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
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Using $\displaystyle f(x|y) = f(x,y) f(y), $ you can calculate the integral $\displaystyle \iint \left(x^2 + y^2 \right) f(x|y)dxdy $ And to find $\displaystyle f(y) $ you have to calculate the integral $\displaystyle \int_o^y f(x,y)dx $
Originally Posted by gustavodecastro Using $\displaystyle f(x|y) = f(x,y) f(y), $ Why is this?
$\displaystyle E(X^2 + Y^2 | Y=y)= E(X^2|Y=y) + y^2$ and $\displaystyle E(X^2|Y=y)=\int x^2 f(x|y)dx$
Last edited by matheagle; Sep 3rd 2009 at 05:10 PM.
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
Originally Posted by gustavodecastro Using $\displaystyle f(x|y) = f(x,y) f(y), $ you can calculate the integral $\displaystyle \iint \left(x^2 + y^2 \right) f(x|y)dxdy $ And to find $\displaystyle f(y) $ you have to calculate the integral $\displaystyle \int_o^y f(x,y)dx $ is it dy or dx in the end of the integral defining f(y)?
dx and $\displaystyle f(x|y) = \frac{f(x,y)}{f(y)}$
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