# Joint CDF's and pdf problem

#### hmm...chocolate..

##### New Member
I am trying to solve part of a question whicg goes as:

Q: Suppose X and Y are two continuous random variables whose joint density function is

f(x,y) = 2exp[-x-y] ( i.e 2e^(-x-y) ) , 0<x<y<+infinity

Find the corresponding Joint Cumulative Distribution Function.

I am having trouble with choosing my limits in the intergration.
obviously for: x<0 or y<0 or both F(x,y)=0 .

And for 0<x<y , x<y<inf (i.e the part above the line y=x in the first quadrant) i used the limits:

u=0 to x and v=0 to y

and this gives me: F(x,y) = 2exp[-x-y] -4exp[-x] +2

is this right?

and it is at this last part where i am having trouble choosing my limits of intergration where: 0<x<y and y<x<inf (the region below the line y=x in the first quadrant).

If anyone could tell me if what i have done so far is correct or not! it will be very helpful.

thanks

E

#### elnaz

##### Guest
Hello
i just say you about base of the way of integrate
to know e.g if you want to compute integral with 0<x<y , x<y<inf
then outer inegral 's limits is from 0 to inf and inner integral 's limits is from 0 to y and you attend to that outer integral is on y and inner integral is on x
about x<0 or y<0 or both F(x,y)=0 , yes its true .

if you dont understand very well , please tell me

Best regards
Elnaz

#### hmm...chocolate..

##### New Member
Hi, thanks for that.

So for the intergration over:

0<x<y and x<y<inf

I get F(x,y) = 2( 1-e^(-y) )

(I think that is wrong though?)

using the limits u= 0 to infinity and v=0 to y. are these the correct limits to use?

E

#### elnaz

##### Guest
Hello
please attend to me , you have several integrals for getting this CDF that one of the integrals is when 0<x<y<inf then integral limits is from 0 to inf for outer integral on y and from 0 to y for inner integral on x, bold words is very important please attend to them and if you do that , your answer will be true .

Best rergards
Elnaz

#### hmm...chocolate..

##### New Member
I'm still having trouble with this integration, thanks for the help so far.

From the question I figure the Cdf is as:

for x<0 or y<0 or both F(x,y)=o

for 0<x<y<inf (above y=x) F(x,y) = ?

for 0<x<inf and y<x (below y=x) F(x,y) = ??

with ? an ?? being two different exprsessions. is this right?

integrating from 0 to y for the inner integral and 0 to inf on the outer integral i keep coming up with F(x,y) = 2(1-exp(-y)) which i think is wrong.

Intergrating from 0 to inf on the outer integral and x to inf I get F(x,y)=1 is this a correct inetegration?

also would F(x,y) = [Int (0 to inf)] of (2exp(-2x)) dx be the expression for '?'

I know that seems long winded but a response would be so helpful.
Thanks

E

#### elnaz

##### Guest
Hello
oh sorry i thought you want to compute density function but for distribution function is diffrent when 0<x<y<inf then integral limits is from 0 to x for outer integral on X and from X to y for inner integral on y.
plz compute it and i hope it will be true.
i hope you will be succeed.

#### hmm...chocolate..

##### New Member
Thanks for that, I think I may now have solved it.

So for the region 0<x<y<inf i used the inetrgration you said which gives me the answer of F(x,y) = e^(-2x) -2e^(-y) +1 .

And I think I computed the last region correctly too, where 0<x<inf and y<x. For this region I used the intergtarion computing the inetegral Y to x for the outter inetrgal on x and 0 to y on the inner integral on y. Is this the correct computation?

With all this i have the following answer:

f(x,y) = 2e^(-x-y) , 0<x<y<inf

then

F(x,y) =

0 for x<0 or y<0 or both

e^(-2x) - 2e^(-y) + 1 for 0<x<y<inf

2[e^(-x-y) + e^(-y) - e^(-2y) - e^(-x)] for 0<x<inf , y<x

Is that correct?

E

Hello