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Hi.

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In this problem, we're going
to look at how to infer a

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discrete random variable from
a continuous measurement.

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And really, what it's going to
give us is some practice

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working with a variation
of Bayes' rule.

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So the problem tells us that
we have a discrete random

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variable x with this PMF.

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It is 1 with probability P,
minus 1 with probability 1

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minus P, and 0 otherwise.

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So here is just a diagram
of this PMF.

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And then we also have another
random variable, y, which is

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continuous.

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And its PDF is given by this.

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It's 1/2 lambda e to the minus
lambda times the absolute

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value of y.

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And so this may look familiar.

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It looks kind of like
an exponential.

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And in fact, it's just a
two-sided exponential.

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That's flattened by
a factor of 1/2.

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And this is what it looks like,
kind of like a tent that

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goes on both ways.

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And then we have a random
variable z, which is equal to

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the sum of x and y.

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And the problem is going to be
figuring out what x is, based

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on an observed value
of what z is.

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So because x is discrete
and y is random--

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sorry, x is discrete, and y is
continuous, z is also going to

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be continuous.

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So our measurement is z,
which is continuous.

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And we want to infer x,
which is discrete.

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So the problem asked us to
find is this what is the

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probability that x equals 1,
given that z is a little z.

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And you can write this another
way, just as a

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conditional PMF as well.

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It's the conditional PMF of
x, given z, evaluated to 1

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conditioned on little z.

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All right, so now let's apply
the correct variation of

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Bayes' rule.

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So remember, it's going to be
this, the probability that x

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equals 1, or the PMF of x
evaluated to 1, times the--

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you flip this conditioning.

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So now it's going to be a
conditional PDF of z, since z

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is continuous.

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It's going to be a conditional
PDF of z, given x, evaluated

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at some little z condition
on x being 1.

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And the bottom is the
conditional PDF--

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or sorry, just the
regular PDF of z.

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And of course, we can rewrite
this denominator.

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Remember, the denominator is
always just-- you can use the

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law of total probability
and rewrite it.

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And one of the terms is
going to be exactly

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the same as the numerator.

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So one of the ways that z can
be some little z is it's in

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combination with x
being equal to 1.

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And the probability of that
is exactly the same

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thing as the numerator.

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And the other way is if x
is equal to negative 1.

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And that gives us this
second term.

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All right.

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And now let's just fill
in what all these

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different terms are.

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So with the PMF of x evaluated
at 1, that is just P. What is

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the conditional PDF of z, given
that x is equal to 1?

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Well, that takes a little
bit more work.

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Given that x is 1, then z
is just going to be--

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so if x equals 1, then z is
just y plus 1, which means

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that you can just imagine taking
y-- this is what y is,

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the distribution of y--

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and just adding 1 to it, which,
in this diagram, would

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amount to shifting
it over by one.

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So now, it would look like
this, the distribution.

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And algebraically, all you would
do is just change this

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thing in the absolute
value to y minus 1.

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That amounts to shifting it
over to the right by one.

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All right.

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So what is that?

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That's just 1/2 lambda,
1/2 lambda, e to the

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minus lambda y--

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sorry, not y, z--

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z minus 1.

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And the denominator, well, the
first term is going to be

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exactly the same.

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It's just also P 1/2 lambda e to
the minus lambda z minus 1.

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What about the second term?

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The second term, first we need
to figure out what is the PMF

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of x evaluated at
a negative 1.

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Or in other words, what's
the probability that

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x is negative 1?

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That is given to
us by the PMF.

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It's 1 minus P.

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And then the second part is,
what is the conditional PDF of

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z, given that x is negative 1?

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Well, we can just do the same
sort of trick here.

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If x is negative 1, then
z is just y minus 1.

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In which case, the PDF of z
would just look like this.

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You're shifted to the
left by one now.

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And now what you have to do is
change this into a plus 1.

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So this conditional PDF would
be 1/2 lambda e to the minus

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lambda z plus 1, absolute
value of z plus 1.

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All right, so this looks
pretty messy.

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And we can try to simplify
things a little bit.

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So we can get rid of
these 1/2 lambdas.

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And then we can multiply the
numerator and the denominator

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by the same term.

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Let's multiply it by e to
the lambda absolute

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value of z minus 1.

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So what we're going to do is
try to cancel out some of

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these exponential terms.

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So that will cancel
out this top term.

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So all we have in the numerator
now is just P. It

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will also cancel out this
exponential in the

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denominator.

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And then we'll have to change
this here, because it'll have

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an extra e to the lambda
absolute value of z minus 1.

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All right, now let's
rewrite this.

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And what we get is plus 1 minus
P e to the minus lambda

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absolute value of z plus
1 minus absolute

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value of z minus 1.

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OK, so that is pretty much as
far as you can go in terms of

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simplifying it.

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And now the question is, are we

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comfortable with this answer?

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And it helps always to try to
interpret it a little bit, to

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make sure that it makes
intuitive sense.

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And one way to do that is to try
to-- some of the limiting

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cases of what some of the
parameters can be.

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So in this case, the parameters
are P and lambda.

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So P is the parameter
related to x.

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And lambda is the parameter
related to y.

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So let's try to see if it
makes sense under some

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limiting cases.

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The first one we want to think
about is when P goes to 0.

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So if P goes to 0, what
happens to our answer?

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Well, the numerator is 0,
this is 0, this is 1.

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But it doesn't matter, because
the numerator is 0.

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So in this case, this
would go to 0.

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Now does that make sense?

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Well, what does that mean
when P goes to 0?

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When P goes to 0, that means
that the probability that x is

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equal to 1 is 0.

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So even without thinking about
y or z, there is already a 0

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probability that x
is equal to 1.

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Now this whole calculation, what
we found is, well, if I

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had some more information, like
what z is, does that help

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me find out what the probability
of x being 1 is?

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Well, no matter what z tells me,
I know for a fact that x

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can't be 1, because P is 0.

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So this posterior, or this
conditional probability,

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should also be 0, because
there's just no way

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that x can be 1.

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So in this case, this formula
does check out.

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Now let's think about another
case where P goes to 1.

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If P goes to 1, that
means that X is for

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sure going to be 1.

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And it can't be anything else.

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In which case, what does
our formula tell us?

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Well, this numerator is 1.

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This term is 1.

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1 minus 1 is 0.

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So the second term gets zeroed
out, and the answer

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is just 1/1 is 1.

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So what does this tell us?

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This tells us that if I know
beforehand that x is for sure

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equal to 1, then, if I now give
myself more information

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and condition on what I observe
for z, that shouldn't

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change anything else.

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I should still know for sure
that x is equal to 1.

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So the probability of this
conditional probability should

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still be equal to 1.

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And it does, so our formula
also works in this case.

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Now let's think about lambda.

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What about when lambda
goes to 0?

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Well, when lambda goes
to 0, that's a

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little harder to visualize.

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But really, what would happen
is that you can imagine this

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distribution getting shallower,
shallower and

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shallower, lower and lower, so
that it's like it is kind of

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flat and goes on forever.

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And so what this tells you
is that, basically, y--

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this is the distribution y-- so
when lambda goes to 0, that

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tells you that y has a really
flat and kind of short

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distribution.

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And so what does our formula
tell us in this case?

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Well, when lambda goes to 0,
this exponent is equal to 0.

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And so e to the 0 is 1.

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So we get P over P plus 1 minus
P, which is just 1.

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So the answer here, our formula
will give us an answer

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of P.

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So what does that tell us?

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That tells us that, in this
case, if lambda goes to 0,

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then our posterior probability,
the probability

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that x equals 1 conditioned
on z being some value,

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conditioned on our continuous
measurement, is still P. So

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the prior, or the original
probability for x being equal

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to 1 is P. And with this
additional continuous

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measurement, our guess of the
probability that x equal to 1

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is still P. So it
hasn't changed.

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So basically, it's telling
us that this additional

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information was not
informative.

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It didn't actually help
us change our beliefs.

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And so why is that?

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Well, one way to think about
it is that, because the

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distribution of y looks like
this, is very flat and it

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could be anything, then, if you
observe some value of z,

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then it could be that that was
due to the fact that it was x

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equal to 1 plus some value
of y that made z

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equal to that value.

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Or it could have just as equally
been likely that x

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equal to negative 1 y equals to
some other value that made

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it equal to z.

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And so, essentially, it's z--

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because y has a shape, it can be
likely to take on any value

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that complements either x being
equal to 1 or x equal

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being to negative 1, to make z
equal to whatever the value it

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is that you observe.

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And so because of that, in
this case, y is not very

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informative.

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And so this probability is
still just equal to P.

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Now the last case is when
lambda goes to infinity.

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And now we have to break
it down into the

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two other cases now.

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The first case is when--

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lets write this over here--

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when lambda goes to infinity.

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The first case, it depends on
what this value is, the sine

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of this value.

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If this value, the absolute
value of z plus 1 minus the

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absolute value of z minus 1,
if that's positive, then,

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because lambda goes to infinity
and you have a

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negative sign, then this
entire exponential

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term will go to 0.

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In which case, the second
term goes to 0.

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And the answer is
P/P, or is 1.

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And so if absolute value of z
plus 1 minus absolute value of

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z minus 1 is greater than
0, then the answer is 1.

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00:14:34,930 --> 00:14:37,480

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But in the other case, if this
term in the exponent, if it's

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actually negative, if it's
negative, then this negative

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sign turns to a positive, and
lambda goes to infinity.

248
00:14:52,370 --> 00:14:56,170
And so this term blows up, and
it dominates everything else.

249
00:14:56,170 --> 00:14:58,620
And so the denominator
goes to infinity.

250
00:14:58,620 --> 00:15:01,520
The numerator is fixed at P,
so this entire expression

251
00:15:01,520 --> 00:15:02,770
would go to 0.

252
00:15:02,770 --> 00:15:05,276

253
00:15:05,276 --> 00:15:09,840
OK, so now let's try to
interpret this case.

254
00:15:09,840 --> 00:15:12,000
Let's start with
the first one.

255
00:15:12,000 --> 00:15:16,310
When is it that absolute value
of z plus 1 minus absolute

256
00:15:16,310 --> 00:15:18,300
value of z minus 1 is
greater than 0?

257
00:15:18,300 --> 00:15:24,450
Or you can also rewrite this as
absolute value of z plus 1

258
00:15:24,450 --> 00:15:27,926
is greater than absolute
value of z minus 1.

259
00:15:27,926 --> 00:15:29,850
Well, when is that case?

260
00:15:29,850 --> 00:15:32,030
Well, it turns out, if you think
about it, this is only

261
00:15:32,030 --> 00:15:35,920
true if z is positive.

262
00:15:35,920 --> 00:15:40,010
If z is positive,
then adding 1--

263
00:15:40,010 --> 00:15:43,430
let me draw a line here, and
if this is 0-- if z is

264
00:15:43,430 --> 00:15:46,770
positive, something here, adding
1 to it and taking the

265
00:15:46,770 --> 00:15:47,790
absolute value--

266
00:15:47,790 --> 00:15:49,240
the absolute value doesn't
do anything--

267
00:15:49,240 --> 00:15:52,010
but you will get something
bigger.

268
00:15:52,010 --> 00:15:55,230
Where subtracting 1 will take
you closer to 0, and so

269
00:15:55,230 --> 00:15:58,490
because of that, the absolute
value, the magnitude, or the

270
00:15:58,490 --> 00:16:01,750
distance from 0 will be less.

271
00:16:01,750 --> 00:16:07,390
Now if you're on the other side,
adding 1 will take you--

272
00:16:07,390 --> 00:16:12,020
if you're on the other side,
adding 1 will take

273
00:16:12,020 --> 00:16:13,940
you closer to 0.

274
00:16:13,940 --> 00:16:16,150
And so this magnitude
would be smaller.

275
00:16:16,150 --> 00:16:18,570
Whereas, subtracting will take
you farther away, so the

276
00:16:18,570 --> 00:16:22,040
absolute value actually
increased the magnitude.

277
00:16:22,040 --> 00:16:28,520
And so this is the same
as z being positive.

278
00:16:28,520 --> 00:16:33,700
And so this is the same
as z being negative.

279
00:16:33,700 --> 00:16:38,530
So what this tells you is that,
if z is positive, then

280
00:16:38,530 --> 00:16:41,410
this probability
is equal to 1.

281
00:16:41,410 --> 00:16:43,610
And if z is negative, this
probability is equal to 0.

282
00:16:43,610 --> 00:16:45,720
Now why does that make sense?

283
00:16:45,720 --> 00:16:49,810
Well, it's because when lambda
goes to infinity, you have the

284
00:16:49,810 --> 00:16:50,920
other case.

285
00:16:50,920 --> 00:16:55,920
Essentially, you pull this all
the way up, really, really

286
00:16:55,920 --> 00:17:08,130
far, and it drops off
really quickly.

287
00:17:08,130 --> 00:17:10,869
And so when you take the limit,
as lambda goes to

288
00:17:10,869 --> 00:17:14,150
infinity, effectively, it just
becomes a spike at 0.

289
00:17:14,150 --> 00:17:19,300
And so, more or less, you're
sure that y is going to be

290
00:17:19,300 --> 00:17:20,420
equal to 0.

291
00:17:20,420 --> 00:17:24,240
And so, effectively, z is
actually going to be equal to

292
00:17:24,240 --> 00:17:26,859
x, effectively.

293
00:17:26,859 --> 00:17:32,930
And because of that, because x
can only be 1 or negative 1,

294
00:17:32,930 --> 00:17:37,278
then, depending on if you get a
z that's positive, then you

295
00:17:37,278 --> 00:17:40,700
know for sure that it must
have been that x

296
00:17:40,700 --> 00:17:42,750
was equal to 1.

297
00:17:42,750 --> 00:17:45,210
And if you get a z that's
negative, you know for sure

298
00:17:45,210 --> 00:17:48,990
that it must have been that
x was equal to negative 1.

299
00:17:48,990 --> 00:17:54,820
And so because of that, you
get this interpretation.

300
00:17:54,820 --> 00:17:59,620
And so we've looked at four
different cases of the

301
00:17:59,620 --> 00:18:00,600
parameters.

302
00:18:00,600 --> 00:18:04,600
And in all four cases, our
answer seems to make sense.

303
00:18:04,600 --> 00:18:08,030
And so we feel more confident
in the answer.

304
00:18:08,030 --> 00:18:12,320
And so to summarize, this whole
problem involved using

305
00:18:12,320 --> 00:18:14,290
Bayes' rule.

306
00:18:14,290 --> 00:18:18,290
You start out with some
distributions, and you apply

307
00:18:18,290 --> 00:18:19,860
Bayes' rule.

308
00:18:19,860 --> 00:18:21,230
And you go through the
steps, and you

309
00:18:21,230 --> 00:18:22,650
plug-in the right terms.

310
00:18:22,650 --> 00:18:25,450
And then, in the end, it's
always helpful to try to check

311
00:18:25,450 --> 00:18:28,270
your answers to make sure
that it makes sense

312
00:18:28,270 --> 00:18:30,140
in some of the limiting.