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Assume $X,Y$ are discrete and have a finite outcome space.

How to evaluate $\mathbb{P}(XY=a)=\mathbb{P}(X=\frac{a}{Y})$, when $X$ or $Y$ take multple values?

Does one write out the result for each output value that $X,Y$ take?

Thus, if $Y \in \{0,1\}$, then

$$\mathbb{P}(XY=a)$$ $$= \mathbb{P}(X=\frac{a}{0}) \text{ or }\mathbb{P}(X=a) $$

out of which the first is undef.

I.e.

$$= \mathbb{P}(X=a) $$

If $Y \in \{0,1,2\}$, then

$$\mathbb{P}(XY=a)= \mathbb{P}(X=a) \text{ or } \mathbb{P}(X=a/2) $$

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    $\begingroup$ Are $X$ and $Y$ continuous or discrete variables? Is the set of possible values for each finite? $\endgroup$ – dlnB Mar 17 at 15:36
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    $\begingroup$ $P(XY=a)$ does not equal $P(X=\frac aY)$ when $Y=0$; just because you are using probability does not mean you are allowed to divide by $0$. $\endgroup$ – Dilip Sarwate Mar 17 at 16:09
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For discrete and finite $X$ and $Y$ you could say

$$\mathbb{P}(XY=a)= \begin{cases} \mathbb{P}(X=a/Y) &\quad \text{if $a \neq 0$}\\ \mathbb{P}(X=0 \lor Y=0) &\quad \text{if $a = 0$} \end{cases}$$

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  • $\begingroup$ My question didn't particularly concern $a$, but $X,Y$. So could assume $a$ fixed. $\endgroup$ – mavavilj Mar 17 at 17:25
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    $\begingroup$ @mavavilj you can replace the condition $a=0$ by $X=0 \lor Y=0$. It just shows that the case $a=0$ (which implies $X=0 \lor Y=0$) is special. When $Y$ is never 0 then $\mathbb{P}(X=0 \lor Y=0) = \mathbb{P}(X=0)$. $\endgroup$ – Martijn Weterings Mar 18 at 7:26
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What you want to do here is use the law of iterated expectation, that is $P(XY=a)=E[P(XY=a|Y)]$, where the expectation is taken over all values of $Y$. This works in the discrete and continuous case. In the discrete case you can write the resulting integral as $P(XY=a)=E[P(XY=a|Y)]=\sum_y P(X=a/y)P(Y=y)$, where the sum is taken over all values that $Y$ takes.

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