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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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 900 sit-ups a day, Claude Leibovici, Jonas Meyer, hardmath, Tunk-Fey
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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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"Please leave a comment ...": We're not supposed to answer such questions. –  Gigili May 26 '12 at 18:25
@Gigili: I think you mean that it is your opinion that we ought not to answer such questions. If that is your opinion, you should feel free to downvote. But I believe that there is no policy, and that in particular "We're not supposed to" is not correct. –  André Nicolas May 26 '12 at 18:45
It's indeed my opinion and many other users. If we answer questions like this, how to punish the user for a poor formatted, multi-time asked question without even paying attention to what is the right way of asking a question? I wound't care if I were the OP because I got the answer I wanted. –  Gigili May 26 '12 at 18:54
@Gigili: Punishment can be a useful tool. I think most people are capable of learning even without punishment. I did do only half of the problem, because I did not find it attractive to be told how to do things. That was, however, not to punish, just a mildly childish reaction on my part. –  André Nicolas May 26 '12 at 19:02
Well, I disagree with you. I don't see any other ways of telling the OP to improve the question to get an answer. Not everything in our everyday life is attractive, we have to learn to do things and we need to be told about "how to"s to learn. –  Gigili May 26 '12 at 19:05