Joint Density expected values

**tttcomrader** · September 1st 2011, 10:46 AM

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

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

My solution:

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

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

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

Any help is appreciated, thank you!

**CaptainBlack** · September 3rd 2011, 01:45 AM

Originally Posted by tttcomrader

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

Is that even a density?

CB

**tttcomrader** · September 3rd 2011, 11:15 AM

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

**CaptainBlack** · September 3rd 2011, 11:29 AM

Originally Posted by tttcomrader

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

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

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

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

CB

**tttcomrader** · September 3rd 2011, 11:38 AM

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

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

**CaptainBlack** · September 3rd 2011, 08:24 PM

Originally Posted by tttcomrader

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

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

Except:

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

CB

**matheagle** · September 4th 2011, 11:32 PM

$f_X(x)=e^{-x}$ an exponential

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

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

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

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

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