Condition for Independence of Discrete Random Variables

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

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

Then $X$ and $Y$ are independent iff there exist functions $f, g: \R \to \R$ such that the joint mass function of $X$ and $Y$ satisfies:
 * $\forall x, y \in \R: p_{X, Y} \left({x, y}\right) = f \left({x}\right) g \left({y}\right)$

Proof
We have by definition of joint mass function that:
 * $x \notin \Omega_X \implies p_{X, Y} \left({x, y}\right) = 0$
 * $y \notin \Omega_Y \implies p_{X, Y} \left({x, y}\right) = 0$

Hence we only need to worry about values of $x$ and $y$ in their appropriate $\Omega$ spaces.

Sufficient Condition
Suppose there exist functions $f, g: \R \to \R$ such that:
 * $\forall x, y \in \R: p_{X, Y} \left({x, y}\right) = f \left({x}\right) g \left({y}\right)$

Then by definition of marginal probability mass function:
 * $\displaystyle p_X \left({x}\right) = f \left({x}\right) \sum_y g \left({y}\right)$
 * $\displaystyle p_Y \left({y}\right) = g \left({y}\right) \sum_x f \left({x}\right)$

Hence:

So it follows that:

Hence the result from the definition of independent random variables.

Necessary Condition
Suppose that $X$ and $Y$ are independent.

Then we can take the variables:
 * $f \left({x}\right) = p_X \left({x}\right)$
 * $g \left({y}\right) = p_Y \left({y}\right)$

and the result follows by definition of independence.