WEBVTT
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We now develop a methodology for
finding the PDF of the sum
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of two independent random
variables, when these random
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variables are continuous
with known PDFs.
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So in that case, Z will
also be continuous and
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so will have a PDF.
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The development is quite
analogous to the one for the
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discrete case.
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And in the discrete
case, we obtained
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this convolution formula.
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This convolution formula
corresponds to a summation
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over all ways that a certain
sum can be realized.
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In this picture, these are all
the ways that the sum of 3 can
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be realized.
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In the continuous case, the
different ways that the
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constant sum can be realized
corresponds to a line.
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So this is a line in which X
plus Y is equal to a constant.
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And we need to somehow add over
all the possible ways
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that the sum can be obtained,
add over all the
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points on this line.
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Now, when we're summing over all
the points of the line we
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really need to employ
an integral.
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And this leads to the following
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guess for the formula.
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Instead of having a summation,
we will have an integral.
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And the integral is over all the
X, Y pairs whose sum is a
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constant number, little z.
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So we have here the
family recipe--
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that sums are replaced by
integrals and PMFs are
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replaced by PDFs.
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So this formula is entirely
plausible.
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And it is called the
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continuous convolution formula.
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What we want to do next is to
actually justify this formula
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more rigorously.
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We will use the following
trick.
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We will first condition on the
random variable X, taking on a
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specific value.
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If we do this conditioning,
then the random variable Z
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becomes little x plus Y. And
to make the argument more
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transparent, we're going to look
first at the special case
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where little x is let's
say, the number 3.
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In which case our random
variable Z is
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equal to Y plus 3.
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Let us now calculate the
conditional PDF of Z in a
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universe in which we are told
that the random variable X
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takes on the value of 3.
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Now, given that X takes on the
value of 3, the random
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variable Z is the same as the
random variable Y plus 3.
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And now we have the conditional
PDF of y plus 3
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given X.
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However, we have assumed that
X and Y are independent.
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So the conditional PDF is going
to be the same as the
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unconditional PDF of Y plus 3.
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And we obtain this expression.
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Now, what is this?
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We know the PDF of Y. But now
we want the PDF of Y plus 3,
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which is a simple version of a
linear function of a single
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random variable Y. For a linear
function of this form,
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we have already derived
a formula.
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In the notation we have used
in the past, if we have a
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random variable X, and we add
the constant to it, the PDF of
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the new random variable is the
PDF of X but shifted by an
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amount equal to b
to the right.
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And that's what the shifting
corresponds to mathematically.
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Now, let's us apply this
formula to the case
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that we have here.
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We need to keep track of
the different symbols.
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So capital Y corresponds to X,
b corresponds to 3, little x
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corresponds to Z. And by using
these correspondences, what we
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obtain is f sub Y of this
argument, which is Z in our
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case minus b, which
is 3 in our case.
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And this is the final form for
the conditional density of Z
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given that X takes
a specific value.
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It's nothing more than the
density of Y, but shifted by 3
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units to the right.
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Let us now generalize this.
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Instead of using X equal to 3,
let us use a general number.
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And this gives us the more
general formula, that the
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conditional PDF of Z given that
X takes on a specific
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value is equal to--
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just use little x here
instead of 3.
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It takes this form.
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So we do have now in our hands
a formula for the conditional
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density of Z given X.
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Since we have the conditional,
and we also know the PDF of X,
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we can use the multiplication
rule to find the joint PDF of
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X and Z. By the multiplication
rule, it is the marginal PDF
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of X times the conditional PDF
of Z given X, which in our
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case takes this particular
form.
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And now that we have the joint
PDF in our hands, we can use
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another familiar formula
that takes us from the
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joint to the marginal.
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It would take the joint PDF and
integrate with respect to
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one argument, we obtain the
marginal PDF of the other
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random variable.
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Using this specific form that
we have for the joint PDF in
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this formula, we have finally
obtained this expression.
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This is the integral of the
joint PDF of X with Z
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integrated over all xs.
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And this proves this convolution
formula.
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In terms of the mechanics of
carrying out the calculation
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of the convolution, the
mechanics are exactly the same
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as in the discrete case.
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If you want to solve a problem
graphically, what you will do
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is to take the PDF of Y, flip
it horizontally, and then
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shift it by an amount of little
z, cross multiply
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terms, and integrate them out.