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Hi.

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In this problem we'll work
through an example of

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calculating a distribution for
a minute variable using the

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method of derived
distributions.

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So in general, the process
goes as follows.

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We know the distribution for
some random variable X and

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what we want is the distribution
for another

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random variable of Y, which is
somehow related to X through

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some function g.

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So Y is a g of X.

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And the steps that we follow--

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we can actually just kind
of summarize them

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using this four steps.

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The first step is to write out
the CDF of Y. So Y is thing

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that we want.

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And what we'll do is we'll
write out the CDF first.

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So remember the CDF is just
capital F of y, y is the

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probability that random variable
Y is less than or

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equal to some value, little y.

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The next thing we'll do is,
we'll use this relationship

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that we know, between Y and X.
And we'll substitute in,

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instead of writing the random
variable Y In here, we'll

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write it in terms of X. So we'll
plug in for-- instead of

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Y, we'll plug-in X. And we'll
use this function g in order

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to do that.

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So what we have now is that up
to here, we would have that

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the CDF of Y is now the
probability that the random

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variable X is less than or equal
to some value, little y.

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Next what we'll do is we'll
actually rewrite this

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probability as a CDF of
X. So the CDF of X,

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remember, would be--

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F of x is that the probability
of X is less than or equal to

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some little x.

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And then once we have that,
if we differentiate this--

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when we differentiate the CDF of
X, we get the PDF of X. And

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what we presume is that we
know this PDF already.

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And from that, what we get is,
when we differentiate this

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thing, we get the PDF of Y. So
through this whole process

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what we get is, we'll get the
relationship between the PDF

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of Y and the PDF of X. So that
is the process for calculating

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the PDF of Y using X.

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So let's go into our
specific example.

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In this case, what we're told
is that X, the one that we

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know, is a standard normal
random variable.

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Meaning that it's mean
0 and variance 1.

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And so we know the
form of the PDF.

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The PDF of x is this, 1 over
square root of 2 pi e to the

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minus x squared over 2.

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And then the next thing that
we're told is this

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relationship between X and Y. So
what we're told is, if X is

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negative, then Y is minus X.
If X is positive, then Y is

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the square root of X. So
this is a graphical its

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representation of the
relationship between X and Y.

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All right, so we have everything
that we need.

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And now let's just go through
this process and calculate

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what the PDF of Y is.

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So the first thing we do is we
write out the PDF of Y. So the

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PDF of Y is what
we've written.

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It's the probability that the
random variable Y is less than

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or equal to some little y.

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Now the next step that we do is
we have to substitute in,

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instead of in terms of Y, we
want to substitute it in terms

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of X. Because we actually know
stuff about X, but we don't

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know anything about Y. So what
is the probability that Y, the

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random variable Y, is
less than or equal

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to some little y?

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Well, let's go back to this
relationship and see if we can

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figure that out.

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So let's pretend that here
is our little y.

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Well, if the random variable
Y is less than or equal to

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little y, it has to
be underneath

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this horizontal line.

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And in order for it to be
underneath this horizontal

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line, that means that X has
to be between this range.

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And what is this range?

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This range goes from minus
Y to Y squared.

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So why is that?

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It's because in this portion X
and Y are related as, Y is

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negative X and here it's Y is
square root of X. So if X is Y

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squared, then Y would be Y. If
X is negative Y, then Y would

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be Y. All right, so this
is the range that

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we're looking for.

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So if Y, the random variable
Y is less than or equal to

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little y, then this is the same
as if the random variable

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X is between negative
Y and Y squared.

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So let's plug that in.

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This is the same as the
probability that X is between

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negative Y and Y squared.

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So those are the first
two steps.

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Now the third step is, we
have to rewrite this

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as the CDF of x.

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So right now we have it in terms
of a probability of some

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event related to X. Let's
actually transform that to be

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explicitly in terms of the CDF
of X. So how do we do that?

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Well, this is just the
probability that X is within

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some range.

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So we can turn that into the
CDF by writing it as a

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difference of two CDFs.

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So this is the same as the
probability that X is less

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than or equal to Y squared minus
the probability that X

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is less than or equal
to negative Y.

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So in order to find the
probability that X is between

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this range, we take the
probability that it's less

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than Y squared, which
is everything here.

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And then we subtract that
probability that it's less

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than Y, negative Y. So what
we're left with is just within

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this range.

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So these actually are now
exactly CDFs of X. So this is

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F of X evaluated at Y squared
and this is F of X evaluated

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at negative Y. So now we've
completed step three.

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And the last step that we need
to do is differentiate.

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So if we differentiate both
sides of this equation with

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respect to Y, we'll get that the
left side would get what

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we want, which is the PDF of
Y. Now we differentiate the

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right side--

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we'll have to invoke
the chain rule.

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So the first thing that we do
is, well, this is a CDF of X.

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So when we differentiate
we'll get the PDF of X.

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But then we also have invoke
the chain rule for this

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argument inside.

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So the derivative of Y
squared would give us

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an extra term, 2Y.

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And then similarly this would
give us the PDF of X evaluated

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at negative Y plus the chain
will give us an extra term of

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negative 1.

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So let's just clean this
up a little bit.

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So it's 2y F X squared plus F
X minus Y. All right, so now

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we're almost done.

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We've differentiated.

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We have the PDF of Y, which
is what we're looking for.

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And we've written it in terms
of the PDF of X. And

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fortunately we know what that
is, so once we plug that in,

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then we're essentially done.

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So what is the PDF?

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Well, the PDF of X evaluated at
Y squared is going to give

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us 1 over square root of
2 pi e to the minus--

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so in this case, X
is Y squared--

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so we get Y to the
fourth over 2.

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And then we get another 1 over
square root of 2 pi e to the

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minus Y squared over 2.

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OK, and now we're almost done.

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The last thing that we
need to take care of

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is, what is the range?

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Now remember, it's important
when you calculate out PDFs to

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always think about the ranges
where things are valid.

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So when we think about this,
what is the range where this

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actually is valid?

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Well, Y, remember is related
to X in this relationship.

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So as we look at this, we see
that Y can never be negative.

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Because no matter what X is, Y
gets transformed into some

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non-negative version.

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So what we know is that this is
now actually valid only for

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Y greater than 0 and for Y less
than 0, the PDF is 0.

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So this gives us the
final PDF of Y.

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All right, so it seems like at
first when you start doing

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these derived restriction
problems

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that it's pretty difficult.

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But if we just remember that
there are these pretty

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straightforward steps that we
follow, and as long as you go

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through these steps and do them
methodically, then you

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can actually come up with
the solution for

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any of these problems.

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And one last thing to remember
is to always think about what

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are the ranges where these
things are valid?

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Because the relationship between
these two random

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variables could be pretty
complicated and you need to

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always be aware of when things
are non-zero and

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when they are 0.

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