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 $$\left({\Omega, \Sigma, \Pr}\right)$$.

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 $$\left({\Omega, \Sigma, \Pr}\right)$$.