00:01
We are given a joint probability distribution for the variables x and y.
00:05
And we are asked a couple things.
00:06
The first thing we have to do is find the value of the constant c so that this is actually a valid density function.
00:14
And remember, one of the axioms of probability is that the integral over the entire domain of the density function has to be equal to 1.
00:31
So when we integrate the density, we should get a 1.
00:34
So we can use that to solve for c.
00:37
Before we do that, i think we should rewrite our density function.
00:41
We can take advantage of its form.
00:46
So we'll still have the 0 down there.
00:48
But this exponential function can be split up, since we have the sum in the exponent there.
00:55
So we can write that as e to the negative x times e to the negative 2y.
01:00
And being able to write it like that makes a lot of things easier for us.
01:03
That actually shows that the variables x and y are independent of each other, which we won't be using in this problem, but it comes in handy for later, things like that.
01:12
And it definitely makes it a bit more straightforward to integrate stuff like that.
01:17
So the domains are the exact same.
01:21
We just rewrote the density function.
01:23
So what does this domain look like? well, it's the support anyway, where the function is non -zero, is for x and y being positive.
01:33
So our domain of integration will be the first quadrant here.
01:38
So let's evaluate this out.
01:41
Again, we're going to say that one is equal to the integral from zero to infinity of the integral from zero to infinity.
01:47
Since this is a rectangular region, the order of integration doesn't particularly matter.
01:57
We can say, d .y, dx or dx, dx, since they're both going zero to infinity.
02:01
I'll say dydx here.
02:04
And then now we need to integrate.
02:06
This and solve the equation for c.
02:08
We can use a property of integration first to split this all up.
02:14
So notice we have the integral with respect to y and the integral with respect to x don't really overlap.
02:20
So we can have e to the negative x d x times the integral of e to the negative 2y, dy.
02:27
We can split it up like that because the density function can be factored as functions of only x and y.
02:34
So now we can treat these integral separately.
02:36
So when we take the any derivative, we get negative e to the negative x evaluated from 0 to infinity times negative 1 half, e to the negative 2y, also evaluated from 0 to infinity.
02:50
So what do we get for those? we can plug in infinity and we'll get 0 minus negative 1 times 0 minus negative 1 half when we evaluate all that out.
03:09
So that tells us we have c divided by 2.
03:16
And since that was equal to 1, that tells us that c equals, so now we have our complete density function of x and y, or at least the non -zero part, 2e to the negative x, e to the negative 2y, over the domains that were given.
03:37
So next, we want to find the cumulative density function.
03:41
So to find the cdf, that's going to be the capital f of x and y, where the density is given by little f, and this is going to be the integral up to the variables x and y.
03:57
The density function will be the same, but we have to change the variables out, since our new variables are x and y.
04:03
We'll use dummy variables.
04:04
So we'll take our density function, and we'll use the variables, i don't know, let's say t and u, either the negative 2u, to replace the x and the y.
04:12
And this will be dt, d ,u.
04:17
And then we just evaluate this out.
04:19
We'll get a function still of x and y, but it'll be the cumulative distribution and not the density.
04:26
So we can do the same tactic where we split up the integration by the variables that are present there.
04:38
So we can have 0 to x, 2 outside, e to the negative tdt, integral 0 to y, e to the negative 2u, and evaluate just like as we did before.
04:57
Negative 1ā2e to the negative to 2 from 0 to y.
05:04
This 2 and this 1ā2 here can cancel.
05:07
To make our calculations a little bit easier.
05:11
And when we evaluate this from 0 to x, we end up with 1 minus e to the negative x and 1 minus e to the negative 2, 1 .0 .1 .0 .2...