Joint Density expected values

• Sep 1st 2011, 10:46 AM
Joint Density expected values
Let X and Y be random variables with $\displaystyle p(x,y)=e^ {-y}$ for $\displaystyle 0<x<y<+ \infty$

Compute $\displaystyle E[X | Y ]$ and $\displaystyle E[Y | X ]$

My solution:

For $\displaystyle E[X | Y ]$, I have $\displaystyle \int x p(x | y) dx = \int x \frac {p(x,y)}{p(y)} dx$

$\displaystyle = \int x \frac {e^{-y}}{p(y)} dx$

But I'm stuck here because I don't know what p(y) is.

Any help is appreciated, thank you!
• Sep 3rd 2011, 01:45 AM
CaptainBlack
Re: Joint Density expected values
Quote:

Let X and Y be random variables with $\displaystyle p(x,y)=e^ {-y}$ for $\displaystyle 0<x<y<+ \infty$

Is that even a density?

CB
• Sep 3rd 2011, 11:15 AM
Re: Joint Density expected values
$\displaystyle p(x,y)=e^ {-y} for 0<x<y<+ \infty$ is the joint pdf, nothing else is given. Is $\displaystyle p(y) = \int ^y _0 p(x,y) dy$ then?
• Sep 3rd 2011, 11:29 AM
CaptainBlack
Re: Joint Density expected values
Quote:

$\displaystyle p(x,y)=e^ {-y} for 0<x<y<+ \infty$ is the joint pdf, nothing else is given. Is $\displaystyle p(y) = \int ^y _0 p(x,y) dy$ then?

I meant is the integral over the 1st quadrant 1? Checking it now the answer appears to be yes.

$\displaystyle p(y)$ is a marginal and so:

$\displaystyle p(y)=\int_0^y e^{-y} \; dx=ye^{-y}$

CB
• Sep 3rd 2011, 11:38 AM
Re: Joint Density expected values
Thank you! Now to find p(x), should I do this:

$\displaystyle \int ^ x _ 0 e^{-y} dy = -e^{-x}-1$?
• Sep 3rd 2011, 08:24 PM
CaptainBlack
Re: Joint Density expected values
Quote:

Thank you! Now to find p(x), should I do this:

$\displaystyle \int ^ x _ 0 e^{-y} dy = -e^{-x}-1$?

Except:

$\displaystyle \int ^ x _ 0 e^{-y}\; dy = \biggl[-e^{-x}\biggr]_0^x=1-e^{-x}$

CB
• Sep 4th 2011, 11:32 PM
matheagle
Re: Joint Density expected values
$\displaystyle f_X(x)=e^{-x}$ an exponential

$\displaystyle f_Y(y)=ye^{-y}$ a gamma

$\displaystyle f_{X|Y}(x,y)=1/y$ but where 0<x<y, which is a uniform

$\displaystyle E(X|Y)=\int_0^y{x\over y}dx={y\over 2}$ the center of that uniform

$\displaystyle E(Y|X)=\int_x^{\infty}ye^{-y}e^xdy=e^x\int_x^{\infty}ye^{-y}dy$