Variance of a Product of Random Variables

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:

Variance of a Product of Random Variables 1

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

Variance of a Product of Random Variables 2

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.

Variance of a Product of Random Variables 3The 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.

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