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Let's consider two independent random variables $X,Y$, where $X\geq0$. We know both probabilty density functions (pdf) and denote them by $f_X$ and $f_Y$. We want to compute the pdf of the random variable $Z:=X\cdot Y$. In order to do that we take a look at the probability $P(X\cdot Y\leq z)$, where $z\in\mathbb{R}$.


After a few steps and assuming that $x>0$ we get $$ \int\limits_{0}^{\infty}\int\limits_{-\infty}^{\frac{z}{x}}f_X(x)f_Y(y)dy~dx. $$ Finally, by taking the derivative we find a pdf of $Z$.

The solution in general is clear to me but what is the correct reasoning to say that we can assume $x>0$ though it is $X\geq0$ in the beginning?


Can I argue like this:

In the context of one dimensional integrals we know that excluding single points doesn't change the integral. I guess that this reasoning can be somehow generalized to two dimensional integrals.

If $x\leq 0$ then we know that $P(X\leq x)=0$ because of $X\geq 0$. Further, we know that $f_X(x)=(P(X\leq x))'$ and since $(P(X\leq X))'=0$ if $x\leq 0$ it follows $f_X(x)=0$ if $x\leq 0$. So the joint pdf $f_X(x)f_Y(y)$ attains $0$ if $x=0$ and it doesn't matter if we assume $x>0$.

Is this correct? Or how would you argue instead if you don't have any further measure theoretic tools at hand?

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Your result holds true only if, for example, $X$ is a continuous random variable. In general, the solution should be: $$\begin{align} \mathbb{P}(X\cdot Y\le z) &=\mathbb{P}(X\cdot Y\le z,X>0)+\mathbb{P}(X\cdot Y\le z, X =0)\\ &=\mathbb{P}(X\cdot Y\le z,X>0)+\mathbb{P}(z \ge 0, X =0)\\ &=\color{red}{\mathbb{P}(X\cdot Y\le z,X>0)}+ \mathbf{1}_{\{z \ge 0 \}}\color{blue}{\mathbb{P}(X =0)}\\ \end{align}$$

The red term is the result in your question : $\int\limits_{0}^{\infty}\int\limits_{-\infty}^{\frac{z}{x}}f_X(x)f_Y(y)dy~dx$.

For a $z \ge 0$, the blue term can be diffrent to $0$ if $X$ is a discrete random variable .

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