WEBVTT

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Hi.

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In this session, we're going to
cover a nice review problem

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that will look at how
to infer one random

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variable based on another.

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And in this problem, we're given
two random variables--

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X and Y--

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and we're also given their joint
pdf, which we're told is

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a constant 2/3 within the
region bounded by

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these orange lines.

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And outside of the region,
the joint pdf is 0.

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So the first thing we're going
to look at, or the first thing

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that we're asked to do, is find
the LMS estimator of Y

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based on X. Now, remember that
LMS estimator is really just a

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conditional expectation.

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So the LMS estimator of Y based
on X is the conditional

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expectation of Y, given X.

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Now, when we have a plot of
the joint pdf and we're

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dealing with these two random
variables, and especially when

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the joint pdf is constant like
this, it's often easy to

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calculate this conditional
expectation visually.

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So what we really need to do
is just say, given any

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particular value of X, what
is the conditional

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expectation of Y?

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So what we can do is we can just
pick some values of X and

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see visually what that initial
expectation is.

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So for example, if X is 1/2,
given that X is 1/2, and since

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this whole joint pdf is uniform,
then the conditional

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slice of Y will be from
here to here.

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And that slice, the conditional
distribution of Y,

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given that X is 1/2, will
also be uniform.

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So it'll be uniform
from here to here.

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And if it's uniform, we know
that the conditional

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expectation will just be
the midpoint here.

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And so that would be what the
conditional expectation of Y

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would be, given that X is 1/2.

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And we could do the same
thing for X equals 1.

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And we'll see that again,
because everything is uniform,

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this slice is also going
to be uniform.

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And so the conditional
expectation will again be the

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midpoint, which is there.

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And then if we just look at it
within this region, it's

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always going to be
the midpoint.

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And so we get that the initial
expectation of Y, given X,

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will just look like that line,
which you can think of it as

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just bisecting this angle formed
by these two parts of

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the region.

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But things are a little bit
different, though, when we

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move to the region where
X is between 1 and 2.

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Between 1 and 2, say
at 1 and 1/2, this

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line doesn't continue.

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Because now, the slice of Y goes
from here to here, and

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again, it's still uniform.

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So the midpoint would
be there.

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And similarly for X equals
2, it would be here.

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And so for X between 1 and 2,
the conditional expectation

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actually looks like this.

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So you see that there's actually
two linear parts of

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it, but there's a kink
at X equals 1.

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And so by looking at this
visually and taking advantage

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of the fact that everything is
uniform, we can pretty easily

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figure out what this conditional
expectation is.

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So now, let's actually just
write it out algebraically.

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So for X between 0 and 1, we
said that it's this line,

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which if we look at it, that's
just 1/2 of X. Now, this is

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for X between 0 and 1.

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And if X is between 1 and 2,
it's going to be this line,

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which is a slope of 1.

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And if we extend this
down, it hits the

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y-axis at negative 1/2.

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So it's X minus 1/2, if
X is between 1 and 2.

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And otherwise, it's undefined.

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So we'll focus on these
two cases here.

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Now, the second part of the
question, now that we know

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what the LMS estimator is, we're
asked to find what is

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the traditional mean squared
error of this estimator?

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So we want to know
how good is it.

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And one way of capturing
that is to look at the

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mean squared error.

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And so recall that the
conditional mean squared error

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is given by this expression.

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So what we're saying is this is
what we estimate Y to be.

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This is what y really is, so
this difference is how wrong,

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or the error in our estimate.

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We square it, because otherwise,
positive and

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negative errors might cancel
each other out,

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so we square it.

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And then this just looking
at each individual

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value of x for now.

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So this is why it's
the conditional

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mean squared error.

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So how do we calculate this?

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Well, remember that this g of
X, we said the LMS estimator

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is just a conditional
expectation.

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So it's just expectation
of Y, given X.

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Well, then if you look at this,
what this reminds you

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of, it reminds you of the
definition of what a

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conditional variance is.

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A variance is just, you take the
random variable, subtract

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its mean, square it, and take
the expectation of that.

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This is no different, except
that everything is now the

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conditional world of X.

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So this is actually the
conditional variance of Y,

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given X is little x.

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What is the conditional
variance of Y, given

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that X is little x?

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Now, we can again go back
to this plot to

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try to help us out.

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We can split this up
into regions again.

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So just take some little x as
an example and see what the

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variance is.

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So if little x is 1/2, then we
know that the conditional

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distribution of Y would
be uniform, we said,

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from 0 up to here.

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Well, that point is this
from 0 to 1/2.

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And remember, the variance of a
uniform distribution is just

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the width of the uniform
distribution

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squared, divided by 12.

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And so in this case, the width
would be 1/2 squared over 12.

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And in general, for the region
of X between 0 and 1, the

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width of the conditional
distribution of Y will always

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be X, because the width will
go from 0 to wherever X is.

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So because of that, the
conditional variance will just

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be X squared, the width squared,
over 12, when X is

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between 0 and 1.

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Now, let's think about the
other case, where X is

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between 1 and 2.

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Well, if X is between 1 and
2, we're over here.

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And now, if we take the
conditional distribution of Y,

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it's again uniform.

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But the width now, instead of
varying with Y, it's always

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going to be the same width.

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Each of these slices have the
same width, and the width goes

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from here-- this is X minus
1, and that's X.

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So if the width is always going
to be a constant of 1.

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And so this variance is
going to be 1/12.

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And from that, we get our answer
for the conditional

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mean squared error.

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Now, part c asks us to find the
mean squared error, which

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is given by this expression.

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And we'll see that it looks very
similar to this, which

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was the conditional mean
squared error.

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And now, given what we
know from part b,

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this is easy to calculate.

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We can just apply total
expectation, because this is

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just equal to the integral
of the conditional

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mean squared error.

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And then we need to also
multiply this by the pf of x,

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and then integrate over X. And
that integral will should be

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from X equals 0 to 2, because
that's the only range that

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applies for x, given
this joint pdf.

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Now, in order to do this first,
though, we need to

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figure out what the
pdf of X is.

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In order to do that, we can go
back to our original joint pdf

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of X and Y and marginalize it.

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So marginalizing, you could
think of it as taking this

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joint pdf and collapsing it onto
the x-axis so that you

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take everything and
integrate out Y.

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Now to do that, let's
do that up here.

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We can split it up into
two sections.

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So the section of X between 0
and 1, we integrate the joint

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pdf from Y equals 0 to Y
equals X, which is this

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portion of this line.

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So we integrate Y. The joint pdf
is 2/3, and we integrate Y

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out from Y equals 0 to X. And
then for the portion of X from

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1 to 2, we again integrate
Y out.

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Now we integrate Y from
X minus 1 up to X.

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So this is X between 0 and 1,
and this is X between 1 and 2.

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So we just do some little bit
of calculus, and we get that

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this is going to be 2/3 X when
X is between 0 and 1.

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And it's going to be 2/3 when
X is between 1 and 2.

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So now that we have what the
marginal pdf of X is, we can

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plug that into this, and plug in
what we had for b, and then

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calculate what this
actually is.

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So remember, we need to take
care of these two cases, these

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two regions--

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X between 0 and 1, and
X between 1 and 2.

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So the conditional mean squared
error for X between 0

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and 1 is X squared over 2.

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So between 0 and 1, this first
part is X squared over 12.

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The pdf of X in that same
region is 2/3 x.

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And we integrate that in the
region from x equals 0 to 1.

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And then, we also have the
second region which

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is X from 1 to 2.

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In that region, the traditional
mean squared error

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from part b is 1/12.

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The marginal pdf of X is 2/3,
and we do this integral.

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And if you just carry out some
calculus here, you'll get that

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the final answer is
equal to 5/72.

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Now, the last part of this
question asks us, is this mean

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squared error the same thing--

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does it equal the expectation
of the conditional variance?

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And it turns out that
yes, it does.

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And to see that, we can just
take this, and use the law of

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iterated expectations, because
iterated expectations tells us

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this is in fact equal to the
expectation of Y minus g of X

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squared, given X. That's just
applying law of iterated

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expectations.

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And then, if we look at this,
this part that's inside is

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exactly equal to the conditional
variance of Y,

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given X. And so these two
are, in fact, the same.

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In part c, we calculated what
the marginal pdf of X is, and

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it'll actually be used later
on in this problem.

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So for future reference, let's
just write it down here in

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this corner.

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Now, so far in this
problem, we've

00:13:24.490 --> 00:13:27.920
looked at the LMS estimator.

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And of course, there are
other estimators that

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you can use as well.

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And now in part d, let's look
at the linear LMS estimator.

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Now remember, the linear LMS
estimator is special, because

00:13:42.810 --> 00:13:48.040
it forces the estimator to have
a linear relationship.

00:13:48.040 --> 00:13:51.310
So the estimator is going to
be a linear function of X.

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Now, compare that
to what the LMS

00:13:53.820 --> 00:13:55.590
estimator was in this case.

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It was two linear pieces,
but there was a kink.

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And so the entire estimator
wasn't actually linear in X.

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Now, the LLMS estimator, or the
linear LMS estimator, will

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give us the linear estimator.

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It's going to be a linear
function of X.

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And we know that we have
a formula for this.

00:14:15.910 --> 00:14:23.930
Is the expectation of Y plus the
covariance of X and Y over

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the variance of X times X minus
expectation of X. All

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right, so in order to calculate
what this is, we

00:14:35.350 --> 00:14:38.810
just need to calculate
what four things are.

00:14:38.810 --> 00:14:43.080
Now, let's start with this last
one, the expected value

00:14:43.080 --> 00:14:47.900
of X. To calculate the expected
value of X, we just

00:14:47.900 --> 00:14:48.590
use a formula.

00:14:48.590 --> 00:14:52.010
And from before, we know
what the pdf of X is.

00:14:52.010 --> 00:14:56.910
And so we know that this
is just going to be X

00:14:56.910 --> 00:15:01.160
times fx of x dx.

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And in particular, this will
give us that from 0 to 1, it's

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going to be X times the
pdf of X is 2/3 X,

00:15:13.200 --> 00:15:16.486
so it's 2/3 X squared.

00:15:19.720 --> 00:15:24.950
And from 1 to 2, it's going to
be X times the pdf of X, which

00:15:24.950 --> 00:15:31.200
is just 2/3, so it's 2/3 X dx.

00:15:31.200 --> 00:15:34.000
And when you calculate
this out, you'll get

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that is equal to 11/9.

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Now, let's calculate the
variance of X next.

00:15:41.250 --> 00:15:43.760
In order to calculate that,
let's use the formula that

00:15:43.760 --> 00:15:46.920
variance is equal to the
expectation of X squared minus

00:15:46.920 --> 00:15:49.220
the expectation of X
quantity squared.

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We had the expectation of X,
so let's calculate what the

00:15:52.440 --> 00:15:56.940
expectation of X squared is.

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Now, it's the same idea.

00:15:59.060 --> 00:16:06.060
Instead, we have X squared
times f of X dx.

00:16:06.060 --> 00:16:08.860
And we'll get the same
sort of formula.

00:16:08.860 --> 00:16:11.880
We'll split it up again into
two different parts from X

00:16:11.880 --> 00:16:14.490
equals 0 to X equals 1.

00:16:14.490 --> 00:16:16.260
It's going to be X squared
times pdf, so

00:16:16.260 --> 00:16:20.430
it's 2/3 X cubed dx.

00:16:20.430 --> 00:16:23.490
And then from X equals 1 to
2, it's going to be X

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squared times 2/3.

00:16:25.060 --> 00:16:28.770
So it's 2/3 X squared dx.

00:16:28.770 --> 00:16:30.890
And when we calculate this
out, we'll get that

00:16:30.890 --> 00:16:32.750
it's equal to 31/18.

00:16:35.900 --> 00:16:38.730
From that, we know that the
variance is going to be equal

00:16:38.730 --> 00:16:44.680
to expectation of X squared
minus expectation of X

00:16:44.680 --> 00:16:46.860
quantity squared.

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Now, expectation of X
squared is 31/18.

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Expectation of X is 11/9.

00:16:55.010 --> 00:16:58.010
And when we calculate this,
we get that the

00:16:58.010 --> 00:17:06.400
variance is equal to 37/162.

00:17:06.400 --> 00:17:08.349
So now we have this,
and we have that.

00:17:08.349 --> 00:17:11.170
Let's calculate what expectation
of Y is.

00:17:11.170 --> 00:17:16.190
Expectation of Y, let's
calculate it using the law of

00:17:16.190 --> 00:17:19.310
iterated expectations.

00:17:19.310 --> 00:17:20.944
The law of iterated expectations
tells us that

00:17:20.944 --> 00:17:25.849
this is equal to the expectation
of Y conditioned

00:17:25.849 --> 00:17:27.700
on X.

00:17:27.700 --> 00:17:29.410
Now, we already know what
expectation of Y

00:17:29.410 --> 00:17:30.430
conditioned on X is.

00:17:30.430 --> 00:17:33.400
That was the LMS estimator that
we calculated earlier.

00:17:33.400 --> 00:17:35.770
It's this.

00:17:35.770 --> 00:17:39.760
So now we just need to
calculate this out,

00:17:39.760 --> 00:17:41.650
and we can do that.

00:17:41.650 --> 00:17:48.210
So we know that in the range
from X between 0 and 1, it's

00:17:48.210 --> 00:17:55.720
equal to 1/2 X. So in the range
from 0 to 1, it's equal

00:17:55.720 --> 00:18:04.130
to 1/2 X. But then, we have to
use total expectation, so we

00:18:04.130 --> 00:18:07.940
have to multiply by the pdf
of X in that region

00:18:07.940 --> 00:18:11.790
which is 2/3 X dx.

00:18:11.790 --> 00:18:16.900
And then in the range from
X equals 1 to 2, this

00:18:16.900 --> 00:18:18.840
conditional expectation
is X minus 1/2.

00:18:22.100 --> 00:18:25.440
And the pdf of X in that
region is 2/3.

00:18:30.200 --> 00:18:35.500
Now, when we calculate out this
value, we'll get that

00:18:35.500 --> 00:18:38.060
it's equal to 7/9.

00:18:40.620 --> 00:18:44.770
And now, the last piece is the
covariance of X and Y.

00:18:44.770 --> 00:18:47.760
Remember, the covariance, we
can calculate that as the

00:18:47.760 --> 00:18:52.810
expectation of X times Y minus
the expectation of X times the

00:18:52.810 --> 00:18:56.030
expectation of Y. We already
know the expectation of X and

00:18:56.030 --> 00:18:58.590
the expectation of Y, so we
just need to calculate the

00:18:58.590 --> 00:19:03.212
expectation of X times Y,
the product of the two.

00:19:03.212 --> 00:19:08.740
And for that, we'll use the
definition, and we'll use the

00:19:08.740 --> 00:19:11.990
joint pdf that we have.

00:19:11.990 --> 00:19:16.610
So this is going to be a double
integral of X times Y

00:19:16.610 --> 00:19:17.860
times the joint pdf.

00:19:22.360 --> 00:19:25.920
And the tricky part here is just
figuring out what these

00:19:25.920 --> 00:19:26.430
limits are.

00:19:26.430 --> 00:19:30.312
So we'll integrate
in this order--

00:19:30.312 --> 00:19:32.590
X and Y.

00:19:32.590 --> 00:19:35.880
Now, let's split this up.

00:19:35.880 --> 00:19:38.450
So let's focus on
splitting X up.

00:19:38.450 --> 00:19:46.990
So for X between 0 and 1, we
just need to figure out what's

00:19:46.990 --> 00:19:51.240
the rate right range of Y to
integrate over such that this

00:19:51.240 --> 00:19:52.120
is actually non-zero.

00:19:52.120 --> 00:19:55.160
Because remember, the
joint pdf is easy.

00:19:55.160 --> 00:19:56.230
It's just a constant 2/3.

00:19:56.230 --> 00:19:58.660
But it's only a constant
2/3 within this region.

00:19:58.660 --> 00:20:01.110
So the difficult part is just
figuring out what the limits

00:20:01.110 --> 00:20:03.100
are in order to specify
that region.

00:20:03.100 --> 00:20:11.410
So for X between 0 and 1, Y
has to be between 0 and X,

00:20:11.410 --> 00:20:18.100
because this line is Y equals
X. So we need to integrate

00:20:18.100 --> 00:20:19.926
from 0 to X--

00:20:19.926 --> 00:20:23.595
X times Y times the joint
pdf, which is 2/3.

00:20:28.260 --> 00:20:34.380
And now, let's do the other
part, which is X from 1 to 2.

00:20:34.380 --> 00:20:38.170
Well, if X is from 1 to 2, in
order to fall into this

00:20:38.170 --> 00:20:43.210
region, Y has to be between X
minus 1 and X. So we integrate

00:20:43.210 --> 00:20:48.110
Y from X minus 1 to X. Against,
it's X times Y times

00:20:48.110 --> 00:20:49.640
the joint pdf, which is 2/3.

00:20:54.070 --> 00:20:57.360
And now, once we have this set
up, the rest of it we can just

00:20:57.360 --> 00:20:58.400
do some calculus.

00:20:58.400 --> 00:21:00.680
And what we find is
that the final

00:21:00.680 --> 00:21:03.940
answer is equal to 41/36.

00:21:07.940 --> 00:21:13.640
Now, what that tells us is that
the covariance of X and

00:21:13.640 --> 00:21:19.430
Y, which is just expectation
of X times Y, the product,

00:21:19.430 --> 00:21:25.080
minus expectation of X times
expectation of Y.

00:21:25.080 --> 00:21:27.640
We know expectation
of X times Y now.

00:21:27.640 --> 00:21:29.440
It's 41/36.

00:21:29.440 --> 00:21:31.920
Expectation of X is 11/9.

00:21:31.920 --> 00:21:34.640
Expectation of Y is 7/9.

00:21:34.640 --> 00:21:37.050
So when we substitute all of
that in, we get that this

00:21:37.050 --> 00:21:43.830
covariance is 61/324.

00:21:43.830 --> 00:21:45.910
All right, so now we have
everything we need.

00:21:45.910 --> 00:21:48.760
Expectation of Y is here.

00:21:48.760 --> 00:21:51.620
Covariance is here.

00:21:51.620 --> 00:21:54.360
Variance of X is here.

00:21:54.360 --> 00:21:57.210
And expectation of X is here.

00:21:57.210 --> 00:22:01.970
So let's substitute that in, and
we can figure out what the

00:22:01.970 --> 00:22:05.410
actual LLMS estimator is.

00:22:05.410 --> 00:22:09.950
So expectation of Y
we know is 7/9.

00:22:09.950 --> 00:22:13.570
Expectation of X is 11/9.

00:22:16.800 --> 00:22:20.980
And when you divide the
covariance, which is 61/324,

00:22:20.980 --> 00:22:32.040
by the variance, which is
37/162, we'll get 61/74.

00:22:32.040 --> 00:22:36.290
And so that is the LLMS
estimator that we calculated.

00:22:36.290 --> 00:22:40.760
And notice that it is,
in fact, linear in X.

00:22:40.760 --> 00:22:44.720
So let's plot that and see
what it looks like.

00:22:44.720 --> 00:22:52.300
So it's going to be a
line, and it's going

00:22:52.300 --> 00:22:54.610
to look like this.

00:22:54.610 --> 00:23:00.700
So at X equals 2, it's actually
a little bit below 1

00:23:00.700 --> 00:23:05.360
and 1/2, which is what the
LMS estimator would be.

00:23:05.360 --> 00:23:13.010
At X equals 1, it's actually a
little bit above 1/2, which is

00:23:13.010 --> 00:23:15.150
what the LMS estimator
would be.

00:23:15.150 --> 00:23:21.120
And then it crosses 0 around
roughly 1/4, and it drops

00:23:21.120 --> 00:23:22.160
actually below 0.

00:23:22.160 --> 00:23:26.040
So if we connect the dots,
it's going to look

00:23:26.040 --> 00:23:28.820
something like this.

00:23:28.820 --> 00:23:32.230
So notice that it's actually not
too far away from the LMS

00:23:32.230 --> 00:23:33.850
estimator here.

00:23:33.850 --> 00:23:38.340
But it doesn't have the kink
because it is a line.

00:23:38.340 --> 00:23:41.700
And note also that it actually
drops below.

00:23:41.700 --> 00:23:45.070
So when X is very small, you
actually estimate negative

00:23:45.070 --> 00:23:51.880
values of Y, which is actually
impossible, given the joint

00:23:51.880 --> 00:23:53.410
pdf distribution that
we're given.

00:23:53.410 --> 00:23:57.950
And that is sometimes a feature
or artifact of the

00:23:57.950 --> 00:24:00.530
linear LMS estimator, that
you'll get values that don't

00:24:00.530 --> 00:24:03.820
necessarily seem
to make sense.

00:24:03.820 --> 00:24:06.720
So now that we've calculated
the linear LMS estimator in

00:24:06.720 --> 00:24:10.740
part d, which is this, and the
LMS estimator in part a, which

00:24:10.740 --> 00:24:14.420
is this, we've also compared
them visually.

00:24:14.420 --> 00:24:17.750
The linear LMS estimator
is the one in pink,

00:24:17.750 --> 00:24:18.680
the straight line.

00:24:18.680 --> 00:24:22.080
And the LMS estimator is the
one in black with the kink.

00:24:22.080 --> 00:24:24.170
It's an interesting question
to now ask, which

00:24:24.170 --> 00:24:25.480
one of these is better?

00:24:25.480 --> 00:24:27.390
And in order to judge that, we
need to come up with some sort

00:24:27.390 --> 00:24:30.060
of criterion to compare
the two with.

00:24:30.060 --> 00:24:34.280
And the one that we're going
to look at in part e is the

00:24:34.280 --> 00:24:35.040
mean squared error.

00:24:35.040 --> 00:24:37.420
Which one gives the lower
mean squared error.

00:24:37.420 --> 00:24:44.530
And so specifically, we're going
to ask ourselves which

00:24:44.530 --> 00:24:49.075
of these two estimators
gives us the smaller

00:24:49.075 --> 00:24:50.325
mean squared error?

00:24:54.000 --> 00:24:57.500
Is it the linear LMS estimator
given by l of X?

00:24:57.500 --> 00:25:02.180
Or is it the LMS estimator,
given by g of X?

00:25:02.180 --> 00:25:08.400
Now, we know that the LMS
estimator is the one that

00:25:08.400 --> 00:25:09.560
actually minimizes this.

00:25:09.560 --> 00:25:13.600
The LMS estimator is designed
to minimize the

00:25:13.600 --> 00:25:14.140
mean squared error.

00:25:14.140 --> 00:25:18.030
And so we know that given any
estimator of X, this one will

00:25:18.030 --> 00:25:20.470
have the smallest mean
squared error.

00:25:20.470 --> 00:25:24.430
And so the linear LMS estimator,
its mean squared

00:25:24.430 --> 00:25:29.480
error has to be at least as
large as the LMS estimators.

00:25:29.480 --> 00:25:32.220
And the last part of the
question now asks us to look

00:25:32.220 --> 00:25:37.680
at a third type of estimator,
which is the MEP estimator.

00:25:37.680 --> 00:25:41.390
Now, we want to ask, why is it
that we haven't been using the

00:25:41.390 --> 00:25:43.260
MEP estimator in this problem?

00:25:43.260 --> 00:25:45.360
Well, remember what the
MEP estimator does.

00:25:45.360 --> 00:25:50.730
In this case, what we would
do is it would take the

00:25:50.730 --> 00:25:54.030
conditional distribution ratio
of Y given any value of X. And

00:25:54.030 --> 00:25:59.200
then it would pick the value
of Y that gives the highest

00:25:59.200 --> 00:26:01.780
value in the conditional
distribution.

00:26:01.780 --> 00:26:04.190
And that would be the
MEP estimate of Y.

00:26:04.190 --> 00:26:08.060
But the problem in this case is
that if you take any slice

00:26:08.060 --> 00:26:11.700
here, so a condition on any
value of X, any of these

00:26:11.700 --> 00:26:18.240
slices, if you just take this
out and look at it, it's going

00:26:18.240 --> 00:26:20.040
to be uniform.

00:26:20.040 --> 00:26:28.910
This is what the conditional
distribution of Y given X is.

00:26:28.910 --> 00:26:33.662
It's going to be uniform between
0 and X. Now, what the

00:26:33.662 --> 00:26:37.610
MEP rule tells us is we're going
to pick the value of Y

00:26:37.610 --> 00:26:43.310
that gives us the highest point
in this conditional

00:26:43.310 --> 00:26:44.190
distribution.

00:26:44.190 --> 00:26:47.380
You can think of it as a
posterior distribution.

00:26:47.380 --> 00:26:49.370
Now, what's the problem here?

00:26:49.370 --> 00:26:52.860
Well, every single point gives
us exactly the same value for

00:26:52.860 --> 00:26:54.140
this conditional distribution.

00:26:54.140 --> 00:26:56.780
And so there's no
unique MEP rule.

00:26:56.780 --> 00:27:02.790
Every single value of Y has
just the same conditional

00:27:02.790 --> 00:27:03.510
distribution.

00:27:03.510 --> 00:27:08.260
So there's no sensible way of
choosing a value based on the

00:27:08.260 --> 00:27:10.490
MEP rule in this case.

00:27:10.490 --> 00:27:16.090
But compare that with the LMS
estimator, which is just get

00:27:16.090 --> 00:27:17.560
conditional expectation.

00:27:17.560 --> 00:27:19.600
In that case, we can always
find a conditional

00:27:19.600 --> 00:27:20.580
expectation.

00:27:20.580 --> 00:27:23.640
In this case, the conditional
expectation is the midpoint,

00:27:23.640 --> 00:27:27.690
which is X/2, just as
had found in part a.

00:27:27.690 --> 00:27:31.640
OK, so in this problem, we
reviewed a bunch of different

00:27:31.640 --> 00:27:35.823
ideas in terms of inference, and
we took a joint pdf of X

00:27:35.823 --> 00:27:39.600
and Y, and we used that to
calculate the LMS estimator,

00:27:39.600 --> 00:27:41.200
the linear LMS estimator.

00:27:41.200 --> 00:27:44.280
We compared the two, and then we
also looked at why in this

00:27:44.280 --> 00:27:47.670
case, the MEP estimator doesn't
really make sense.

00:27:47.670 --> 00:27:49.330
All right, so I hope that
was helpful, and we'll

00:27:49.330 --> 00:27:50.580
see you next time.