Condition for Independence from Product of Expectations/Corollary/Converse

From ProofWiki
Jump to navigation Jump to search

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

Proof by Counterexample:

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.

$\blacksquare$


Sources