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We will now go through an
example that brings together
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all of the concepts that
we have introduced.
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We have a stick of length l.
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And we break that stick at some
random location, which
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corresponds to a random
variable, X.
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And we assume that this random
variable is uniform over the
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length of the stick.
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So its PDF has this
particular shape.
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And for the PDF to integrate to
1, the height of this PDF
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must be equal to 1 over l.
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Then we take the piece of the
stick that we are left with,
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which has length X, and we break
it at a random location,
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which we call Y. And we assume
that this location Y is
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uniformly distributed over the
length of the stick that we
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were left with.
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What does this assumption
mean?
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It means that if the first break
was at some particular
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value, x, then the random
variable Y has a conditional
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distribution, which is uniform
over the interval from 0 to x.
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So the conditional
PDF is uniform.
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A conditional PDF, like
any other PDF, must
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integrate to 1.
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So the height of this
conditional PDF is
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equal to 1 over x.
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Are X and Y independent?
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No.
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One way to see it is that if
you change little x, the
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conditional PDF of Y would have
been something different.
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Whereas if we have independence,
all the
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conditional PDFs have to be the
same when you change the
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value of little x.
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Another way to see it is that
if I tell you that x is 0.5,
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this gives you lots of
information about Y. It tells
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you that Y has to be less
than or equal to 0.5.
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So the value of the random
variable X gives you plenty of
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information about the other
random variable.
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And so we do not have
independence.
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Notice that in this example,
instead of starting with a
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full description of the random
variables in terms of a joint
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PDF, we use a marginal PDF and
then a conditional PDF to
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construct our model.
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Of course, with these two pieces
of information, we can
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reconstruct the joint PDF using
the multiplication rule.
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The marginal is 1 over l.
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The conditional is 1 over x.
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So the joint is equal
to 1 over lx.
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But for which values of x and
y is this the correct
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expression?
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It is correct only for those
values that are possible.
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So 0 has to be less than y,
less than x, less than l.
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This is the range of values
that are possible in this
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particular experiment.
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And we can visualize
those values.
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They are those that correspond
to this shaded triangle here.
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x and y are less than
or equal to l.
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And y has to be less
than or equal to x.
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If you try to visualize the
joint PDF, notice that since
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it only depends on x not on y,
if you fix a value of x and
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you look at the slice of the
joint PDF, the value of the
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joint PDF is going to be a
constant on that slice.
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On this slice, it's going to be
another constant, actually
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a bigger one.
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On that slice, an even
bigger constant.
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And actually, this constant is
bigger and bigger and goes to
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infinity as we approach 0.
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Of course, the fact that the
slice is constant is just a
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reflection of the fact that
the conditional PDF is
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constant over the range of
values that the random
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variable can take.
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Let us now continue with
some calculations.
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Let us find the marginal PDF
of Y. How do we do it?
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Since we have in our hands the
joint PDF, we can find the
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marginal by integrating
the joint.
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And in our case, the joint
is equal to 1 over lx.
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And we integrate over all x's.
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Now, what is the range
of the integration?
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If we fix a certain value of y,
the joint PDF is actually 0
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in this region and
in that region.
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So we should only integrate over
x's that correspond to
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this interval.
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What is that interval?
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It's the interval
that ends at l.
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And because this is a line
of slope 1, this value
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here is also y.
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So we integrate over
an interval where x
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ranges from y to l.
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In fact, this is just the range
of x's that are possible
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for a given value of y.
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x must always be larger
than or equal to y.
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Now, the integral of 1 over
x is a logarithm.
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And using this fact, we can
evaluate this integral.
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And it's 1 over l times the
logarithm of l over y.
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For what y's is this a
correct expression?
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Well, it makes sense only for
those y's that are possible in
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this experiment.
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And that's the range
from 0 to l.
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When y is equal to l, we have
the logarithm of 1, which is
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equal to 0.
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So the value of the
PDF is 0 here.
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As y decreases, this ratio
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increases and goes to infinity.
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So the log of that also
blows up to infinity.
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And we get a shape of this form,
where the function that
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we're dealing with goes to
infinity as we approach 0.
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Is this a problem having a PDF
that blows up to infinity?
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Not really.
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As long as the area under this
PDF is equal to 1, it's still
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a legitimate PDF.
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And blowing up to infinity
is not an issue.
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Let us now calculate the
expected value of Y. One way
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of doing this is by using the
definition of the expectation.
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It's the integral of y times the
density of y, which is 1
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over l times the log
of l over y.
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And the range of integration
has to be those values for
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which we have a non-zero
density.
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So we integrate from 0 to l,
which are the possible values
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of the random variable Y.
This is an integral
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that's pretty messy.
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One can actually integrate it
using integration by parts.
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But the calculation
is a bit tedious.
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So let us look for
an alternative
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and more clever approach.
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The idea is to divide
and conquer.
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We're going to use the total
expectation theorem, where
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we're going to condition on X.
The total expectation theorem
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tells us that the expected value
of Y is the integral
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over all possible values of the
random variable X, which
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is from 0 to l.
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The density of X, which is 1
over l, times the conditional
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expectation of Y given that X
is equal to some little x.
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And we integrate over all x's.
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Why is this simpler?
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When we condition on X taking
a specific value, Y has a
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uniform distribution
between 0 and x.
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And therefore, this conditional
expectation is the
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expectation of a uniform,
which is 1/2 the
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range of that uniform.
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So we obtain the integral
from 0 to l.
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1 over l times x over 2, dx.
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And finally, that's an
integral that we
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can evaluate easily.
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Or we can think even
in a simpler way.
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This expression here is
the density of x.
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This is x itself.
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So the integral of this times x
gives us the expected value
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of X. And there's only
a factor of 1/2
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that's left out there.
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So we obtain 1/2 the expected
value of X. But now, X itself
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is uniform on an interval
that has length l.
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And therefore, the expected
value of x is l over 2.
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And so we get the final answer,
which is 1/2 times l
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over 2, which is l over four.
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This answer makes
intuitive sense.
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If we break a stick once, the
expected value or what we're
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left with is half of what
we started with.
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But if we break it once more,
then we expect it on the
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average to be cut by a
factor again of 1/2.
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And so we expect to be left with
a stick that has length
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1/4 of what we started with.
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So this example is a
particularly nice one, because
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we used all of the concepts
that we have introduced--
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marginal PDFs, joint PDFs,
conditional PDFs, and the
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relations between them, as
well as expectations,
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calculations of expectations,
and conditional expectations,
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as well as the total probability
theorem.