Joint distribution

**Lononeer** · September 3rd 2009, 03:59 AM

The joint distribution of X and Y is given by f(x,y)= (exp(-y)/y)
0<x<y<infinity
I need to compute E(X^2 + Y^2 | Y=y)

can anyone please guide me on how to do this.

Thank You very much for Your help

**gustavodecastro** · September 3rd 2009, 04:47 AM

Using $f(x|y) = f(x,y) f(y),$ you can calculate the integral

$\iint \left(x^2 + y^2 \right) f(x|y)dxdy$

And to find $f(y)$ you have to calculate the integral

$\int_o^y f(x,y)dx$

**pedrosorio** · September 3rd 2009, 05:53 AM

Originally Posted by gustavodecastro

Using $f(x|y) = f(x,y) f(y),$

Why is this?

**matheagle** · September 3rd 2009, 04:57 PM

$E(X^2 + Y^2 | Y=y)= E(X^2|Y=y) + y^2$

and $E(X^2|Y=y)=\int x^2 f(x|y)dx$

**Lononeer** · September 4th 2009, 04:38 AM

is f(y) then =exp(-y) or Ei(-y)?
and if f(y) is equal to exp(-y)
is then
E(X^2+Y^2|Y0Y)
equal to [(exp(-2y)x^3)/3y]+[y^2] ?

really appreciate the help

**sssouljah** · September 5th 2009, 10:06 AM

Originally Posted by gustavodecastro

Using $f(x|y) = f(x,y) f(y),$ you can calculate the integral

$\iint \left(x^2 + y^2 \right) f(x|y)dxdy$

And to find $f(y)$ you have to calculate the integral

$\int_o^y f(x,y)dx$

is it dy or dx in the end of the integral defining f(y)?

**pedrosorio** · September 5th 2009, 10:49 AM

dx

and

$f(x|y) = \frac{f(x,y)}{f(y)}$

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