An independent collection, where the pairwise products are also independent

Question

Can you give me a non-trivial example of a sequence $\{X_k\}_{k=1}^n$ of independent, and identically distributed random variables, such that, the pairwise products, $\{X_kX_{\ell}\}_{1\leq k<\ell \leq n}$ are all independent?

Of course, if $\{k,\ell\}\cap \{k',\ell'\}=\varnothing$, we automatically have this, and since $(k,\ell)\neq \{k',\ell'\}$, it boils down to giving an example where $X_kX_\ell$ and $X_kX_{\ell'}$ are always independent, whenever $\ell\neq \ell'$ and $\ell,\ell'\neq k$.

Answer 1

I don't know what you mean by non-trivial but consider this example: $P(X = 1) = P(X = -1) = \frac 1 2$

Knowing that the product is -1 or 1 doesn't change the probability for other products. $P(X_i = 1|X_i.X_j = 1) = \frac 1 2$ and $P(X_i = -1|X_i.X_j = 1) = \frac 1 2$ because 1*1 and -1*-1 happen with the same probability.

Then we can write $$P(X_i.X_k = 1|X_i.X_j = 1) = P(X_k = 1|X_i.X_j = 1, X_i=1 ).P(X_i = 1|X_i.X_j = 1) + P(X_k = -1|X_i.X_j = 1, X_i=-1 ).P(X_i = -1|X_i.X_j = 1) = \frac 1 2 $$

$$P(X_i.X_k = 1|X_i.X_j = -1) = P(X_k = 1|X_i.X_j = -1, X_i=1 ).P(X_i = 1|X_i.X_j = -1) + P(X_k = -1|X_i.X_j = -1, X_i=-1 ).P(X_i = -1|X_i.X_j = -1) = \frac 1 2 $$

$$P(X_i.X_k = 1) = P(X_k = 1| X_i=1 ).P(X_i = 1) + P(X_k = -1|X_i=-1 ).P(X_i = -1) = \frac 1 2 $$

Since the conditional probability is the same as unconditional, you can say they are independent.

Hope it helps

Answer 2

@Ofya, thanks for your answer, I was seeking for something where I can control the support, and I guess, I've found my answer.

EDIT (Extra question) Suppose that, $X_i$ are independent, standard normal random variables. Is it true that, $(X_iX_j)_{1\leq i <j\leq n}$ forms an independent collection of random variables?

Take $X_i\sim {\rm Unif}\{1,2,\dots,p-1\}$ with $p$ being a prime, and consider $X_iX_j$ as $X_iX_j\pmod{p}$. We have, $$ \mathbb{P}(X_iX_j = k,X_iX_k\ell)=\sum_{t=1}^{p-1}\mathbb{P}(X_iX_j=k,X_iX_k=\ell|X_i=t)\mathbb{P}(X_i=t)=\sum_{t=1}^{p-1}\mathbb{P}(X_j=t^{-1}k,X_k=t^{-1}\ell)\mathbb{P}(X_i=t), $$ which evaluates as, $\frac{1}{(p-1)^2}$. Finally, it is easy to see that, for any pair $(i ,j)$ and any $k\in\{1,2,\dots,p-1\}$, $\mathbb{P}(X_iX_j=k)=\frac{1}{p-1}$, proving the independence.