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Hi.
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In this problem, we're going
to get a bunch of practice
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working with multiple random
variables together.
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And so we'll look at joint
PDFs, marginal PDFs,
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conditional PDFs, and also get
some practice calculating
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expectations as well.
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So the problem gives us a pair
of random variables-- x and y.
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And we're told that the joint
distribution is uniformly
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distributed on this triangle
here, with the vertices being
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0, 0 1, 0, and 0, 1.
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So it's uniform in
this triangle.
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And the first part of the
problem is just to figure out
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what exactly is disjoint PDF of
the two random variables.
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So in this case, it's pretty
easy to calculate, because we
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have a uniform distribution.
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And remember, when you have a
uniform distribution, you can
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just imagine it being
a sort of plateau
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coming out of the board.
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And it's flat.
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And so the height of the
plateau, in order to calculate
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it, you just need to figure
out what the area of this
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thing is, of this triangle is.
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So remember, when you had single
random variables, what
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we had to do was calculate, for
uniform distribution, we
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had to integrate to 1.
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So you took the length, and you
took 1 over the length was
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the correct scaling factor.
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Here, you take the area.
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And the height has to make it so
that the entire volume here
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integrates to 1.
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So the joint PDF is just
going to be 1 over
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whatever this area is.
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And the area is pretty
simple to calculate.
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It's 1/2 base times height.
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So it's 1/2.
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And so what we have is
that the area is 1/2.
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And so the joint PDF of x and
y is going to equal 2.
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But remember, you always have
to be careful when writing
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these things to remember
the ranges when
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these things are valid.
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So it's only 2 within
this triangle.
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And outside of the
triangle, it's 0.
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So what exactly does inside
the triangle mean?
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Well, we can write it
more mathematically.
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So this diagonal line, it's
given by x plus y equals 1.
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So everything in the triangle
is really x plus y is less
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than or equal to 1.
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It means everything under
this triangle.
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And so we need x plus y to be
less then or equal to 1 and
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also x to be non-negative and
y to be non-negative.
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So with these inequalities,
that captures everything
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within this triangle.
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And otherwise, the joint
PDF is going to be 0.
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The next part asks us to find,
using this joint PDF, the
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marginal of y.
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And remember, when you have
a joint PDF of two random
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variables, you essentially have
everything that you need,
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because from this joint PDF, you
can calculate marginals,
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you can calculate from the
margins, you can calculate
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conditionals.
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The joint PDF captures
everything that there is to
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know about this pair of
random variables.
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Now, to calculate a marginal PDF
of y, remember a marginal
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really just means collapsing
the other
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random variable down.
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And so you can just imagine
taking this thing and
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collapsing it down
onto the y-axis.
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And mathematically, that is just
saying that we integrate
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out the other random variable.
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So the other random variable
in this case will be x.
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We take x and we get rid of
it by integrating out from
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negative infinity to infinity.
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Of course, this joint PDF
is 0 in a lot of places.
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And so a lot of these
will be 0.
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And only for a certain range
of x's will this integral
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actually be non-zero.
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And so again, the other time
when we have to be careful is
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when we have these limits of
integration, we need to make
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sure that we have the
right limits.
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And so we know that the
joint PDF is 2.
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It's nonzero only within
this triangle.
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And so it's only 2 within
this triangle, which
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means what for x?
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Well, depending on what
x and y are, this will
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be either 2 or 0.
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So let's just fix
some value of y.
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Pretend that we've picked some
value y, let's say here.
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We want this value of y.
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Well, what are the values of x
such that the joint PDF for
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that value y is actually
nonzero, it's actually 2?
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Well, it's everything from
x equals 0 to whatever
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x value this is.
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But this x value, actually, if
you think about it, is just 1
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minus y, because this line
is x plus y equals 1.
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So whatever y is, x is going
to be 1 minus that.
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And so the correct limits
would actually be from
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0 to 1 minus y.
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And then the rest of that
is pretty simple.
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You integrate this.
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This is a pretty simple
integral.
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And you get that it's actually
two times 1 minus y.
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That's a y.
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But of course, again, we need to
make sure that we have the
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right regions.
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So this is not always true
for y, of course.
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This is only true for
y between 0 and 1.
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And otherwise, it's actually 0,
because when you take a y
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down here, well, there's no
values of x that will give you
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a nonzero joint PDF.
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And if you take a value of y
higher than this, the same
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thing happens.
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So we can actually draw
this out and see
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what it looks like.
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So let's actually draw
a small picture here.
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Here's y.
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Here's the marginal PDF of y.
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And here's 2.
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And it actually looks
like this.
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It's a triangle and a 0
outside this range.
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So does that make sense?
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Well, first of all, you see
that actually does in fact
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integrates to 1,
which is good.
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And the other thing we notice
is that there is a higher
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density for smaller
values of y.
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So why is that?
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Why are smaller values
of y more likely than
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larger values of y?
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Well, because when you have
smaller values of
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y, you're down here.
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And it's more likely because
there are more values of x
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that go along with it that
make that value of y more
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likely to appear.
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Say you have a large
value of y.
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Then you're up here
at the tip.
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Well, there aren't very many
combinations of x and y that
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give you that large
a value of y.
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And so that large value of
y becomes less likely.
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Another way to think about it
is, when you collapse this
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down, there's a lot more stuff
to collapse down its base.
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There's a lot of x's
to collapse down.
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But up here, there's only a
very little bit of x to
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collapse down.
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And the PDF of y becomes
more skewed towards
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smaller values of y.
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So now, the next thing that we
want to do is calculate the
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conditional PDF of x, given y.
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Well, let's just recall
what that means.
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This is what we're looking for--
the conditional PDF of
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x, given y.
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And remember, this is calculated
by taking the joint
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and dividing by the
marginal of y.
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So we actually have the
top and the bottom.
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We have to joint PDF from part
A. And from part B, we
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calculated the marginal
PDF of y.
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So we have both pieces.
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So let's actually
plug them in.
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Again, the thing that you have
to be careful here is about
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the ranges of x and y where
these things are valid,
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because this is only non-zero
when x and y
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fall within this triangle.
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And this is only non-zero when
y is between 0 and 1.
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So we need to be careful.
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So the top, when it's
non-zero, it's 2.
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And the bottom, when it's
non-zero, it's 2
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times 1 minus y.
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So we can simplify that to
be 1 over 1 minus y.
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And when is this true?
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Well, it's true when x and y are
in the triangle and y is
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between 0 and 1.
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So put another way, that means
that this is valid when y is
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between 0 and 1 and x is between
0 and 1 minus y,
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because whatever x has to be,
it has to be such that they
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actually still fall within
this triangle.
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And outside of this, it's 0.
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So let's see what this
actually looks like.
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So this is x, and this is the
conditional PDF of x, given y.
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Let's say this is
1 right here.
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Then what it's saying is, let's
say we're given that y
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is some little y.
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Let's say it's somewhere here.
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Then it's saying that the
conditional PDF of x given y
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is this thing.
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But notice that this value,
1 over 1 minus y, does not
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depend on x.
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So in fact, it actually
is uniform.
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So it's uniform between
0 and 1 minus y.
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And the height is something
like 1 over 1 minus y.
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And this is so that the scaling
makes it so that
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actually is a valid PDF, because
the integral is to 1.
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So why is the case?
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Why is that when you condition
on y being some value, you get
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that the PDF of x is
actually uniform?
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Well, when you look over here,
let's again just pretend that
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you're taking this value of y.
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Well, when you're conditioning
on y being this value, you're
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basically taking a slice of this
joint PDF at this point.
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But remember, the original
joint PDF was uniform.
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So when you take a slice of a
uniform distribution, joint
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uniform distribution,
you still get
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something that is uniform.
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Just imagine that you have
a cake that is flat.
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Now, you take a slice
at this level.
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Then whatever slice you have
is also going to be imagine
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being a flat rectangle.
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So it's still going
to be uniform.
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And that's why the conditional
PDF of x
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given y is also uniform.
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Part D now asks us to find a
conditional expectation of x.
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So we want to find the
expectation of x, given that y
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is some little y.
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And for this, we can
use the definition.
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Remember, expectations are
really just weighted sums.
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Or in the [? continuous ?]
case, it's an integral.
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So you take the value.
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And then you weight
it by the density.
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And in this case, because we're
taking conditional a
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expectation, what we weight it
by is the conditional density.
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So it's the conditional
density of x given
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that y is little y.
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We integrate with
respect to x.
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And fortunately, we know what
this conditional PDF is,
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because we calculated it earlier
in part C. And we know
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that it's this--
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1 over 1 minus y.
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But again, we have to be
careful, because this formula,
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1 over 1 minus y, is only
valid certain cases.
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So let's think about
this first.
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Let's think about some
extreme cases.
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What if y, little
y, is negative?
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If little y is negative,
we're conditioning on
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something over here.
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And so there is no density for
y being negative or for y,
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say, in other cases when
y is greater than 1.
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And so in those cases, this
expectation is just undefined,
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because conditioning on that
doesn't really make sense,
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because there's no density
for those values of y.
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Now, let's consider the case
that actually makes, sense
253
00:14:06,560 --> 00:14:09,120
where y is between 0 and 1.
254
00:14:09,120 --> 00:14:12,500
Now, we're in business, because
that is the range
255
00:14:12,500 --> 00:14:15,010
where this formula is valid.
256
00:14:15,010 --> 00:14:17,870
So this formula is valid,
and we can plug it in.
257
00:14:17,870 --> 00:14:21,760
So it's 1 over 1 minus y dx.
258
00:14:21,760 --> 00:14:24,665
And then the final thing that we
again need to check is what
259
00:14:24,665 --> 00:14:27,270
the limits of this
integration is.
260
00:14:27,270 --> 00:14:29,610
So we're integrating
with respect to x.
261
00:14:29,610 --> 00:14:33,810
So we need to write down what
values of x, what ranges of x
262
00:14:33,810 --> 00:14:36,560
is this conditional PDF valid.
263
00:14:36,560 --> 00:14:39,070
Well, luckily, we specified
that here.
264
00:14:39,070 --> 00:14:43,072
x has to be between
0 and 1 minus y.
265
00:14:43,072 --> 00:14:46,200
266
00:14:46,200 --> 00:14:52,310
So let's actually calculate
this integral.
267
00:14:52,310 --> 00:14:56,660
This 1 over 1 minus y is a
constant with respect to x.
268
00:14:56,660 --> 00:14:58,090
You can just pull that out.
269
00:14:58,090 --> 00:15:01,130
And then now, you're really
just integrating x from
270
00:15:01,130 --> 00:15:02,970
0 to 1 minus y.
271
00:15:02,970 --> 00:15:06,530
So the integral of x is
[? 1 ?], 1/2x squared.
272
00:15:06,530 --> 00:15:10,960
So you get a 1/2x squared, and
you integrate that from
273
00:15:10,960 --> 00:15:12,580
0 to 1 minus y.
274
00:15:12,580 --> 00:15:14,600
And so when you plug in
the limits, you'll
275
00:15:14,600 --> 00:15:18,200
get a 1 minus y squared.
276
00:15:18,200 --> 00:15:20,550
That will cancel out the
1 over 1 minus y.
277
00:15:20,550 --> 00:15:24,570
And what you're left with is
just 1 minus y over 2.
278
00:15:24,570 --> 00:15:28,600
279
00:15:28,600 --> 00:15:31,780
And again, we have to specify
that this is only true for y
280
00:15:31,780 --> 00:15:34,650
between 0 and 1.
281
00:15:34,650 --> 00:15:39,720
Now, we can again actually
verify that this makes sense.
282
00:15:39,720 --> 00:15:42,450
What we're really looking for is
the conditional expectation
283
00:15:42,450 --> 00:15:44,870
of x given some value of y.
284
00:15:44,870 --> 00:15:47,750
And we already said that
condition on y being some
285
00:15:47,750 --> 00:15:50,800
value of x is uniformly
distributed between
286
00:15:50,800 --> 00:15:52,700
0 and 1 minus y.
287
00:15:52,700 --> 00:15:55,190
And so remember for our uniform
distribution, the
288
00:15:55,190 --> 00:15:56,140
expectation is simple.
289
00:15:56,140 --> 00:15:57,310
It's just the midpoint.
290
00:15:57,310 --> 00:16:00,550
So the midpoint of 0
and 1 minus y is
291
00:16:00,550 --> 00:16:02,350
exactly 1 minus y/2.
292
00:16:02,350 --> 00:16:05,340
So that's a nice way of
verifying that this answer is
293
00:16:05,340 --> 00:16:06,590
actually correct.
294
00:16:06,590 --> 00:16:09,450
295
00:16:09,450 --> 00:16:15,740
Now, the second part of
part D asks us to do
296
00:16:15,740 --> 00:16:16,990
a little bit more.
297
00:16:16,990 --> 00:16:20,060
298
00:16:20,060 --> 00:16:26,000
We have to use the total
expectation theorem in order
299
00:16:26,000 --> 00:16:29,510
to somehow write the expectation
of x in terms of
300
00:16:29,510 --> 00:16:30,760
the expectation of y.
301
00:16:30,760 --> 00:16:33,560
302
00:16:33,560 --> 00:16:36,510
So the first thing we'll
do is use the
303
00:16:36,510 --> 00:16:37,880
total expectation theorem.
304
00:16:37,880 --> 00:16:41,630
So the total expectation theorem
is just saying, well,
305
00:16:41,630 --> 00:16:44,970
we can take these conditional
expectations.
306
00:16:44,970 --> 00:16:54,420
And now, we can integrate this
by the marginal density of y,
307
00:16:54,420 --> 00:16:58,080
then we'll get the actual
expectation of x.
308
00:16:58,080 --> 00:17:02,115
You can think of it as just kind
of applying the law of
309
00:17:02,115 --> 00:17:03,365
iterated expectations as well.
310
00:17:03,365 --> 00:17:06,490
311
00:17:06,490 --> 00:17:12,990
So this integral is going
to look like this.
312
00:17:12,990 --> 00:17:17,329
You take the conditional
expectation.
313
00:17:17,329 --> 00:17:22,569
So this is the expectation of x
if y were equal to little y.
314
00:17:22,569 --> 00:17:26,260
And now, what is that
probability?
315
00:17:26,260 --> 00:17:30,370
Well, now we just multiply that
by the density of y at
316
00:17:30,370 --> 00:17:32,830
that actual value of little y.
317
00:17:32,830 --> 00:17:34,310
And we integrate with
respect to y.
318
00:17:34,310 --> 00:17:37,470
319
00:17:37,470 --> 00:17:39,540
Now, we've already calculated
what this conditional
320
00:17:39,540 --> 00:17:40,700
expectation is.
321
00:17:40,700 --> 00:17:42,100
It's 1 minus y/2.
322
00:17:42,100 --> 00:17:45,430
So let's plug that in.
323
00:17:45,430 --> 00:17:49,680
1 minus y/2 times the
marginal of y.
324
00:17:49,680 --> 00:17:55,540
325
00:17:55,540 --> 00:17:58,010
There's a couple ways of
attacking this problem now.
326
00:17:58,010 --> 00:18:00,310
One way is, we can actually
just plug in
327
00:18:00,310 --> 00:18:01,490
that marginal of y.
328
00:18:01,490 --> 00:18:06,710
We've already calculated that
out in part B. And then we can
329
00:18:06,710 --> 00:18:09,570
do this integral and calculate
out the expectation.
330
00:18:09,570 --> 00:18:13,230
But maybe we don't really want
to do so much calculus.
331
00:18:13,230 --> 00:18:15,190
So let's do what the
problem says and
332
00:18:15,190 --> 00:18:16,710
try a different approach.
333
00:18:16,710 --> 00:18:20,870
So what the problem suggests is
to write this in terms of
334
00:18:20,870 --> 00:18:22,450
the expectation of y.
335
00:18:22,450 --> 00:18:23,960
And what is the expectation
of y?
336
00:18:23,960 --> 00:18:28,370
Well, the expectation of y is
going to look something like
337
00:18:28,370 --> 00:18:33,010
the integral of y times
the marginal of y.
338
00:18:33,010 --> 00:18:35,220
So let's see if we can identify
something like that
339
00:18:35,220 --> 00:18:36,500
and pull it out.
340
00:18:36,500 --> 00:18:38,630
Well, yeah, we actually
do have that.
341
00:18:38,630 --> 00:18:42,070
We have y times the marginal
of y, integrated.
342
00:18:42,070 --> 00:18:44,700
So let's isolate that.
343
00:18:44,700 --> 00:18:48,660
So besides that, we
also have this.
344
00:18:48,660 --> 00:18:57,930
We have the integral of the
first term, is 1/2 times the
345
00:18:57,930 --> 00:18:59,320
marginal of y.
346
00:18:59,320 --> 00:19:04,320
And then the second term is
minus 1/2 times the integral
347
00:19:04,320 --> 00:19:10,580
of y of dy.
348
00:19:10,580 --> 00:19:13,000
This is just me splitting
this integral up into
349
00:19:13,000 --> 00:19:15,980
two separate integrals.
350
00:19:15,980 --> 00:19:17,690
Now, we know what this is.
351
00:19:17,690 --> 00:19:18,950
The 1/2 we can pull out.
352
00:19:18,950 --> 00:19:21,870
And then the rest of it is
just the integral of a
353
00:19:21,870 --> 00:19:24,540
marginal of a density from minus
infinity to infinity.
354
00:19:24,540 --> 00:19:27,810
And by definition, that
has to be equal to 1.
355
00:19:27,810 --> 00:19:32,250
So this just gives us a 1/2.
356
00:19:32,250 --> 00:19:33,820
And now, what is this?
357
00:19:33,820 --> 00:19:35,130
We get a minus 1/2.
358
00:19:35,130 --> 00:19:39,290
And now this, we already said
that is the expectation of y.
359
00:19:39,290 --> 00:19:44,060
So what we have is the
expectation of y.
360
00:19:44,060 --> 00:19:49,500
So in the second part of this
part D, we've expressed the
361
00:19:49,500 --> 00:19:53,050
expectation of x in terms
of the expectation of y.
362
00:19:53,050 --> 00:19:56,520
Now, maybe that seems like
that's not too helpful,
363
00:19:56,520 --> 00:19:59,710
because we don't know what
either of those two are.
364
00:19:59,710 --> 00:20:04,170
But if we think about this
problem, and as part E
365
00:20:04,170 --> 00:20:07,190
suggests, we can see that
there's symmetry in this
366
00:20:07,190 --> 00:20:12,420
problem, because x and y are
essentially symmetric.
367
00:20:12,420 --> 00:20:16,740
So imagine this is x equals y.
368
00:20:16,740 --> 00:20:20,430
There's symmetry in this
problem, because if you were
369
00:20:20,430 --> 00:20:24,530
to swap the roles of x and y,
you would have exactly the
370
00:20:24,530 --> 00:20:26,510
same joint PDF.
371
00:20:26,510 --> 00:20:31,660
So what that suggests is that
by symmetry then, it must be
372
00:20:31,660 --> 00:20:36,060
that the expectation of x and
the expectation of y are
373
00:20:36,060 --> 00:20:39,230
exactly the same.
374
00:20:39,230 --> 00:20:41,080
And that is using the
symmetry argument.
375
00:20:41,080 --> 00:20:45,850
And that helps us now, because
we can plug that in and solve
376
00:20:45,850 --> 00:20:46,920
for expectation of x.
377
00:20:46,920 --> 00:20:54,340
So expectation of x is 1/2 minus
1/2 expectation of x.
378
00:20:54,340 --> 00:21:00,860
So we have 3/2 expectation
of x equals 1/2.
379
00:21:00,860 --> 00:21:05,810
So expectation of
x equals 1/3.
380
00:21:05,810 --> 00:21:09,770
And of course, expectation
of y is also 1/3.
381
00:21:09,770 --> 00:21:14,970
And so it turns out that the
expectation is around there.
382
00:21:14,970 --> 00:21:17,760
383
00:21:17,760 --> 00:21:21,220
So this problem had
several parts.
384
00:21:21,220 --> 00:21:26,120
And it allowed us to start
out from just a raw joint
385
00:21:26,120 --> 00:21:28,840
distribution, calculate
marginals, calculate
386
00:21:28,840 --> 00:21:31,650
conditionals, and then from
there, calculate all kinds of
387
00:21:31,650 --> 00:21:34,530
conditional expectations
and expectations.
388
00:21:34,530 --> 00:21:39,020
And a couple of important points
to remember are, when
389
00:21:39,020 --> 00:21:42,620
you do these joint
distributions, it's very
390
00:21:42,620 --> 00:21:47,230
important to consider where
values are valid.
391
00:21:47,230 --> 00:21:50,340
So you have to keep in mind
when you write out these
392
00:21:50,340 --> 00:21:55,620
conditional PDFs and joint PDFs
and marginal PDFs, what
393
00:21:55,620 --> 00:21:59,590
ranges the formulas you
calculated are valid for.
394
00:21:59,590 --> 00:22:02,880
And that also translates to
when you're calculating
395
00:22:02,880 --> 00:22:03,930
expectations and such.
396
00:22:03,930 --> 00:22:06,805
When you have integrals, you
need to be very careful about
397
00:22:06,805 --> 00:22:08,590
the limits of your integration,
to make sure that
398
00:22:08,590 --> 00:22:12,290
they line up with the range
where the values
399
00:22:12,290 --> 00:22:13,860
are actually valid.
400
00:22:13,860 --> 00:22:17,180
And the last thing, which is
kind of unrelated, but it is
401
00:22:17,180 --> 00:22:20,890
actually a common tool that's
used in a lot of problems is,
402
00:22:20,890 --> 00:22:25,170
when you see symmetry in these
problems, that can help a lot,
403
00:22:25,170 --> 00:22:28,590
because it will simplify things
and allow you to use
404
00:22:28,590 --> 00:22:31,330
facts like these to help
you calculate what the
405
00:22:31,330 --> 00:22:32,190
final answer is.
406
00:22:32,190 --> 00:22:35,520
Of course, this is also comes
along with practice.
407
00:22:35,520 --> 00:22:37,620
You may not immediately see that
there could be a symmetry
408
00:22:37,620 --> 00:22:39,120
argument that will help
with this problem.
409
00:22:39,120 --> 00:22:42,680
But with practice, when you do
more of these problems, you'll
410
00:22:42,680 --> 00:22:44,440
eventually build up
that kind of--
411
00:22:44,440 --> 00:22:45,690