Help setting integral limits for P(X1 <X3 -X2)

Let (X1 ,X2, X3) be i.i.d. r.v.'s with common probability density function

f(x)= e^(-x), 0 <x and 0 otherwise

Find P(X1 <X3 -X2)

I'm having a difficult time setting up my integral

I know f(x1,x2,x3) = f(x1)*f(x2)*f(x3)= e^(-x1-x2-x3)

I can't figure out the limits.

Any help would certainly be appreciated

Re: Help setting integral limits for P(X1 <X3 -X2)

If the pdf was supported between zero and one, say, then the space would be a unit cube, and the region satsifying X_{1} < X_{3} - X_{2} ...

... or let's call it z < x - y ...

... would be the tetrahedron underneath z = x - y ...

http://www.ballooncalculus.org/draw/misc/tetra.png

Which would be...

$\displaystyle \displaystyle{\int_0^1\ \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }$

... but extended becomes...

$\displaystyle \displaystyle{\int_0^{\infty} \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }$

Re: Help setting integral limits for P(X1 <X3 -X2)

Quote:

Originally Posted by

**tom@ballooncalculus** If the pdf was supported between zero and one, say, then the space would be a unit cube, and the region satsifying X

_{1} < X

_{3} - X

_{2} ...

... or let's call it z < x - y ...

... would be the tetrahedron underneath z = x - y ...

http://www.ballooncalculus.org/draw/misc/tetra.png
Which would be...

$\displaystyle \displaystyle{\int_0^1\ \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }$

... but extended becomes...

$\displaystyle \displaystyle{\int_0^{\infty} \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }$

I really appreciate you breaking it down for me. That's what I ended up "guessing" but this was very helpful. When I solved the integral I got 1/4