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rLlZpnT02ZU | So just let's look at back into
our favorite example, which |
rLlZpnT02ZU | is the average of
Bernoulli random variables, |
rLlZpnT02ZU | so we studied that maybe
that's the third time already. |
rLlZpnT02ZU | So the sample average, Xn
bar, is a strongly consistent |
rLlZpnT02ZU | estimator of p. |
rLlZpnT02ZU | That was one of the
properties that we wanted. |
rLlZpnT02ZU | Strongly consistent means
that as n goes to infinity, |
rLlZpnT02ZU | it converges almost surely
to the true parameter. |
rLlZpnT02ZU | That's the strong
law of large number. |
rLlZpnT02ZU | It is consistent also, because
it's strongly consistent, |
rLlZpnT02ZU | so it also converges
in probability, |
rLlZpnT02ZU | which makes it consistent. |
rLlZpnT02ZU | It's unbiased. |
rLlZpnT02ZU | We've seen that. |
rLlZpnT02ZU | We've actually computed
its quadratic risk. |
rLlZpnT02ZU | And now what I have
is that if I look at-- |
rLlZpnT02ZU | thanks to the central limit
theorem, we actually did this. |
rLlZpnT02ZU | We built a confidence interval
at level 1 minus alpha-- |
rLlZpnT02ZU | asymptotic level, sorry,
asymptotic level 1 minus alpha. |
rLlZpnT02ZU | And so here, this
is how we did it. |
rLlZpnT02ZU | Let me just go through it again. |
rLlZpnT02ZU | So we know from the
central limit theorem-- |
rLlZpnT02ZU | so the central limit
theorem tells us |
rLlZpnT02ZU | that Xn bar minus p divided
by square root of p1 minus p, |
rLlZpnT02ZU | square root of n converges
in distribution as n |
rLlZpnT02ZU | goes to infinity to some
standard normal distribution. |
rLlZpnT02ZU | So what it means is that if
I look at the probability |
rLlZpnT02ZU | under the true p, that's
square root of n, Xn bar |
rLlZpnT02ZU | minus p divided by square
root of p1 minus p, |
rLlZpnT02ZU | it's less than Q alpha
over 2, where this is |
rLlZpnT02ZU | the definition of the quintile. |
rLlZpnT02ZU | Then this guy-- and I'm actually
going to use the same notation, |
rLlZpnT02ZU | limit as n goes to infinity,
this is the same thing. |
rLlZpnT02ZU | So this is actually going to
be equal to 1 minus alpha. |
rLlZpnT02ZU | That's exactly what
I did last time. |
rLlZpnT02ZU | This is by definition of the
quintile of a standard Gaussian |
rLlZpnT02ZU | and of a limit in distribution. |
rLlZpnT02ZU | So the probabilities computed on
this guy in the limit converges |
rLlZpnT02ZU | to the probability
computed on this guy. |
rLlZpnT02ZU | And we know that this
is just the probability |
rLlZpnT02ZU | that the absolute
value of sum n 0, 1 |
rLlZpnT02ZU | is less than Q alpha over 2. |
rLlZpnT02ZU | And so in particular,
if it's equal, |
rLlZpnT02ZU | then I can put some
larger than or equal to, |
rLlZpnT02ZU | which guarantees my
asymptotic confidence level. |
rLlZpnT02ZU | And I just solve for p. |
rLlZpnT02ZU | So this is equivalent
to the limit |
rLlZpnT02ZU | as n goes to infinity
of the probability |
rLlZpnT02ZU | that theta is between
Xn bar minus Q |
rLlZpnT02ZU | alpha over 2 divided by-- |
rLlZpnT02ZU | times square root of p1 minus p
divided by square root of n, Xn |
rLlZpnT02ZU | bar plus q alpha over 2,
square root of p1 minus p |
rLlZpnT02ZU | divided by square root of
n is larger than or equal |
rLlZpnT02ZU | to 1 minus alpha. |
rLlZpnT02ZU | And so there you go. |
rLlZpnT02ZU | I have my confidence interval. |
rLlZpnT02ZU | Except that's not, right? |
rLlZpnT02ZU | We just said that the bounds
of a confidence interval |
rLlZpnT02ZU | may not depend on the
unknown parameter. |
rLlZpnT02ZU | And here, they do. |
rLlZpnT02ZU | And so we actually
came up with two ways |
rLlZpnT02ZU | of getting rid of this. |
rLlZpnT02ZU | Since we only need this thing--
so this thing, as we said, |
rLlZpnT02ZU | is really equal. |
rLlZpnT02ZU | Every time I'm going to
make this guy smaller |
rLlZpnT02ZU | and this guy larger,
I'm only going |
rLlZpnT02ZU | to increase the probability. |
rLlZpnT02ZU | And so what we do is
we actually just take |
rLlZpnT02ZU | the largest possible
value for p1 minus |
rLlZpnT02ZU | p, which makes the interval
as large as possible. |
rLlZpnT02ZU | And so now I have this. |
rLlZpnT02ZU | I just do one of the two tricks. |
rLlZpnT02ZU | I replace p1 minus p by their
upper bound, which is 1/4. |
rLlZpnT02ZU | As we said, p1 minus p, the
function looks like this. |
rLlZpnT02ZU | So I just take the
value here at 1/2. |
rLlZpnT02ZU | Or, I can use Slutsky and say
that if I replace p by Xn bar, |
rLlZpnT02ZU | that's the same as just
replacing p by Xn bar here. |
rLlZpnT02ZU | And by Slutsky, we know that
this is actually converging |
rLlZpnT02ZU | also to some standard Gaussian. |
rLlZpnT02ZU | We've seen that when we
saw Slutsky as an example. |
rLlZpnT02ZU | And so those two
things-- actually, |
rLlZpnT02ZU | just because I'm
taking the limit |
rLlZpnT02ZU | and I'm only caring about the
asymptotic confidence level, |
rLlZpnT02ZU | I can actually just plug in
consistent quantities in there, |
rLlZpnT02ZU | such as Xn bar where
I don't have a p. |
rLlZpnT02ZU | And that gives me another
confidence interval. |
rLlZpnT02ZU | All right. |
rLlZpnT02ZU | So this by now, hopefully
after doing it three times, |
rLlZpnT02ZU | you should really, really be
comfortable with just creating |
rLlZpnT02ZU | this confidence interval. |
rLlZpnT02ZU | We did it three times in class. |
rLlZpnT02ZU | I think you probably did
it another couple times |
rLlZpnT02ZU | in your homework. |
rLlZpnT02ZU | So just make sure you're
comfortable with this. |
rLlZpnT02ZU | All right. |
rLlZpnT02ZU | That's one of the basic
things you would want to know. |
rLlZpnT02ZU | Are there any questions? |
rLlZpnT02ZU | Yes. |
rLlZpnT02ZU | AUDIENCE: So Slutsky holds
for any single response set p. |
rLlZpnT02ZU | But Xn converges [INAUDIBLE]. |
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