Probability of product of two random variables Suppose you are given 2 random variables $X$ and $Y$ following some distribution (say geometric, $p_1$ and $p_2$ respectively). Can anyone give a hint how to calculate $P(XY = z)$? By which I mean the probability of $X$ multiplied by $Y$ equals $z$, for some $z$.
 A: Since the distribution of $XY$ is characterised by its cdf, you want to compute $\mathbb{P}(XY<z)$ for an arbitrary value $z$. Let us assume first that $Y$ is always positive. Then 
$$\eqalign{\mathbb{P}(XY<z)&=\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]\\&=\mathbb{E}[\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]|Y]]\\&=\mathbb{E}[\mathbb{P}(XY<z|Y)]\\&=\mathbb{E}[F_X(z/Y)|Y]}$$
If $Y$ can take both positive and negative values,
$$\eqalign{\mathbb{P}(XY<z)&=\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]\\&=\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)\mathbb{I}_{(-\infty,0)}(Y)]+\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)\mathbb{I}_{(0,\infty)}(Y)]\\&=\mathbb{E}[\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]|Y]\mathbb{I}_{(-\infty,0)}(Y)]+\mathbb{E}[\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]|Y]\mathbb{I}_{(0,\infty)}(Y)]\\&=\mathbb{E}[\mathbb{P}(XY<z|Y)\mathbb{I}_{(-\infty,0)}(Y)]+\mathbb{E}[\mathbb{P}(XY<z|Y)\mathbb{I}_{(0,\infty)}(Y)]\\&=\mathbb{E}[\mathbb{I}_{(-\infty,0)}(Y)F_X(z/Y)]+\mathbb{E}[\mathbb{I}_{(0,\infty)}(Y)\{1-F_X(z/Y)\}]}$$
Depending on the setting, the density of $XY$ can then be obtained by derivation of the above.
A: $ P(XY=k)= \sum_{t}P(X=t,Y=k/t)=\sum_{t}P(Y=k/t|X=t)P(X=t)$
Now,if $X$ and $Y$ are independent,then $P(Y=k/t|X=t)=P(Y=k/t)$ 
