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Hi.
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In this problem, we're going to
be dealing with a variation
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of the usual coin-flipping
problem.
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But in this case, the bias
itself of the coin
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is going to be random.
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So you could think of it as, you
don't even know what the
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probability of heads
for the coin is.
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So as usual, we're still taking
one coin and we're
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flipping it n times.
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But the difference here is that
the bias is because it
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was random variable Q. And
we're told that the
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expectation of this bias is some
mu and that the variance
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of the bias is some sigma
squared, which
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we're told is positive.
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And what we're going to be
asked is find a bunch of
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different expectations,
covariances, and variances.
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And we'll see that this problem
gives us some good
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exercise in a few concepts, a
lot of iterated expectations,
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which, again, tells you that
when you take the expectation
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of a conditional expectation,
it's just the expectation of
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the inner random variable.
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The covariance of two random
variables is just the
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expectation of the product
minus the product of the
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expectations.
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Law of total variance is the
expectation of a variance, of
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a conditional variance plus the
variance of a conditional
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expectation.
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And the last thing, of course,
we're dealing with a bunch of
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Bernoulli random variables,
coin flips.
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So as a reminder, for a
Bernoulli random variable, if
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you know what the bias is, it's
some known quantity p,
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then the expectation of the
Bernoulii is just p, and the
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variance of the Bernoulli
is p times 1 minus p.
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So let's get started.
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The problem tells us that we're
going to define some
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random variables.
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So xi is going to be a Bernoulli
random variable for
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the i coin flip.
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So xi is going to be 1 if the i
coin flip was heads and 0 if
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it was tails.
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And one very important thing
that the problem states is
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that conditional on Q, the
random bias, so if we know
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what the random bias is, then
all the coin flips are
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independent.
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And that's going to be important
for us when we
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calculate all these values.
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OK, so the first thing that we
need to calculate is the
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expectation of each of these
individual Bernoulli random
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variables, xi.
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So how do we go about
calculating what this is?
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Well, the problem
gives us a int.
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It tells us to try using the law
of iterated expectations.
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But in order to use it, you need
to figure out what you
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need the condition on.
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What this y?
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What takes place in y?
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And in this case, a good
candidate for what you
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condition on would be
the bias, the Q that
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we're unsure about.
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So let's try doing that
and see what we get.
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So we write out the law of
iterated expectations with Q.
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So now hopefully, we can
simplify it with this
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inter-conditional
expectation is.
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Well, what is it really?
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It's saying, given what Q is,
what is the expectation of
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this Bernoulli random
interval xi?
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Well, we know that if we knew
what the bias was, then the
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expectation is just
the bias itself.
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But in this case, the
bias is random.
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But remember a conditional
expectation is
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still a random variable.
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And so in this case, this
actually just simplifies into
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Q. So whatever the bias is, the
expectation is just equal
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to the bias.
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And so that's what
it tells us.
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And this part is easy because
we're given that the
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expectation of q is mu.
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And then the problem also
defines the random variable x.
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X is the total number of heads
within the n tosses.
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Or you can think of it as a sum
of all these individual xi
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Bernoulli random variables.
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And now, what can
we do with this?
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Well we can remember that
linearity of expectations
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allows us to split
up this sum.
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Expectation of a sum, we could
split up into a sum of
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expectations.
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So this is actually just
expectation of x1 plus dot dot
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dot plus all the way to
expectation of xn.
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All right.
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And now, remember that we're
flipping the same coin.
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We don't know what the bias is,
but for all the n flips,
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it's the same coin.
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And so each of these
expectations of xi should be
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the same, no matter
what xi is.
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And each one of them is mu.
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We already calculated
that earlier.
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And there's 10 of them, so the
answer would be n times mu.
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So let's move on to part B.
Part B now asks us to find
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what the covariance is
between xi and xj.
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And we have to be a little bit
careful here because there are
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two different scenarios, one
where i and j are different
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indices, different tosses,
and another where i
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and j are the same.
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So we have to consider both
of these cases separately.
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Let's first do the case where
x and i are different.
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So i does not equal j.
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In this case, we can just apply
the formula that we
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talked about in the beginning.
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So this covariance is just equal
to the expectation of xi
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times xj minus the expectation
of xi times expectation of xj.
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All right, so we actually know
what these two are, right?
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Expectation of xi is mu.
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Expectation of xj is also mu.
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So this part is just
mu squared.
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But we need to figure out
what this expectation
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of xi times xj is.
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Well, the expectation of xi
times xj, we can again use the
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law of iterated expectations.
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So let's try conditioning
on cue again.
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And remember we said
that this second
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part is just mu squared.
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All right, well, how can
we simplify this
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inner-conditional expectation?
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Well, we can use the fact that
the problem tells us that,
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conditioned on Q, the tosses
are independent.
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So that means that, conditioned
on Q, xi and xj
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are independent.
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And remember, when random
variables are independent, the
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expectation of product, you
could simplify that to be the
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product of the expectations.
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And because we're in the
condition world on Q, you have
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to remember that it's going
to be a product of two
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conditional expectations.
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So this will be expectation of
xi given Q times expectation
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of xj given Q minus
mu squared still.
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All right, now what is this?
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Well the expectation of xi given
Q, we already argued
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earlier here that it should just
be Q. And then the same
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thing for xj.
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That should also be Q. So this
is just expectation of Q
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squared minus mu squared.
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All right, now if we look at
this, what is the expectation
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of Q squared minus mu squared?
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Well, remember mu is just,
we're told that mu is the
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expectation of Q. So what we
have is the expectation of Q
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squared minus the quantity
expectation of Q squared.
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And what is that, exactly?
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That is just the formula or
the definition of what the
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variance of Q should be.
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So this is, in fact, exactly
equal to the variance of Q,
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which we're told is
sigma squared.
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All right, so what we found is
that for i not equal to j, the
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coherence of xi and
xj is exactly
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equal to sigma squared.
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And remember, we're told that
sigma squared is positive.
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So what does that tell us?
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That tells us that xi and xj, or
i not equal to j, these two
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random variables
are correlated.
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And so, because they're
correlated, they can't be
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independent.
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Remember, if two intervals are
independent, that means
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they're uncorrelated.
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But the converse isn't true.
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But if we do know that two
random variables are
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correlated, that means that
they can't be independent.
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And now let's finish this by
considering the second case.
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The second case is when i
actually does equal j.
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And in that case, well, the
covariance of xi and xi is
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just another way of writing
the variance of xi.
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So covariance, xi, xi, it's
just the variance of xi.
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And what is that?
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That is just the expectation
of xi squared minus
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expectation of xi quantity
squared.
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And again, we know what
the second term is.
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The second term is expectation
of xi quantity squared.
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Expectation of xi we know from
part A is just mu, right?
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So that's just second term
is just mu squared.
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But what is the expectation
of xi squared?
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Well, we can think about
this a little bit more.
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And you can realize that xi
squared is actually exactly
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the same thing as just xi.
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And this is just a special case
because xi is a Bernoulli
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random variable.
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Because Bernoulli is
either 0 or 1.
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And if it's 0 and you square
it, it's still 0.
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And if it's 1 and you square
it, it's still 1.
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So squaring it doesn't
really doesn't
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actually change anything.
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It's exactly the same thing as
the original random variable.
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And so, because this is a
Bernoulli random variable,
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this is exactly just the
expectation of xi.
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And we said this part
is just mu squared.
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So this is just expectation of
xi, which we said was mu.
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So the answer is just
mu minus mu squared.
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OK, so this completes part B.
And the answer that we wanted
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was that in fact, xi and xj are
in fact not independent.
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Right.
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So let's write down some facts
that we'll want to remember.
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One of them is that expectation
of xi is mu.
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And we also want to remember
what this covariance is.
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The covariance of xi and xj is
equal to sigma squared when i
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does not equal j.
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So we'll be using these
facts again later.
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And the variance of xi is equal
to mu minus mu squared.
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So now let's move on to the last
part, part C, which asks
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us to calculate the variance
of x in two different ways.
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So the first way we'll
do it is using the
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law of total variance.
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So the law of total variance
will tell us that we can write
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the variance of x as a sum
of two different parts.
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So the first is variance of x
expectation of the variance of
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x conditioned on something
plus the variance of the
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initial expectation of x
conditioned on something.
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And as you might have guessed,
what we're going to condition
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on is Q.
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Let's calculate what these
two things are.
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So let's do the two
terms separately.
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What is the expectation
of the conditional
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variance of x given Q?
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Well, what is--
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this, we can write out x.
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Because x, remember, is just
the sum of a bunch of these
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Bernoulli random variables.
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And now what we'll do was, well,
again, use the important
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fact that the x's, we're told,
are conditionally independent,
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conditional on Q.
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And because they're independent,
remember the
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variance of a sum is not the
sum of the variance.
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It's only the sum of the
variance if the terms in the
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sum are independent.
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In this case, they are
conditionally independent
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given Q. So we can in fact split
this up and write it as
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the variance of x1 given Q
plus all the way to the
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variance of xn given Q.
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And in fact, all these
are the same, right?
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So we just have n copies of the
variance of, say, x1 given
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00:14:39,530 --> 00:14:43,310
Q. Now, what is the variance
of x1 given Q?
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00:14:43,310 --> 00:14:46,770
Well, x1 is just a Bernoulli
random variable.
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00:14:46,770 --> 00:14:51,620
But the difference is that for
x, we don't know what the bias
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00:14:51,620 --> 00:14:54,060
or what the Q is.
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00:14:54,060 --> 00:14:57,910
Because it's some
random bias Q
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00:14:57,910 --> 00:15:01,010
But just like we said earlier
in part A, when we talked
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00:15:01,010 --> 00:15:07,640
about the expectation of x1
given Q, this is actually just
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00:15:07,640 --> 00:15:13,250
Q times 1 minus Q. Because if
you knew what the bias were,
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00:15:13,250 --> 00:15:14,810
it would be p times 1 minus p.
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00:15:14,810 --> 00:15:16,860
So the bias times 1
minus the bias.
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00:15:16,860 --> 00:15:19,190
But you don't know what it is.
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00:15:19,190 --> 00:15:21,060
But if you did, it
would just be q.
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00:15:21,060 --> 00:15:23,870
So what we do is we just plug
in Q, and you get Q
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00:15:23,870 --> 00:15:26,770
times 1 minus 2.
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00:15:26,770 --> 00:15:36,110
All right, and now this
is expectation of n.
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00:15:36,110 --> 00:15:38,960
I can pull out the n.
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00:15:38,960 --> 00:15:43,470
So it's n times the expectation
of Q minus Q
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00:15:43,470 --> 00:15:51,090
squared, which is just n times
expectation Q, we can use
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00:15:51,090 --> 00:15:55,450
linearity of expectations again,
expectation of Q is mu.
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00:15:55,450 --> 00:16:00,540
And the expectation of Q 2
squared is, well, we can do
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00:16:00,540 --> 00:16:01,230
that on the side.
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00:16:01,230 --> 00:16:08,840
Expectation of Q squared is
the variance of Q plus
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00:16:08,840 --> 00:16:14,230
expectation of Q quantity
squared.
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00:16:14,230 --> 00:16:22,120
So that's just sigma squared
plus mu squared.
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00:16:22,120 --> 00:16:27,810
And so this is just going to
be then minus sigma squared
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00:16:27,810 --> 00:16:29,060
minus mu squared.
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00:16:29,060 --> 00:16:32,080
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00:16:32,080 --> 00:16:33,820
All right, so that's
the first term.
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00:16:33,820 --> 00:16:35,950
Now let's do the second term.
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00:16:35,950 --> 00:16:43,720
The variance the conditional
expectation of x given Q. And
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00:16:43,720 --> 00:16:52,740
again, what we can do is we can
write x as the sum of all
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00:16:52,740 --> 00:16:55,435
these xi's.
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00:16:55,435 --> 00:16:59,270
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00:16:59,270 --> 00:17:04,730
And now we can apply linearity
of expectations.
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00:17:04,730 --> 00:17:08,705
So we would get n times one
of these expectations.
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00:17:08,705 --> 00:17:13,440
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00:17:13,440 --> 00:17:18,530
And remember, we said earlier
the expectation of x1 given Q
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00:17:18,530 --> 00:17:23,720
is just Q. So it's the variance
of n times Q.
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00:17:23,720 --> 00:17:26,375
And remember now, n is just--
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00:17:26,375 --> 00:17:27,460
it's not random.
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00:17:27,460 --> 00:17:29,680
It's just some number.
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00:17:29,680 --> 00:17:32,070
So when you pull it out of a
variance, you square it.
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00:17:32,070 --> 00:17:36,290
So this is n squared times
the variance of Q.
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00:17:36,290 --> 00:17:39,130
And the variance of Q we're
given is sigma squared.
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00:17:39,130 --> 00:17:42,660
So this is n squared times
sigma squared.
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00:17:42,660 --> 00:17:45,280
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00:17:45,280 --> 00:17:47,860
So the final answer is
just a combination
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00:17:47,860 --> 00:17:49,250
of these two terms.
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00:17:49,250 --> 00:17:54,290
This one and this one.
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00:17:54,290 --> 00:17:56,010
So let's write it out.
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00:17:56,010 --> 00:17:59,295
The variance of x, then,
is equal to--
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00:17:59,295 --> 00:18:02,790
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00:18:02,790 --> 00:18:04,580
we can combine terms
a little bit.
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00:18:04,580 --> 00:18:08,010
So the first one, let's
take the mus and
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00:18:08,010 --> 00:18:08,730
we'll put them together.
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00:18:08,730 --> 00:18:11,325
So it's n mu minus mu squared.
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00:18:11,325 --> 00:18:15,830
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00:18:15,830 --> 00:18:22,660
And then we have n squared times
sigma squared from this
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00:18:22,660 --> 00:18:28,520
term and minus n times sigma
squared from this term.
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00:18:28,520 --> 00:18:34,450
So it would be n squared minus
n times sigma squared, or n
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00:18:34,450 --> 00:18:38,400
times n minus 1 times
sigma squared.
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00:18:38,400 --> 00:18:40,970
So that is the final answer
that we get for
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00:18:40,970 --> 00:18:42,220
the variance of x.
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00:18:42,220 --> 00:18:45,030
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00:18:45,030 --> 00:18:47,450
And now, let's try doing
it another way.
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00:18:47,450 --> 00:18:51,800
315
00:18:51,800 --> 00:18:53,960
So that's one way of doing it.
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00:18:53,960 --> 00:18:57,140
That's using the law of total
expectations and conditioning
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00:18:57,140 --> 00:19:05,880
on Q. Another way of finding
the variance of x is to use
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00:19:05,880 --> 00:19:11,330
the formula involving
covariances, right?
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00:19:11,330 --> 00:19:18,652
And we can use that because x is
actually a sum of multiple
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00:19:18,652 --> 00:19:23,590
random variables
x1 through xn.
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00:19:23,590 --> 00:19:40,780
And the formula for this is, you
have n variance terms plus
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00:19:40,780 --> 00:19:44,110
all these other ones.
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00:19:44,110 --> 00:19:48,140
Where i is not equal to j, you
have the covariance terms.
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00:19:48,140 --> 00:19:51,770
And really, it's just, you can
think of it as a double sum of
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00:19:51,770 --> 00:19:59,150
all pairs of xi and xj where if
i and j happen just to be
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00:19:59,150 --> 00:20:02,710
the same, that it simplifies
to be just the variance.
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00:20:02,710 --> 00:20:06,240
Now, so we pulled theses n terms
out because they are
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00:20:06,240 --> 00:20:10,770
different than these because
they have a different value.
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00:20:10,770 --> 00:20:14,060
And now fortunately, we've
already calculated what these
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00:20:14,060 --> 00:20:16,690
values are in part B. So we
can just plug them them.
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00:20:16,690 --> 00:20:18,890
All the variances
are the same.
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00:20:18,890 --> 00:20:21,300
And there's n of them,
so we get n times the
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00:20:21,300 --> 00:20:22,260
variance of each one.
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00:20:22,260 --> 00:20:26,960
The variance of each one we
calculated already was mu
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00:20:26,960 --> 00:20:29,790
minus mu squared.
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00:20:29,790 --> 00:20:32,630
And then, we have all
the terms were i is
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00:20:32,630 --> 00:20:34,210
not equal to j.
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00:20:34,210 --> 00:20:39,650
Well, there are actually n
squared minus n of them.
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00:20:39,650 --> 00:20:44,040
So because you can take any one
of the n's to be the first
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00:20:44,040 --> 00:20:48,110
to be i, any one of
the n to be j.
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00:20:48,110 --> 00:20:49,890
So that gives you
n squared pairs.
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00:20:49,890 --> 00:20:52,590
But then you have to subtract
out all the ones where i and j
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00:20:52,590 --> 00:20:53,190
are the same.
344
00:20:53,190 --> 00:20:54,320
And there are n of them.
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00:20:54,320 --> 00:20:59,250
So that leaves you with n
squared minus n of these pairs
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00:20:59,250 --> 00:21:01,600
where i is not equal to j.
347
00:21:01,600 --> 00:21:04,130
And the coherence for this case
where i is not equal to
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00:21:04,130 --> 00:21:08,176
j, we also calculated in part B.
That's just sigma squared.
349
00:21:08,176 --> 00:21:13,050
All right, and now if we compare
these two, we'll see
350
00:21:13,050 --> 00:21:15,610
that they are proportionally
exactly the same.
351
00:21:15,610 --> 00:21:18,510
352
00:21:18,510 --> 00:21:23,700
So we've use two different
methods to calculate the
353
00:21:23,700 --> 00:21:27,510
variance, one using this
summation and one using the
354
00:21:27,510 --> 00:21:29,860
law of total variance.
355
00:21:29,860 --> 00:21:33,040
So what do we learn
from this problem?
356
00:21:33,040 --> 00:21:37,430
Well, we saw that first of all,
in order to find some
357
00:21:37,430 --> 00:21:40,940
expectations, it's very useful
to use law of iterated
358
00:21:40,940 --> 00:21:41,700
expectations.
359
00:21:41,700 --> 00:21:44,620
But the trick is to figure out
what you should condition on.
360
00:21:44,620 --> 00:21:47,780
And that's kind of an
art that you learn
361
00:21:47,780 --> 00:21:49,230
through more practice.
362
00:21:49,230 --> 00:21:52,920
But one good rule of thumb is,
when you have kind of a
363
00:21:52,920 --> 00:21:57,650
hierarchy or layers of
randomness where one layer of
364
00:21:57,650 --> 00:22:00,640
randomness depends
on the randomness
365
00:22:00,640 --> 00:22:01,960
of the layer above--
366
00:22:01,960 --> 00:22:05,780
so in this case, whether or
not you get heads or tails
367
00:22:05,780 --> 00:22:09,600
depends on, that's random, but
that depends on the randomness
368
00:22:09,600 --> 00:22:12,040
on the level above, which
was the random
369
00:22:12,040 --> 00:22:14,150
bias of the coin itself.
370
00:22:14,150 --> 00:22:19,410
So the rule of thumb is, when
you want to calculate the
371
00:22:19,410 --> 00:22:23,360
expectations for the layer where
you're talking about
372
00:22:23,360 --> 00:22:27,710
heads or tails, it's useful to
condition on the layer above
373
00:22:27,710 --> 00:22:30,590
where that is, in this case,
the random bias.
374
00:22:30,590 --> 00:22:34,430
Because once you condition on
the layer above, that makes
375
00:22:34,430 --> 00:22:36,210
the next level much simpler.
376
00:22:36,210 --> 00:22:39,830
Because you kind of assume that
you know what all the
377
00:22:39,830 --> 00:22:42,650
previous levels of randomness
are, and that helps you
378
00:22:42,650 --> 00:22:47,480
calculate what the expectation
for this current level.
379
00:22:47,480 --> 00:22:52,180
And the rest of the problem was
just kind of going through
380
00:22:52,180 --> 00:22:54,160
exercises of actually
applying the--
381
00:22:54,160 --> 00:22:55,410