# Condition for Independence from Product of Expectations/Corollary/Converse

## Converse of Corollary to Condition for Independence from Product of Expectations

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

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

$\expect {X Y} = \expect X \expect Y$

Then it is not necessarily the case that $X$ and $Y$ are independent.

## Proof

Let $X$ be a discrete random variable whose distribution is defined as:

$\map {p_X} {-1} = \map {p_X} 0 = \map {p_X} 1 = \dfrac 1 3$

Let $Y$ be the discrete random variable defined as:

$Y = \begin{cases} 0 & : X = 0 \\ 1 & : X \ne 0 \end{cases}$

We have:

 $\ds \map {p_{X, Y} } {0, 1}$ $=$ $\ds 0$ $\ds \map {p_X} 0 \map {p_Y} 1$ $=$ $\ds \frac 1 3 \cdot \frac 2 3 = \frac 2 9$

So $X$ and $Y$ are dependent.

But:

 $\ds \expect {X, Y}$ $=$ $\ds \sum_{x, y} x y \map {p_{X,Y} } {x, y}$ $\ds$ $=$ $\ds \paren {-1} \frac 1 3 + 0 \cdot \frac 1 3 + 1 \cdot \frac 1 3$ $\ds$ $=$ $\ds 0$ $\ds \expect X \expect Y$ $=$ $\ds 0 \cdot \frac 2 3$ $\ds$ $=$ $\ds 0$

So $\expect {X, Y} = \expect X \expect Y$ but $X$ and $Y$ are not independent.

$\blacksquare$