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The joint density function of $X$ and $Y$ is given by $f(x,y)=xe^{-x(y+1)}$ for $x>0,y>0$. find the conditional density of $Y$ given $X$.

I am close to get the answer but with a little problem
$\displaystyle f_{Y|X}(y|x)=\frac{f(x,y)}{f_x(x)} =\frac{xe^{-x(y+1)}}{\int_0^{\infty} xe^{-x(y+1)}dy} =\frac{e^{-xy}}{\int_0^\infty e^{-xy}dy}$

but when it goes infinite it will approach 0 and the final conditional density wouldn't make any sense.....hope someone can guide me to the right approach....thanks

universalset
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natsu
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  • You should use TeX (MathJax) to format, see here: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – universalset Dec 05 '13 at 02:33

1 Answers1

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It seems like your integral and simplification is incorrect.

For now, I would leave $f(x,y)=xe^{-x(y+1)}=xe^{-xy}e^{-x} = \frac{x}{e^{xy}e^{x}}$

Then, $f_x(x) = \int_0^\infty xe^{-x(y+1)}dy = \int_0^\infty xe^{-xy}e^{-x}dy = \frac{-xe^{-xy}e^{-x}}{x}|_0^\infty = -e^{-xy}e^{-x}|_0^\infty = \frac{-1}{e^\infty} - \frac{-1}{e^x} = \frac{1}{e^x}$

And now your fraction is:

$f_{Y|X}(y|x) = \frac{\frac{x}{e^{xy}e^{x}}}{\frac{1}{e^x}} = \frac{xe^x}{e^{xy}e^{x}} = \frac{x}{e^{xy}}$

I'm not sure if that is correct, but based on the information given, it would appear to be.

Nathan Towle
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