Sum and 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:

• $\forall \omega \in \Omega: U \left({\omega}\right) = X \left({\omega}\right) + Y \left({\omega}\right)$
• $\forall \omega \in \Omega: V \left({\omega}\right) = X \left({\omega}\right) Y \left({\omega}\right)$

Then both $U$ and $V$ are also 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:

$(1)$ The image of $U$ and $V$ are countable subsets of $\R$;

$(2)$ $\forall x \in \R: \left\{{\omega \in \Omega: U \left({\omega}\right) = x}\right\} \in \Sigma$ and $\left\{{\omega \in \Omega: V \left({\omega}\right) = x}\right\} \in \Sigma$.

Proof for Addition

First we consider any $U_u = \left\{{\omega \in \Omega: U \left({\omega}\right) = u}\right\}$ such that $U_u \ne \varnothing$.

We have that $U_u = \left\{{\omega \in \Omega: X \left({\omega}\right) + Y \left({\omega}\right) = u}\right\}$.

Consider any $\omega \in U_u$.

Then: $\omega \in X_x \cap Y_x$ where $X_x = \left\{{\omega \in \Omega: X \left({\omega}\right) = x}\right\}, Y_x = \left\{{\omega \in \Omega: Y \left({\omega}\right) = u - x}\right\}$.

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 $U_u = \bigcup_{x \in \R} \left({X_x \cap Y_x}\right)$.

That is, it is the union of all such intersections of sets whose discrete random variables add up to $u$.

As $X_x$ is a countable set it follows that $U_u$ is a Union of Countable Sets and therefore itself a countable set.

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

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

$\blacksquare$

Proof for Multiplication

The argument here is exactly the same as for addition.

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

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 Union of Countable Sets and therefore itself 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 $\left({\Omega, \Sigma, \Pr}\right)$.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 2.1$: Exercise $1$