Function of Two Discrete Random Variables

Theorem
Let $$\left({\Omega, \Sigma, \Pr}\right)$$ be a probability space.

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

Let $$g: \R^2 \to \R$$ be a real-valued function.

Then $$Z = g \left({X, Y}\right)$$, defined as:
 * $$\forall \omega \in \Omega: Z \left({\omega}\right) = g \left({X \left({\omega}\right), Y \left({\omega}\right)}\right)$$

is also a discrete random variable.

Corollary
Let $$\left({\Omega, \Sigma, \Pr}\right)$$ be a probability space.

Let $$X_1, X_2, \ldots, X_n$$ be discrete random variables on $$\left({\Omega, \Sigma, \Pr}\right)$$.

Then $$Z = g \left({X_1, X_2, \ldots, X_n}\right)$$, defined as:
 * $$\forall \omega \in \Omega: Z \left({\omega}\right) = g \left({X_1 \left({\omega}\right), X_2 \left({\omega}\right), \ldots, X_n \left({\omega}\right)}\right)$$

is also a discrete random variable.

Proof
As $$\operatorname{Im} \left({X}\right)$$ and $$\operatorname{Im} \left({Y}\right)$$ are countable, then so is $$\operatorname{Im} \left({g \left({X, Y}\right)}\right)$$.

Now consider $$g^{-1} \left({Z}\right)$$.

We have that:
 * $$\forall x \in \R: X^{-1} \left({x}\right) \in \Sigma$$.
 * $$\forall y \in \R: Y^{-1} \left({x}\right) \in \Sigma$$.

We also have that $$\forall z \in \R: g^{-1} \left({z}\right) = \bigcup_{\left({x, y}\right): g \left({x, y}\right) = z} \left\{{\left({x, y}\right)}\right\}$$.

But $$\Sigma$$ is a sigma-algebra and therefore closed for unions.

Thus $$\forall z \in \R: g^{-1} \left({z}\right) \in \Sigma^2$$.

Now consider $$\left({v, w}\right) \in g^{-1} \left({Z}\right)$$.

As $$\Sigma$$ is a sigma-algebra it follows directly that:
 * $$\forall x \in v: X^{-1} \left({x}\right) \in \Sigma$$
 * $$\forall y \in w: Y^{-1} \left({y}\right) \in \Sigma$$

Hence the result.

Proof of Corollary
Follows straightforwardly by induction from the main result.