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