Hey, guys
in this video, I'm going to do this statistics and probability problem.
In this problem, we're given a joint pdf, which is for x1 and x2, and we're asked to
find the joint pdf of y1 and y2, where y1 and y2 are linear transformations.
So you have two random variables, and then a transform to two new random variables and
you're asked to find the joint probability of the new two random variables.
This is a simple, introductory example for a problem like this.
So let's go ahead and get started.
There's a formula for transforming the pdf of x1 and x2 to a new two variables, so let
me go ahead and write the formula before we continue.
That is, f_y1,y2 is equal to this long thing here.
This is basically telling you that you do the change of variables.
We're going to solve for x1 and x2 using these equations here that we're given.
We're going to plug that back in to the original function that we're given as well and find
new ranges and also multiply this function by the Jacobian and that will give you the
joint pdf for y1 and y2.
To get started, I'm going to solve for x1 and x2 given the y1 and y2 equations.
So we have y1 is equal to x2 minus x1.
Y2 is equal to x1.
This is going to just be simply that x1 is equal to y2.
And that solves for x1.
This, we're going to want to solve for x2 because we already have x1.
X2 is equal to y1 plus x1, and we have already that x1 is equal to y2.
So we have, x2 is equal to y1 plus y2.
This is basically our--in our equation, it's w1 and w2.
So we have that ready.
And we can also find the support and the Jacobian.
So let me go ahead and do the Jacobian first.
If we use x1 first, we're going to do partial y1 for the first element and then partial
y2 for the second.
It doesn't matter the order as long as you keep it consistent.
That's going to be zero, one.
And then for x2, as long as we're consistent, it's going to be one and one.
So the formula here, it's actually the absolute value of the Jacobian.
So the Jacobian is, you take the determinant of this matrix.
That's going to be zero minus one.
So the absolute value of negative one is one.
So in this case we just have a very simple Jacobian that's not going to affect the final
answer and so let's going ahead and plug in now to the formula.
We have f_y1,y2 is going to be equal to--.
We're going to take the original function of x1 and x2 and then use the new equations
for x1 and x2 that we solved for anywhere we see x1 and x2.
This is going to be 2e^-(y2 + y1+y2) and then this all times the Jacobian, which is just
one.
If we simplify this, that's going to be 2e^-(y1+2y2).
So that's our new pdf and let's also find the new ranges.
We have the ranges for x1 and x2 and now let's convert that to the new ranges for y1 and
y2.
The way we do that is we just break up what we're given and then solve for y1 and y2.
So we have zero is less than x1.
X1 is less than x2.
So again, this is kind of a simple example.
This just transforms into, if x1 is greater than zero, then y2 is greater than zero and
this implies that y2, which is x1, is less than x2, which is y1 plus y2 and this implies
that therefore y1 is greater than zero.
And these are both--you can think of this as less than infinity, so these are basically
are greater than zero and going toward infinity.
So for our final answer, this is our joint pdf and the support of that is over y1 > 0
and y2 > 0.
And we are done with this problem!
*chiptune music*