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February 10th, 2017, 10:14 PM   #1
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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?
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February 10th, 2017, 10:41 PM   #2
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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.
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February 11th, 2017, 11:53 AM   #3
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Oh that makes sense why you would use the compliment . Thanks!
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February 11th, 2017, 12:16 PM   #4
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Mmmmmmm

Last edited by Rramos2; February 11th, 2017 at 12:35 PM.
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February 11th, 2017, 12:38 PM   #5
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Quote:
Originally Posted by Rramos2 View Post
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
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February 11th, 2017, 12:38 PM   #6
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Quote:
Originally Posted by Rramos2 View Post
Mmmmmmm
care to elaborate on that? :P
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