WEBVTT
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In this video, we're going to
do an example in which we
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derive the probability density
function of the sum of two
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random variables.
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The problem tells us
the following.
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We're given that X and Y are
independent random variables.
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X is a discrete random
variable with PMF Px.
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Y is continuous with PDF Fy.
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And we'd like to compute the PDF
of Z which is equal to X
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plus Y. We're going to use the
standard approach here--
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compute the CDF of Z
and then take the
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derivative to get the PDF.
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So in this case, the CDF, which
is Fz, by definition is
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the random variable Z being
less than little z.
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But Z is just X plus Y. So now,
we'd actually like to,
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instead of having to deal with
two random variables, X and Y,
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we'd like to deal with
one at a time.
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And the total probability
theorem allows us to do this
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by conditioning on one of the
two random variables.
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Conditioning on Y here is a
bit tricky, because Y is
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continuous, and you have to be
careful with your definitions.
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So conditioning on X seems
like the way to go.
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So let's do that.
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This is just the probability
that X equals little x, which
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is exactly equal to the PMF
of X evaluated at x.
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Now we're given we're fixing
X equal to little x.
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So we can actually replace every
instance of the random
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variable with little x.
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And now I'm going to just
rearrange this so that it
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looks a little nicer.
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So I'm going to have Y on the
left and say Y is less than z
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minus x, where z minus
x is just a constant.
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Now, remember that X and
Y are independent.
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So telling us something about X
shouldn't change our beliefs
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about Y. So in this case,
we can actually drop the
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conditioning.
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And this is exactly the CDF of
Y evaluated at z minus x.
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So now we've simplified
as far as we could.
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So let's take the derivative and
see where that takes us.
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So the PDF of Z is, by
definition, the derivative of
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the CDF, which we just
computed here.
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This is sum over x
Fy z minus x Px.
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What next?
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Interchange the derivative
and the summation.
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And a note of caution here.
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So if x took on a finite number
of values, you'd have a
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finite number of terms here.
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And this would be completely
valid.
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You can just do this.
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But if x took on, for
example, a countably
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infinite number of values--
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a geometric random variable,
for example--
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this would actually require
some formal justification.
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But I'm not going to
get into that.
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So here, the derivative with
respect to z-- this is
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actually z--
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is you use chain rule here.
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Px doesn't matter, because
it's not a function of z.
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So we have Fy evaluated at z
minus x according to the chain
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rule, and then the derivative of
the inner quantity, z minus
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x, which is just 1.
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So we don't need to put
anything there.
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And we get Px of x.
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So there we go.
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We've derived the PDF of z.
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Notice that this looks quite
similar to the convolution
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formula when you assume that
both X and Y are either
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continuous or discrete.
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And so that tells us that
this looks right.
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So in summary, we've basically
computed the PDF of X plus Y
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where X is discrete and
Y is continuous.
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And we've used the standard
two-step approach--
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compute the CDF and then take
the derivative to get the PDF.