Are two uncorrelated Rademacher rv always independent?

Question

Let $X$ and $Y$ be uncorrelated Rademacher random variables

Then,

$Cov(X,Y)=E[XY]-E[X]E[Y]=0$

But $E[X]=E[Y]=0$

Then $E[XY]=P(X=1,Y=1)+P(X=-1,Y=-1)-P(X=1,Y=-1)-P(X=-1,Y=1)=0$

I know that $P(X=1)P(Y=1) = 1/4$

How to show that it is equal or not to $P(X=1,Y=1)$ from what I have ?

I'm not sure how you could have Rademacher variables that weren't independent unless they were equal to each other (or equal to the negative of each other), given that there are not actually any parameters in the Rademacher distribution. — jbowman, Dec 14 '19 at 16:45
Because questions about correlation do not depend on how the variables might be shifted or scaled, your question would be the same upon replacing "Rademacher" by "Bernoulli$(1/2).$" If we were to generalize from $1/2$ to arbitrary $p$ the question would be made more interesting: and the solution is given at https://stats.stackexchange.com/a/285008/919. — whuber, Dec 14 '19 at 16:48

score 0 · Accepted Answer · answered Dec 16 '19 at 20:49

You could summarize your $E[XY] = 0$ result with $P(X=Y) = P(X\neq Y) = 0.5$.

You could proceed with:

\begin{align*} 2P(X=1,Y=1) &= P(X=1) + P(Y=1) - P(X\neq Y) \\ &= 0.5 + 0.5 - 0.5 \\ &= 0.5 \end{align*}

Now you can conclude that $P(X=1)P(Y=1) = P(X=1,Y=1)$.

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