Product of Discrete Random Variables

Theorem
Let $X$ and $Y$ be discrete random variables on the probability space $\left({\Omega, \Sigma, \Pr}\right)$.

Let $U: \Omega \to \R$ and $V: \Omega \to \R$ be defined as:
 * $\forall \omega \in \Omega: V \left({\omega}\right) = X \left({\omega}\right) Y \left({\omega}\right)$

Then $V$ is also a discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$.

Proof
To show that $U$ and $V$ are discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$, we need to show that:


 * $(1): \quad $The image of $U$ and $V$ are countable subsets of $\R$;
 * $(2): \quad \forall x \in \R: \left\{{\omega \in \Omega: U \left({\omega}\right) = x}\right\} \in \Sigma$ and $\left\{{\omega \in \Omega: V \left({\omega}\right) = x}\right\} \in \Sigma$.

First we consider any $V_V = \left\{{\omega \in \Omega: V \left({\omega}\right) = v}\right\}$ such that $V_v \ne \varnothing$.

We have that:
 * $V_v = \left\{{\omega \in \Omega: X \left({\omega}\right) Y \left({\omega}\right) = v}\right\}$

Consider any $\omega \in V_v$.

If $v = 0$ the result follows immediately, so we assume that $v \ne 0$.

Then:
 * $\omega \in X_x \cap Y_x$

where:
 * $X_x = \left\{{\omega \in \Omega: X \left({\omega}\right) = x}\right\}, Y_x = \left\{{\omega \in \Omega: Y \left({\omega}\right) = \dfrac v x}\right\}$

Both $X_x \in \Sigma$ and $Y_x \in \Sigma$ (as $X$ and $Y$ are discrete random variables.

As $\left({\Omega, \Sigma, \Pr}\right)$ is a probability space then $X_x \cap Y_x \in \Sigma$.

Now note that $V_v = \bigcup_{x \in \R} \left({X_x \cap Y_x}\right)$.

That is, it is the union of all such intersections of sets whose discrete random variables multiply together to make $v$.

As $X_x$ is a countable set it follows that $V_v$ is a countable union of countable sets.

From Countable Union of Countable Sets is Countable it follows that $X_x$ is a countable set.

And, by dint of $\left({\Omega, \Sigma, \Pr}\right)$ being a probability space, $V_v \in \Sigma$.

Thus $V$ is a discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$.