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:
 * $$p_X \left({x}\right) = f \left({x}\right) \sum_y g \left({y}\right)$$
 * $$p_Y \left({y}\right) = g \left({y}\right) \sum_x f \left({x}\right)$$

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.