# does uncorrelated imply independence for 2 valued random variables?

Let $(X_n)_{n \geq 1}$ be identically distributed random variables such that $\mathbb{P}[X_1 = 1] = \mathbb{P}[X_1 = -1] = \frac{1}{2}$ and for any $i_1< \cdots < i_p, \mathbb{E}[X_{i_1} \cdots X_{i_p}] = 0$ for any $p \in \mathbb{N}.$ Can I conclude that $(X_n)_{n \geq 1}$ are i.i.d.? In other words, does uncorrelated imply independence in this case? I know that in this case they are pairwise independent.

I think I solved it. First of all we shift and rescale the random variables, so we denote by $Y_i = \frac{X_i + 1}{2}.$ It is not difficult to check that $\mathbb{E}[Y_{i_1} \cdots Y_{i_p}] = \mathbb{E}[Y_{i_1}] \cdots \mathbb{E}[Y_{i_p}]$ and $\mathbb{P}[Y_i = 0] = \mathbb{P}[Y_i = 1] = \frac{1}{2}.$
But on the other hand, $\mathbb{P} [Y_{i_1} = 1, \dots, Y_{i_p} = 1] = \mathbb{E}[Y_{i_1} \cdots Y_{i_p}] = \mathbb{E}[Y_{i_1}] \cdots \mathbb{E}[Y_{i_p}] = \mathbb{P} [Y_{i_1} = 1] \cdots \mathbb{P} [Y_{i_p} = 1].$
Now to show independence in general case, we do induction. In particular, for $p=2$, the result follows obvoiusly from the last equation. Next, we suppose that any $p=n$ random variables are independent and we show for $p=n+1.$ For example, we can show $$\mathbb{P} [Y_1 = 0, Y_2 = 0, \dots, Y_{n+1} = 0] = \\ \mathbb{P} [Y_1 = 0, Y_2 = 0, \dots, Y_{n} = 0] - \mathbb{P} [Y_1 = 0, Y_2 = 0, \dots, Y_n = 0, Y_{n+1} = 1] \\= \mathbb{P} [Y_1 = 0] \mathbb{P} [ Y_2 = 0] \cdots \mathbb{P} [ Y_{n} = 0] - \mathbb{P} [Y_1 = 0] \mathbb{P}[Y_2 = 0] \cdots \mathbb{P}[Y_{n-1} = 0] \mathbb{P}[ Y_{n+1} = 1] + \mathbb{P} [Y_1 = 0, Y_2 = 0, \dots, Y_{n-1} = 0, Y_n = 1, Y_{n+1} = 1].$$ And we continue this process.
I believe this result is also true also when $\mathbb{P}[X_1 = -1] = q = 1 - \mathbb{P}[X_1 = 1] = 1-p.$
• You don't need the induction. The result follows immediately by symmetry from the one in your second paragraph for all values $1$. – joriki Jan 13 at 14:04