My Math Forum Finding a probability density function from a distribution function and using it.

 February 10th, 2017, 09:14 PM #1 Newbie   Joined: Feb 2017 From: Corpus Christi Texas Posts: 6 Thanks: 0 Finding a probability density function from a distribution function and using it. F(x,y)=1-e^[-x]-e^[-y]+e^[-x-y] So f(x,y)= e^[-x-y] if I did both partials over. I'm told to find P(X+Y>3) I can't quite get the bounds right. It's a double integral of f(x,y). Are the founds (3-y) to (infinite) for dx And 0 to (infinite) for dy?
 February 10th, 2017, 09:41 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,128 Thanks: 1108 the pdf is actually $f_{XY}(x,y) = e^{-(x+y)},~x,y \geq 0$ so $P[X+Y>3] = 1-P[X+Y\leq 3] = 1-\displaystyle{\int_0^\infty \int_0^{3-x}}~e^{-(x+y)}~dx =$ $1 -\displaystyle{\int_0^3 \int_0^{3-x}}~e^{-(x+y)}~dx$ I leave it to you to compute this.
 February 11th, 2017, 10:53 AM #3 Newbie   Joined: Feb 2017 From: Corpus Christi Texas Posts: 6 Thanks: 0 Oh that makes sense why you would use the compliment . Thanks!
 February 11th, 2017, 11:16 AM #4 Newbie   Joined: Feb 2017 From: Corpus Christi Texas Posts: 6 Thanks: 0 Mmmmmmm Last edited by Rramos2; February 11th, 2017 at 11:35 AM.
February 11th, 2017, 11:38 AM   #5
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Quote:
 Originally Posted by Rramos2 Hmm the book says its (e^(-2)-e^(-3))^(2)
do they show how they derive this?

you can do the integral directly as

$P[X+Y>3] = \displaystyle{\int_0^3 \int_{3-x}^\infty}~e^{-(x+y)}~dx + \displaystyle{\int_3^\infty \int_{0}^\infty}~e^{-(x+y)}~dx$

and come up with the same answer as in post #2

February 11th, 2017, 11:38 AM   #6
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Quote:
 Originally Posted by Rramos2 Mmmmmmm
care to elaborate on that? :P

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