# Condition for Independence of Discrete Random Variables

## Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.

Then $X$ and $Y$ are independent if and only if there exist functions $f, g: \R \to \R$ such that the joint mass function of $X$ and $Y$ satisfies:

- $\forall x, y \in \R: \map {p_{X, Y} } {x, y} = \map f x \map g y$

## Proof

We have by definition of joint mass function that:

- $x \notin \Omega_X \implies \map {p_{X, Y} } {x, y} = 0$
- $y \notin \Omega_Y \implies \map {p_{X, Y} } {x, y} = 0$

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

### Necessary Condition

Suppose there exist functions $f, g: \R \to \R$ such that:

- $\forall x, y \in \R: \map {p_{X, Y} } {x, y} = \map f x \map g y$

Then by definition of marginal probability mass function:

- $\displaystyle \map {p_X} x = \map f x \sum_y \map g y$
- $\displaystyle \map {p_Y} y = \map g y \sum_x \map f x$

Hence:

\(\displaystyle 1\) | \(=\) | \(\displaystyle \sum_{x, y} \map {p_{X, Y} } {x, y}\) | Definition of Joint Probability Mass Function | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{x, y} \map f x \map g y\) | by hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_x \map f x \sum_y \map g y\) |

So it follows that:

\(\displaystyle \map {p_{X, Y} } {x, y}\) | \(=\) | \(\displaystyle \map f x \map g y\) | by hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map f x \map g y \sum_x \map f x \sum_y \map g y\) | from above | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\map f x \sum_y \map g y} \paren {\map g y \sum_x \map f x}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {p_X} x \map {p_Y} y\) | from above |

Hence the result from the definition of independent random variables.

$\blacksquare$

### Sufficient Condition

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

Then we can take the variables:

- $\map f x = \map {p_X} x$
- $\map g y = \map {p_Y} y$

and the result follows by definition of independence.

$\blacksquare$

## Sources

- 1986: Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*... (previous) ... (next): $\S 3.3$: Independence of discrete random variables: Theorem $3 \ \text{B}$