If *X*(1), *X*(2), …, *X*(*n*) are independent random variables, not necessarily with the same distribution, what is the variance of *Z* = *X*(1) *X*(2) … *X*(*n*)? It turns out that the computation is very simple:

In particular, if all the expectations are zero, then the variance of the product is equal to the product of the variances. See here for details.

Even more surprising, if *Z* = *X*(1) + *X*(1) *X*(2) + *X*(1) *X*(2) *X*(3) + … and all the *X*(*k*)’s are independent and have the same distribution, then we have

The proof is more difficult in this case, and can be found here. Interestingly, in this case, *Z* has a geometric distribution of parameter of parameter 1 – *p* if and only if the *X*(*k*)’s have a Bernouilli distribution of parameter *p*. Also, *Z* has a uniform distribution on [-1, 1] if and only if the *X*(*k*)’s have the following distribution: P( *X*(*k*) = -0.5 ) = 0.5 = P( *X*(*k*) = 0.5 ). The proof can be found here. If you slightly change the distribution of *X*(*k*), to say P( *X*(*k*) = -0.5) = 0.25 and P( *X*(*k*) = 0.5 ) = 0.75, then Z has a singular, very wild distribution on [-1, 1]. Its percentile distribution is pictured below.

The details can be found in the same article, including the connection to the binary digits of a (random) number in the base-2 numeration system.

Credit: Data Science Central By: Vincent Granville