WEBVTT
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Hi, In this problem, we'll
be looking at the PDF the
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absolute value of x.
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So if we know a random variable,
x, and we know it's
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PDF, how can we use that
information to help us find
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the PDF of another random
variable-- the
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absolute value of x?
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And so throughout this problem,
we'll define a new
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random variable called y.
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And we'll define that y to be
equal to the absolute value of
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x, just to make things
simpler.
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So we'll do a couple of concrete
examples, and then
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we'll try to generalize
at the end.
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The first example that we'll
deal with in part A is
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this PDF for x.
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So we're told that the PDF of
x is 1/3 between negative 2
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and 1, and 0 otherwise.
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And here's a picture of
what it looks like.
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It's just a rectangle from
negative 2 to 1.
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So now we want to find out
what is the PDF of the
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absolute value of x, which
we've called y?
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And at this point, it may be
helpful to step back and think
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about this problem from the
discrete point of view again.
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So if x were a discrete random
variable, the problem would
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be, what is the probability that
the absolute value of x
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is equal to, say, 1/2?
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Well, the probability that
the absolute value of
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x is equal to 1/2--
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that can occur in two
different ways.
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One is that x itself is 1/2.
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Or x could be negative 1/2, in
which case, the absolute value
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of x would still be 1/2.
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So those two events are
mutually exclusive.
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And so the probability of either
one of them happening
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is you can just add them up.
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And so the probability of the
absolute value of x being 1/2
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would have two contributions,
one from x being 1/2, and one
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from x being negative 1/2.
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The analogous idea carries over
to the continuous case,
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when you have a PDF.
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So now let's say that we're
interested in the case where
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we want to know the
PDF of y at 1/2.
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Well, that again, is going to
have two contributions, one
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from where x is 1/2, and one
from where x is minus 1/2.
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And so you can just imagine that
each one of these values
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for y-- and remember, y has to
be non-negative, because it's
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an absolute value--
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has two contributions, one from
the right side of 0, and
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one from the left, or
negative, side of 0.
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So you can come up and write
an algebraic expression for
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this, and we'll do that in Part
C. But you can also look
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at this from a visual
point of view.
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And you can take the PDF diagram
itself and imagine
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transforming it to find out what
the PDF of the absolute
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value of x would look like.
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So the way to do it would be
you take what's on the
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negative side.
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You flip it over and take the
mirror image, and then you
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stack it on top of what you have
on the right-hand side,
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or the positive side.
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So take this, flip it over,
and stack it on top.
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You can imagine just taking this
block, flipping it over.
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And just think of it as like a
Tetris block that's falling
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down from above.
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And it stacks on top of
wherever it lands.
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So it'll turn it into something
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that looks like this.
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So there's already a block of
height 1/3 from 0 to 1.
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That's from the original 1.
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And now we take this, and flip
it over, and drop it on top.
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Well, this part is going to
fall on top of the segment
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from 0 to 1.
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And then this part gets
flipped over and
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dropped over here.
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And it falls down here.
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And so the final PDF actually
looks like this kind of
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staircase, where this is 2/3
now, because this has two
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contributions of 1/3 each,
and this is 1/3.
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So that is the graphical way
of approaching this.
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And the PDF for completeness,
the PDF of y would be 2/3 for
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y between 0 and 1, 1/3 for y
from 1 to 2, and 0 otherwise.
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All right, so let's move
on to part B, and
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get some more practice.
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Part B, we're given that this
PDF of x now is 2 times e to
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the negative 2x for x positive,
and 0 otherwise.
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Now you may just recognize this
as an exponential random
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variable with a parameter
of 2.
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And again, we can graph this
and see what it looks like.
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And it turns out that it's going
to start out at 2 and
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fall off exponentially.
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So in this case, this is
actually quite simple.
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Because if you look at it,
x is already positive.
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It doesn't have any
negative parts.
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So in fact, the absolute value
of x is the same as x itself,
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because x is never negative.
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And so y is just the
same thing as x.
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And so in this case, actually,
the PDF of y is exactly the
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same as the PDF of x.
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It's just 2e to the
minus 2y, for y
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positive and zero otherwise.
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Now you can also see this
graphically also, because to
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the left of 0, the negative
part, there is no PDF.
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The PDF is 0.
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And so if you were to take this,
flip it over, and drop
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it on top, you wouldn't get
anything, because there's
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nothing there.
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And so the entire PDF, even
after you take the absolute
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value, is just the
original one.
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So to generalize, what I said
at the beginning was that,
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remember, the probability in
the discrete case, if you
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wanted the probability that the
absolute value of a random
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variable equals something,
that would just be the
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probability that the random
variable equals that value of
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little x, or the random
variable equals
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negative little x.
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In either of those two
cases, the absolute
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value would equal x.
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So you get those two
contributions.
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And so to generalize in the
continuous case with PDFs, you
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get something that looks
very similar.
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So in this case, the PDF or y
is just the PDF of x at y.
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So this is the case where x is
just equal to y, plus the PDF
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of x evaluated negative y.
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So you, again, have both of
these two contributions.
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And we can rewrite this
top one to make
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it look more similar.
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So the PMF of some discrete
[? number ?] y, where this is
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a discrete random variable
that's equal to the absolute
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value of x, would be the PMF of
x evaluated at y, plus the
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PMF of x evaluated
at negative y.
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So in both the discrete and
continuous cases, you have the
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same thing.
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So the overall summary of this
problem is that, when you take
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a transformation--
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in this case, an absolute
value--
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you can reason about it and
figure out how to decompose
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that into arguments about the
original random variable, just
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plain old x.
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And for the specific case of
the absolute value, it just
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becomes taking a mirror image
and popping it on top of what
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you originally had.
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So remember, you always have
these two contributions.
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And so if you ever have a random
variable that you need
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to take an absolute value of,
you don't have to be scared.
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All you have to do is consider
both of these contributions
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and add them up, and you have
the PDF that you want.
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So I'll see you next time.