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

(a) Find the conditional density of $X$, given $Y = y$, and that of $Y$, given $X = x$.

(b) Find the density function of $Z = XY$.


In the first place, we should determine the marginal distributions: \begin{align*} f_{X}(x) = \int_{0}^{\infty}f_{X,Y}(x,y)\mathrm{d}y = \int_{0}^{\infty}xe^{-x(y+1)}\mathrm{d}y = xe^{-x} \end{align*}

where $x > 0$. Analogously, we have \begin{align*} f_{Y}(y) = \int_{0}^{\infty}f_{X,Y}(x,y)\mathrm{d}x = \int_{0}^{\infty}xe^{-x(y+1)}\mathrm{d}x = \frac{1}{(y+1)^{2}} \end{align*}

where $y > 0 $.

(a) Based on the previous results, it comes

\begin{cases} f_{X|Y}(x|y) = \displaystyle\frac{f_{X,Y}(x,y)}{f_{Y}(y)} = x(y+1)^{2}e^{-x(y+1)}\\\\ f_{Y|X}(y|x) = \displaystyle\frac{f_{X,Y}(x,y)}{f_{X}(x)} = e^{-xy} \end{cases}

(b) Finally, we have \begin{align*} F_{Z}(z) & = \textbf{P}(XY \leq z) = \int_{0}^{\infty}\int_{0}^{z/x}f_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x = \int_{0}^{\infty}\int_{0}^{z/x}xe^{-x(y+1)}\mathrm{d}y\mathrm{d}x\\\\ & = \int_{0}^{\infty}(e^{-x} - e^{-x-z})\mathrm{d}x = 1 - e^{-z} \Rightarrow f_{Z}(z) = \frac{\mathrm{d}}{\mathrm{d}z}(1-e^{-z}) = e^{-z} \end{align*}

where $z > 0$.


1 Answer 1


The joint density $f(x,y)$ factors in the following ways:



From $(1)$ it follows that the conditional density of $Y\mid X$ is

$$f_{Y\mid X=x}(y)=xe^{-xy}\mathbf1_{y>0}$$

, and the marginal density of $X$ is $$f_X(x)=e^{-x}\mathbf1_{x>0}$$

From $(2)$, your answer for the conditional and marginal densities is correct.

So your answer for $f_X(x)$, though seemingly okay as it integrates to unity, is not correct. As a result, $f_{Y\mid X=x}(y)$ (the variable is $y$ here because it is a function of $y$ for a given $x$, not $(x,y)$ as you have written) is also out of line. The rest looks okay.

  • $\begingroup$ In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$. $\endgroup$
    – user0102
    Feb 3, 2019 at 16:34
  • $\begingroup$ That is the notation used in books, to emphasize the "given $x$" part. $\endgroup$ Feb 3, 2019 at 16:36
  • 3
    $\begingroup$ (+1) A minor quibble: the notation $f_{Y\mid X}(y\mid x)$ is common, one could even argue it is preferable to $f_{Y\mid X=x}(y)$. $\endgroup$
    – Did
    Feb 3, 2019 at 16:36

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