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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