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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)=1e^[x]e^[y]+e^[xy] So f(x,y)= e^[xy] 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 (3y) to (infinite) for dx And 0 to (infinite) for dy? 
February 10th, 2017, 09:41 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664 
the pdf is actually $f_{XY}(x,y) = e^{(x+y)},~x,y \geq 0$ so $P[X+Y>3] = 1P[X+Y\leq 3] = 1\displaystyle{\int_0^\infty \int_0^{3x}}~e^{(x+y)}~dx =$ $1 \displaystyle{\int_0^3 \int_0^{3x}}~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 
Senior Member Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664  do they show how they derive this? you can do the integral directly as $P[X+Y>3] = \displaystyle{\int_0^3 \int_{3x}^\infty}~e^{(x+y)}~dx + \displaystyle{\int_3^\infty \int_{0}^\infty}~e^{(x+y)}~dx$ and come up with the same answer as in post #2 
February 11th, 2017, 11:38 AM  #6 
Senior Member Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664  

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density, distribution, finding, function, probability 
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