, to the random variable

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Z, what is going to happen?

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The random variable Y now
will take values from

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minus 2 plus 3--

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this is plus 1--

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all the way up to 2 plus
3, which is plus 5.

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Values in the range from 1 to 3
correspond to values of Z in

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the range from minus 2 to 0.

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These values are all, in some
sense, equally likely.

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So they should also be
equally likely here.

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And by a similar argument, these
values in the range from

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3 to 5 should also be
equally likely.

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This rectangle corresponds
to this rectangle here.

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So the area should
be the same.

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And therefore, the height
should also be the same.

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Therefore, the height
here should be 1/6.

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And by the same argument,
the height here

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should be equal to 1/3.

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So what happens here is that
when we add 3 to a random

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variable, the PDF just gets
shifted by 3 but otherwise

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retains the same shape.

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So the story is entirely similar
to what happened in

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the discrete case.

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We start with a PDF of X. We
stretch it horizontally by a

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factor of 2.

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And then we shift it
horizontally by 3.

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The only difference is that here
in the continuous case,

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we also need to scale the plot
in the vertical dimension by a

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factor of 2.

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Actually, make it smaller
by a factor of 2.

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And this needs to be done in
order to keep the total area

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under the PDF equal to 1.

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Let us now go through a
mathematical argument with the

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purpose of also finding a
formula that represents what

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we just did in our
previous example.

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Let Y be equal to aX plus b.

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Here, X is a random variable
with a given PDF.

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a and b are given constants.

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Now, if a is equal to 0, then
Y is identically equal to b.

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So it is a constant random
variable and

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does not have a PDF.

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So let us exclude this case and
start by assuming that a

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is a positive number.

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We can try to work, as in the
discrete case, and try

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something like the following.

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The probability that Y takes
on a specific value is the

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same as the probability that aX
plus b takes on a specific

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value, which is the same as the
probability that X takes

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on the specific value, y
minus b divided by a.

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This equality was useful
in the discrete case.

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Is it useful here?

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Unfortunately not.

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When we're dealing with
continuous random variables,

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the probability that the
continuous random variable is

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exactly equal to a given number,
this probability is

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going to be equal to 0.

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And the same applies to
this side as well.

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So we have that 0
is equal to 0.

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And this is uninformative, and
we have not made any progress.

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So instead of working with
probabilities of individual

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points which will always
be 0, we will work with

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probabilities of intervals that
generally have non-zero

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probability.

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The trick is to work
with CDFs.

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So let us try to find the CDF
of Y. The CDF of the random

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variable Y is defined as the
probability that the random

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variable is less than or equal
to a certain number.

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Now, in our case,
Y is aX plus b.

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We move b to the other side of
the inequality and then divide

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both sides of the
inequality by a.

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And we get that this is the same
as the probability that X

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is less than or equal to y minus
b divided by a, which is

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the same as the CDF of X
evaluated at y minus b over a.

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So we have a formula for the CDF
of Y in terms of the CDF

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of X.

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How can we find the PDF?

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Simply by differentiating.

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We differentiate both sides
of this equation.

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The derivative of
a CDF is a PDF.

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And therefore, the PDF of Y is
going to be equal to the

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derivative of this side.

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Here we need to use
the chain rule.

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First, we take the derivative
of this function.

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And the derivative of the CDF
is a PDF, so the PDF of X

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evaluated at this particular
number.

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But then we also need to take
the derivative of the argument

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inside with respect to y.

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And that derivative
is equal to 1/a.

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And this gives us a formula for
the PDF of Y in terms of

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the PDF of X.

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How about the case where
a is less than 0?

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What is going to change?

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The first step up to
here remains valid.

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But now when we divide both
sides of the inequality by a,

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the direction of the inequality
gets reversed.

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So we obtain instead the
probability that X is larger

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than or equal to y minus
b divided by a.

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And this is 1 minus the
probability that X is less

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than y minus b over a.

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Now, X is a continuous random
variable, so the probability

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is not going to change if here
we make the inequality to be a

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less than or equal sign.

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And what we have here is 1 minus
the CDF of X evaluated

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at y minus b over a.

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We use the chain rule once more,
and we obtain that the

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PDF of Y, in this case, is equal
to minus the PDF of X

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evaluated at y minus
b over a times 1/a.

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Now, when a is positive,
a is the same as the

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absolute value of a.

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When a is negative and we have
this formula, we have here a

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minus a, which is the same as
the absolute value of a.

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So we can unify these two
formulas by replacing the

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occurrences of a and that minus
sign by just using the

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absolute value.

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00:10:04,670 --> 00:10:10,360
And this gives us this formula
for the PDF of Y in terms of

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the PDF of X. And it is a
formula that's valid whether a

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is positive or negative.

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What this formula represents
is the following.

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Because of the factor of a that
we have here, we take the

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PDF of X and scale it
horizontally by a factor of a.

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00:10:32,482 --> 00:10:37,040
Because of the term b that we
have here, the PDF also gets

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shifted horizontally by b.

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And finally, this term here
corresponds to a vertical

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scaling of the plot
that we have.

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And the reason that this term is
present is so that the PDF

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of Y integrates to 1.

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It is interesting to also
compare with the corresponding

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discrete formula that
we derived earlier.

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The discrete formula has exactly
the same appearance

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except that the scaling
factor is not present.

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So for the case of continuous
random variables, we need to

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00:11:10,340 --> 00:11:12,880
scale vertically the PDF.

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But in the discrete case, such
a scaling is not present.