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Hi.

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In this problem, we'll get some
practice working with

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PDFs and also using PDFs
to calculate CDFs.

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So the PDF that we're given
in this problem is here.

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So we have a random variable,
z, which is a continuous

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random variable.

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And we're told that the PDF of
this random variable, z, is

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given by gamma times 1 plus z
squared in the range of z

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between negative 2 and 1.

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And outside of this
range, it's 0.

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All right, so first thing we
need to do and the first part

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of this problem is we need to
figure out what gamma is

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because it's not really a
fully specified PDF yet.

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We need to figure out exactly
what the value gamma is.

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And how do we do that?

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Well, we've done analogous
things before for

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the discrete case.

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So the tool that we use
is that the PDF must

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integrate to 1.

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So in the discrete case, the
analogy was that the PMF had

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to sum to 1.

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So what do we know?

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We know that when you integrate
this PDF from

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negative infinity to infinity,
fz of z, it has to equal 1.

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All right, so what
do we do now?

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Well, we know what
the PDF is--

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partially, except for gamma--
so let's plug that in.

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And the first thing that we'll
do is we'll simplify this

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because we know that the PDF is
actually only non-zero in

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the range negative 2 to 1.

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So instead of integrating from
negative infinity to infinity,

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we'll just integrate from
negative 2 to 1.

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And now let's plug in
this gamma times 1

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plus z squared dc.

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And now the rest of the problem
is just applying

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calculus and integrating this.

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So let's just go through
that process.

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So we get z plus 1/3 z cubed
from minus 2 to 1.

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And now we'll plug
in the limits.

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And we get gamma, and that's 1
plus 1/3 minus minus 2 plus

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1/3 times minus 2 cubed.

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And then if we add this all up,
you get 4/3 plus 2 plus

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8/3, which will give you 6.

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So what we end up with
in the end is that 1

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is equal to 6 gamma.

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So what does that tell us?

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That tells us that, in this
case, gamma is 1/6.

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OK, so we've actually figured
out what this PDF really is.

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And let's just substitute
that in.

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So we know what gamma is.

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So it's 1/6.

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So from this PDF, we can
calculate anything

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that we want to.

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This PDF, basically, fully
specifies everything that we

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need to know about this
random variable, z.

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And one of the things that
we can calculate from

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the PDF is the CDF.

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So the next part of the
problem asks us to

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calculate the CDF.

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So remember the CDF, we use
capital F. And the definition

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is that you integrate from
negative infinity to this z.

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And what do you integrate?

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You integrate the PDF.

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And all use some dummy variable,
y, here in the

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integration.

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So what is it really doing?

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It's basically just taking the
PDF and taking everything to

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the left of it.

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So another way to think about
this-- this is the probability

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that the random variable
is less than or equal

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to some little z.

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It's just accumulating
probability as you go from

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left to right.

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So the hardest part about
calculating the CDFs, really,

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is actually just keeping track
of the ranges, because unless

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the PDF is really simple, you'll
have cases where the

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PDF cold be 0 in some ranges and
non-zero in other ranges.

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And then what you really have
to keep track of is where

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those ranges are and where you
actually have non-zero

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probability.

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So in this case, we actually
break things down into three

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different ranges because
this PDF actually looks

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something like this.

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So it's non-zero between
negative 2 and 1, and it's 0

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everywhere else.

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So then what that means is
that our job is a little

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simpler because everything to
the left of negative 2, the

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CDF will be 0 because there's
no probability

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density to the left.

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And then everything to the
right of 1, well we've

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accumulated all the probability
in the PDF because

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we know that when you integrate
from negative 2 to

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1, you capture everything.

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So anything to the right of
1, the CDF will be 1.

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So the only hard part is
calculating what the CDF is in

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this intermediate range, between
negative 2 and 1.

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So let's do that case first--

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so the case of z is between
negative 2 and 1.

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So what is the CDF
in that case?

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Well, the definition is to
integrate from negative

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infinity to z.

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But we know that everything
to the left of negative 2,

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there's no probably density.

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So we don't need to
include that.

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So we can actually change this
lower limit to negative 2.

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And the upper limit is
wherever this z is.

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So that becomes our integral.

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And the inside is
still the PDF.

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So let's just plug that in.

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We know that it's 1/6 1 plus--

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we'll make this y squared--

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by.

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And now it's just
calculus again.

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And in fact, it's more or less
the same integral, so what we

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get is y plus 1/3 y cubed
from negative 2 to z.

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Notice the only thing that's
different here is that we're

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integrating from negative 2 to
z instead of negative 2 to 1.

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And when we calculate this out,
what we get is z plus 1/3

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z cubed minus minus 2 plus 1/3
minus 2 cubed, which gives us

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1/6 z plus 1/3 z cubed plus plus
2 plus 8/3 gives us 14/3.

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So that actually is our CDF
between the range of

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negative 2 to 1.

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So for full completeness, let's
actually write out the

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entire CDF, because there's two
other parts in the CDF.

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So the first part is that
it's 0 if z is less

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than negative 2.

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And it's 1 if z is
greater than 1.

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And in between, it's this
thing that we've just

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calculated.

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So it's 1/6 z plus 1/3 z cubed
plus 14/3 if z is between

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minus 2 and 1.

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So that is our final answer.

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So the main point of this
problem was to drill a little

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bit more the concepts
of PDFs and CDFs.

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So for the PDF, the important
thing to remember is that in

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order to be a valid PDF, the
PDF has to integrate to 1.

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And you can use that fact to
help you calculate any unknown

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constants in the PDF.

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And then to calculate the CDF,
it's just integrating the PDF

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from negative infinity to
whatever point that you want

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to cut off at.

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And the tricky part, as I said
earlier, was really just

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keeping track of the ranges.

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In this case, we've broke it
down into three ranges.

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If we had a slightly more
complicated PDF, then you

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would have to keep track
of even more ranged.

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All right, so I hope that
was helpful, and we'll

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see you next time.

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