Hey guys, in this video I'm going to go over this problem.
You would usually find a problem like this in a statistics and probability class.
It says let U be a uniform random variable and then it gives you the PDF and
the CDF. Then it asks you to consider the function or the transform Y equal to U squared and it asks you to find the
probability distribution of Y. So if you're asked to get a probability distribution,
you can provide three
answers or you can basically solve the problem three ways. You can give the PDF of Y, which would be f_Y(y).
You can give the CDF, which is capital F_Y(y).
Or you can give the moment-generating function, which is one-to-one with a PDF.
That's also a unique solution. And so for this problem
I'm going to use the CDF approach to provide the probability distribution of Y.
So the CD--to get the CDF of Y,
we're going to use what the transform is, which is what we're given.
There's a formula basically that says that the CDF of Y,
written like this, is equal to the probability that Y is less than or equal to y and
we know that Y is equal to U squared so usually you'll--when you learn about it,
you'll learn that Y can be
considered a function of X. In this case our X is U, so Y is equal to U squared.
This is the same as saying U squared is less than or equal to Y and if we solve for Y,
this is the same as saying the probability
of U is less than or equal to the square root of y. And now that we have U is less than or equal to
some value,
this probability is the same as the CDF of X
when you have X equal to the square root of y and in this case our X is U. So we go from
F_Y of y to F_U of a value where you plug in y.
So once you get from here to here,
we already have F_U of u and now
to get the CDF for F of Y, all we have to do is plug in the square root of y for u in
this distribution.
So I'm going to go ahead and do that. So F_Y(y) is equal to zero. And again,
we're going to plug in the square root of y for u.
And if we square that, we still get y less than zero for the range. For u,
we're going to plug in the square root of y.
And then this is going to be 1. If we have 0
less than or equal to the square root of y less than 1, if we square that on both sides
we're still going to get 0 less than or equal to y less than 0--less than 1, excuse me--and the same thing here.
So now we officially have the CDF of Y. The CDF of Y is a unique solution,
so it's sufficient to just list the CDF of Y if you have a question asking what the probability distribution of Y is.
But if there's a follow-up question, for example, that asks what the PDF of Y is
you can also get that solution. So the PDF of Y, which is written
f_Y(y), is equal to the derivative of the CDF.
If we have that the
CDF function is the square root of y, that's equal to y to the 1/2. If you take the derivative, that's y to the negative
1/2, which is equal to 1 over 2 times the square root of y. So for f_Y(y), we have
1 over 2 times square root of y. Y cannot be 0 because then the denominator
for this fraction will be 0. So y has to be greater than 0 and we have from our CDF of Y
that this goes to 1, and then it's going to be 0 otherwise. And with that we are done with this problem.
*chiptune music*