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.

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.

$\blacksquare$

1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 3.3$: Independence of discrete random variables: Example $13$