Given a sequence of real numbers in $[0,1]$, is there a sequence of Bernoulli random variables with probability of success being that sequence?

Question

Let $\{p_n\}_{n\ge 1} $ be a sequence of numbers in $[0,1]$.

Does there exist a sequence of independent Bernoulli random variables $\{B_n\}_{n\ge 1}$ with probability of success for $B_n$ being $p_n$ ?

Answer 1

Yes, certainly. The Kolmogorov extension theorem (take $T = \mathbb{N}$) states more generally that given any family of finite-dimensional joint distributions which is consistent, there exists a family (here a sequence) of random variables having those joint distributions.

The probability space where they all live can be taken to be the infinite product $\{0,1\}^{\mathbb{N}}$ (aka Cantor space) equipped with the Borel $\sigma$-algebra, and the probability measure on this sample space is the one whose existence is asserted by Kolmogorov's theorem.

In this special case, you can also try to explicitly construct a random variable $X$ taking values in $[0,1]$, such that if we let $B_n$ be the $n$th binary digit then they have the desired distributions. You can work out what the cdf $F(x) = P(X \le x)$ must be in terms of the $p_n$, at least for $x$ a dyadic rational, and check that $F$ extends to all $x$ by continuity. Then by letting $U \sim U(0,1)$ and $X = F^{-1}(U)$ there does exist a random variable with this cdf.