Condition for Independence of Discrete Random Variables

From ProofWiki
Jump to navigation Jump to search


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:

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


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:

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


\(\displaystyle 1\) \(=\) \(\displaystyle \sum_{x, y} \map {p_{X, Y} } {x, y}\) 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.


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.