Product of Discrete Random Variables

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.

Let $V: \Omega \to \R$ be defined as:
 * $\forall \omega \in \Omega: \map V \omega = \map X \omega \map Y \omega$

Then $V$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Proof
To show that $V$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:


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

First we consider any $V_V = \set {\omega \in \Omega: \map V \omega = v}$ such that $V_v \ne \O$.

We have that:
 * $V_v = \set {\omega \in \Omega: \map X \omega \map Y \omega = v}$

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:
 * $X_x = \set {\omega \in \Omega: \map X \omega = x}, Y_x = \set {\omega \in \Omega: \map Y \omega = \dfrac v x}$

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

As $\struct {\Omega, \Sigma, \Pr}$ is a probability space then $X_x \cap Y_x \in \Sigma$.

Now note that $\ds V_v = \bigcup_{x \mathop \in \R} \paren {X_x \cap Y_x}$.

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 $\struct {\Omega, \Sigma, \Pr}$ being a probability space, $V_v \in \Sigma$.

Thus $V$ is a discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.