Probability Theory: distribution of Z=XY [on hold]

The joint distribution function of $X$ and $Y$ is given by:

$$f(x,y)= \begin{cases} x e ^ {-x(y+1)} & x>0,y>0\\ 0 &\text{otherwise} \end{cases}$$

Find the distribution of $Z=XY$ using the following methods:

• distribution function
• transformation
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put on hold as off-topic by 900 sit-ups a day, Claude Leibovici, Jonas Meyer, hardmath, Tunk-Fey2 days ago

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How to ask a homework question? –  Gigili May 26 '12 at 15:22
–  Did May 26 '12 at 15:35
@user32268: When you are asking a question, it is best not to specify the method that answers should use. Then you may get various answers using various approaches, giving you a chance to learn more. –  André Nicolas May 26 '12 at 17:17
@André Of course you may be right on this, but this is reallly not the first observation I would make about this user. –  Did May 28 '12 at 12:19

We do the first method only. We want the cumulative distribution function $F_Z(z)$ of $Z$. This is $P(XY \le z)$, There is nothing interesting for $z\le 0$. So take $z>0$.

Let $R$ be the part of the first quadrant which is below the hyperbola $xy=z$. Then $P(XY \le z)$ is the integral of our joint density over the region $R$.

Now we have a pure integration question. It is (much) more pleasant to integrate first with respect to $y$. So we want $$\int_{x=0}^\infty \left(\int_{y=0}^{z/x} xe^{-x(y+1)}\, dy \right)dx.$$

The inner integral is easy. An antiderivative is $-e^{-x(y+1)}$. Substitute the endpoints. We get $e^{-x} -e^{-(z+x)}$. Now integrate with respect to $x$. It is best to simplify a bit first.

You will end up with something very familiar. That may help in finding a less computational way to solve the same problem,

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