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Welcome back guys.
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Today we're going to work on
a problem that tests your
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knowledge of joint PMFs.
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And we're also going to get
some practice computing
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conditional expectations and
conditional variances.
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So in this problem, we
are given a set of
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points in the xy plane.
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And we're told that these points
are equally likely.
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So there's eight of them.
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And each point has a
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probability of 1/8 of occurring.
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And we're also given this
list of questions.
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And we're going to work
through them together.
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So in part a, we are asked to
find the values of x that
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maximize the conditional
expectation of y given x.
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So jumping right in, this
is the quantity
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we're interested in.
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And so this quantity
is a function of x.
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You plug-in various
values of x.
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And then this will spit
out a scalar value.
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And that value will correspond
to the conditional expectation
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of y conditioned on the value
of x that you put in.
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So let's see, when x is equal
to 0, for instance, let's
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figure out what this value is.
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Well, when x is equal to 0
we're living in a world,
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essentially, on this line.
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So that means that
only these two
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points could have occurred.
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And in particular, y can only
take on the values of 1 and 3.
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Now, since all these points in
the unconditional universe
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were equally likely, in the
conditional universe they will
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still be equally likely.
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So this happens with
probability 1/2.
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And this happens with
probability 1/2.
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And therefore, the expectation
would just be 3/2 plus 1/2
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which is 4/2, or 2.
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But a much faster way of seeing
this-- and it's the
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strategy that I'm going to use
for the rest of the problem--
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is to remember that
expectation acts
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like center of mass.
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So the center of mass, when
these two points are equally
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likely, is just the midpoint,
which of course is 2.
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So we're going to use that
intuition on the other ones.
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So I'm skipping to x is
equal to 2 because 1
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and 3 are not possible.
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So when x is equal to 2,
y can only take on the
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values of 1 or 2.
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Again, they're equally likely.
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So the center of mass is in the
middle which happens at
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1.5 or 3/2.
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Similarly, x is equal to 4.
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We're living in this conditional
universe, where y
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can take on of these
four points with
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probability 1/4 each.
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And so again, we expect
the center of mass to
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be at 1.5 or 3/2.
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And this quantity is undefined
otherwise.
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OK, so we're almost done.
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Now we just need to find which
value of x maximizes this.
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Well, let's see, 2 is the
biggest quantity out of all of
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these numbers.
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So the maximum is 2.
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And it occurs when
x is equal to 0.
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So we come over here.
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And we found our answer.
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x is equal to 0 is the value,
which maximizes the
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conditional expectation
of y given x.
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So part b is very similar
to part a.
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But there is slightly more
computation involved.
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Because now we're dealing with
the variance and not an
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expectation.
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And variance is usually a little
bit tougher to compute.
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So we're going to start
in the same manner.
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But I want you guys to see if
you can figure out intuitively
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what the right value is.
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I'm going to do the entire
computation now.
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And then you can compare whether
your intuition matches
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with the real results.
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So variance of x conditioned
on a particular value of y,
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this is now a function of y.
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For each value of y you plug in
you're going to get out a
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scalar number.
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And that number represents the
conditional variance of x when
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you condition on the value
of y that you plugged in.
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So let's see, when y is equal
to 0 we have a nice case.
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If y is equal to 0 we have no
freedom about what x is.
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This is the only point that
could have occurred.
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Therefore, x definitely
takes on a value of 4.
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And there's no uncertainty
left.
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So in other words, the
variance is 0.
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Now, if y is equal to 1, x can
take on a value of 0, a value
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of 2 or a value of 4.
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And these all have the same
probability of occurring, of
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1/3,
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And again, the reasoning behind
that is that all eight
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points were equally likely in
the unconditional universe.
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If you condition on y being
equal to 1 these outcomes
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still have the same relative
frequency.
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Namely, they're still
equally likely.
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And since there are three
of them they now have a
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probability of 1/3 each.
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So we're going to go ahead and
use a formula that hopefully,
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you guys remember.
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So in particular, variance is
the expectation of x squared
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minus the expectation
of x all squared,
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the whole thing squared.
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So let's start by computing
this number first.
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So conditioned on y
is equal to 1--
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so we're in this line--
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the expectation of x
is just 2, right?
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The same center-of-mass
to argument.
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So this, we have a minus
2 squared over here.
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Now, x squared is only slightly
more difficult.
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With probability 1/3,
x squared will take
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on a value of 0.
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With probability 1/3,
x squared will take
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on a value of 4.
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I'm just doing 2 squared.
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And with probability 1/3, x
squared takes on a value of 4
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squared or 16.
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So writing down when I just
said, we have 0 times 1/3
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which is 0.
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We have 2 squared, which
is 4 times 1/3.
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And then we have 4 squared,
which is 16 times 1/3.
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And then we have our minus
4 from before.
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So doing this math out, we get,
let's see, 20/3 minus
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12/3, which is equal
to 8/3, or 8/3.
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So we'll come back up
here and put 8/3.
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So I realize I'm going through
this pretty quickly.
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Hopefully this step didn't
confuse you.
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Essentially, when I was doing
is, if you think of x squared
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as a new random variable, x
squared, the possible values
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that it can take on are 0,
4, and 16 when you're
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conditioning on y
is equal to 1.
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And so I was simply saying
that that random variable
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takes on those values with
equal probability.
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So let's move on to
the next one.
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So if we condition on y is equal
to 2 we're going to do a
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very similar computation.
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Oops, I shouldn't have
erased that.
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OK, so we're going to use the
same formula that we just
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used, which is the expectation
of x given y is equal to 2.
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Sorry, x squared minus the
expectation of x conditioned
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on y is equal to
2, all squared.
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So conditioned on y
is equal to 2, the
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expectation of x is 3.
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Same center of mass argument.
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So 3 squared is 9.
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And then x squared can
take on a value of 4.
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Or it can take on
a value of 16.
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And it does so with
equal probability.
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So we get 4/2, 4
plus 16 over 2.
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So this is 2 plus 8, which
is 10, minus 9.
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That'll give us 1.
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So we get a 1 when
y is equal to 2.
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And last computation and
then we're done.
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I'm still recycling
the same formula.
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But now we're conditioning
on y is equal to 3.
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And then we'll be done with
this problem, I promise.
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OK, so when y is equal to 3 x
can take on the value of 0.
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Or it can take on
the value of 4.
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Those two points happen with
probability 1/2, 1/2.
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So the expectation is right
in the middle which is 2.
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So we get a minus 4.
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And similarly, x squared can
take on the value of 0.
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When x takes on the value of
0-- and that happens with
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probability 1/2--
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similarly, x squared can take
on the value of 16 when x
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takes on the value of 4.
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And that happens with
probability 1/2.
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So we just have 0/2
plus 16/2 minus 4.
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And this gives us 8 minus
4, which is simply 4.
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So finally, after all that
computation, we are done.
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We have the conditional
variance of x given y.
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Again, we're interested in when
this value is largest.
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And we see that 4 is the biggest
value in this column.
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And this value occurs when
y takes on a value of 3.
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So our answer, over here,
is y is equal to 3.
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All right, so now we're going to
switch gears in part c and
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d a little bit.
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And we're going to
be more concerned
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with PMFs, et cetera.
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So in part c, we're given a
random variable called r which
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is defined as the minimum
of x and y.
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So for instance, this
is the 0.01.
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The minimum of 0 and 1 is 0.
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So r would have a
value of 0 here.
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Now, we can be a little bit
smarter about this.
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If we plot the line,
y is equal to x.
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So that looks something
like this.
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We see that all of the points
below this line satisfy y
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being less or equal to x.
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And all the points above this
line have y greater than or
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equal to x.
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So if y is less than or equal to
x, you hopefully agree that
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here the min, or r,
is equal to y.
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But over here, the min, r, is
actually equal to x, since x
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is always smaller.
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So now we can go
ahead quickly.
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And I'm going to write the value
of r next each point
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using this rule.
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So here, r is the value
of y, which is 1.
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Here, r is equal to 0.
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Here r is 1.
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Here r is 2.
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Here r is 3.
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Over here, r is the
value of x.
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So r is equal to 0.
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And r is equal to 0 here.
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And so the only point we didn't
handle is the one that
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lies on the line.
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But in that case it's easy.
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Because x is equal to 2.
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And y is equal to 2.
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So the min is simply 2.
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So with this information
I claim we're now done.
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We can just write down
what the PMF of r is.
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So in particular, r takes
on a value of 0.
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When this point happens,
this point happens,
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or this point happens.
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And those collectively have a
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probability of 3/8 of occurring.
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r can take on a value of 1
when either of these two
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points happen.
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So that happens with
probability 2/8.
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r is equal to 2.
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This can happen in two ways.
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So we get 2/8.
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And r equal to 3 can happen
in only one way.
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So we get 1/8.
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Quick sanity check, 3 plus
2 is 5, plus 2 is
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7, plus 1 is 8.
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So our PMF sums to 1.
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And to be complete, we
should sketch it.
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Because the problem asks
us to sketch it.
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So we're plotting PR
of r, 0, 1, 2, 3.
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So here we get, let's
see, 1, 2, 3.
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For 0 we have 3/8.
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For 1 we have 2/8.
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For 2 we have 2/8.
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And for 3 we have 1/8.
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So this is our fully labeled
sketch of Pr of r.
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And forgive me for erasing so
quickly, but you guys can
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pause the video, presumably,
if you need more time.
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Let's move on to part d.
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So in part d we're given an
event named a, which is the
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event that x squared is greater
than or equal to y.
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And then we're asked to find the
expectation of xy in the
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unconditional universe.
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And then the expectation of x
times y conditioned on a.
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So let's not worry about the
conditioning for now.
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Let's just focus on
the unconditional
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expectation of x times y.
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So I'm just going to
erase all these r's
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so I don't get confused.
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But we're going to follow a very
similar strategy, which
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is at each point I'm going to
label what the value of w is.
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And we'll find the expectation
of w that way.
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So let's see, here,
we have 4 times 0.
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So w is equal to 0.
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Here we have 4 times 1.
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w is equal to 4.
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4 times 2, w is equal to 8.
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4 times 3, w is equal to 12.
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w is equal to 2.
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w is equal to 4.
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w is equal to 0.
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w is equal to 0.
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OK, so that was just algebra.
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And now, I claim again, we can
just write down what the
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expectation of x times y is.
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And I'm sorry, I didn't
announce my notation.
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I should mention that now.
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I was defining w to be the
random variable x times y.
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And that's why I labeled
the product of x
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times y as w over here.
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My apologies about not defining
that random variable.
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So the expectation of w, well,
w takes on a value of 0.
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When this happens, this happens
or that happens.
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00:14:23,680 --> 00:14:28,920
And we know that those
three points occur
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with probability 3/8.
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So we have 0 times 3/8.
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I'm just using the normal
formula for expectation.
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00:14:38,080 --> 00:14:40,815
w takes on a value of 2
with probability 1/8.
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00:14:40,815 --> 00:14:43,590
Because this is the lead
point in which it
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00:14:43,590 --> 00:14:46,220
happens, 2 times 1/8.
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Plus it can take on the value
of 4 with probability
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2/8, 4 times 2/8.
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And 8, with 1/8 probability.
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And similarly, 12 with
1/8 probability.
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So this is just algebra.
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The numerator sums up to 30.
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00:15:06,674 --> 00:15:08,445
Yes, that's correct.
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So we have 30/8, which
is equal to 15/4.
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00:15:13,920 --> 00:15:18,390
So this is our first
answer for part d.
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And now we have to do this
slightly trickier one, which
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is the conditional expectation
of x times y, or w
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conditioned on a.
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00:15:26,110 --> 00:15:31,360
So similar to what I did in part
c, I'm going to draw the
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line y equals x squared.
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00:15:34,510 --> 00:15:37,870
So y equals x squared
is 0 here, 1 here.
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00:15:37,870 --> 00:15:41,070
And at 2, it should take
on a value of 4.
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00:15:41,070 --> 00:15:45,760
So the curve should look
something like this.
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This is the line y is
equal to x squared.
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So we know all the points below
this line satisfy y less
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than or equal to x squared.
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And all the points above this
line have y greater than or
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equal to x squared.
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00:15:59,170 --> 00:16:02,870
And a is y less than or
equal to x squared.
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So we are in the conditional
universe where only points
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below this line can happen.
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So that one, that one, that
one, that one, that
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00:16:11,240 --> 00:16:12,190
one and that one.
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So there are six of them.
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And again, in the unconditional
world, all of
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the points were equally
likely.
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00:16:17,190 --> 00:16:20,040
So in the conditional world
these six points are still
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00:16:20,040 --> 00:16:21,040
equally likely.
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00:16:21,040 --> 00:16:24,630
So they each happen with
probability 1/6.
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00:16:24,630 --> 00:16:33,310
So in this case, the expectation
of w is simply 2
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00:16:33,310 --> 00:16:38,050
times 1/6 plus 0 times 1/6.
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00:16:38,050 --> 00:16:38,930
But that's 0.
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00:16:38,930 --> 00:16:41,260
So I'm not going to write it.
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00:16:41,260 --> 00:16:52,250
4 times 2/6 plus 4 times 2/6
plus 8 times 1/6, plus 12
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00:16:52,250 --> 00:16:54,820
times 1 over 6.
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00:16:54,820 --> 00:16:56,650
And again, the numerator
summed to 30.
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00:16:56,650 --> 00:16:58,780
But this time our denominator
is 6.
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00:16:58,780 --> 00:17:01,210
So this is simply 5.
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00:17:01,210 --> 00:17:04,230
So we have, actually, finished
the problem.
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00:17:04,230 --> 00:17:07,940
Because we've computed this
value and this value.
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00:17:07,940 --> 00:17:12,369
And so the important takeaways
of this problem are,
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00:17:12,369 --> 00:17:16,089
essentially, honestly, just to
get you comfortable with
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00:17:16,089 --> 00:17:19,410
computing things involving
joint PMFs.
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00:17:19,410 --> 00:17:23,109
We talked a lot about finding
expectations quickly by
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00:17:23,109 --> 00:17:25,180
thinking about center
of mass and the
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00:17:25,180 --> 00:17:27,030
geometry of the problem.
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00:17:27,030 --> 00:17:31,461
We've got practice computing
conditional variances.
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00:17:31,461 --> 00:17:34,330
And we did some derived
distributions.
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00:17:34,330 --> 00:17:36,170
And we'll do a lot more
of those later.
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00:17:36,170 --> 00:17:37,420