Probability Question on jointly distributed function

**DCU** · May 18th 2009, 07:20 AM

I'm having problems figuring out the limits for this. The exponential is e^-(x + y) apologies about the lack of clarity on it.

**Moo** · May 18th 2009, 07:44 AM

Hello,

You must have $\iint_{\mathbb{R}^2} f_{X,Y}(x,y) ~dxdy=1$

That is to say, by writing the boundaries:

$\int_0^\infty \int_0^y cxe^{-(x+y)} ~dxdy=1$

maybe reversing the order of integration will help :
$=\int_0^\infty \int_x^\infty cxe^{-(x+y)} ~dydx=c\int_0^\infty x \int_x^\infty e^{-(x+y)} ~dydx=1$

and that's all

**DCU** · May 18th 2009, 09:40 AM

For part (c) here would the limits be 0,inf for y and 0, 2-y for x dxdy?

**Moo** · May 18th 2009, 09:51 AM

Originally Posted by DCU

For part (c) here would the limits be 0,inf for y and 0, 2-y for x dxdy?

integral of the pdf, yup that's correct

**DCU** · May 18th 2009, 09:59 AM

Thanks, I think I might pass this exam!

**Moo** · May 18th 2009, 10:04 AM

Originally Posted by DCU

Thanks, I think I might pass this exam!

Sure ! Just follow your intuition, it doesn't look so bad

and if you have any other question, you know where you can get answers/help...

**matheagle** · May 18th 2009, 02:24 PM

I'd go one further and also pull out the $e^{-x}$ , as I did below.

Originally Posted by Moo

Hello,

You must have $\iint_{\mathbb{R}^2} f_{X,Y}(x,y) ~dxdy=1$

That is to say, by writing the boundaries:

$\int_0^\infty \int_0^y cxe^{-(x+y)} ~dxdy=1$

maybe reversing the order of integration will help :
$1=\int_0^\infty \int_x^\infty cxe^{-(x+y)} ~dydx=c\int_0^\infty x e^{-x}\int_x^\infty e^{-y} ~dydx$

and that's all

$1=c\int_0^\infty x e^{-x}e^{-x} dx=c\int_0^\infty x e^{-2x}dx$

Which can either be solved by parts or by recognizing the constant in the gamma density
with $\alpha=2$ and what I call $\beta =.5$ .

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