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A month ago, I answered a question that asked, in a nutshell, whether for a given finite set of dependent Bernoulli-distributed random variables, you could construct a set of independent Bernoulli random variables that carried the same information. I formalized the idea like this:

Let $(Ω, Σ, μ)$ be a probability space. If $X_1, X_2, …, X_n$ are Bernoulli random variables on $Ω$, then there exist independent Bernoulli random variables $Y_1, Y_2, …, Y_m$ on $Ω$ and a function $f : \{0, 1\}^m → \{0, 1\}^n$ such that for all $ω ∈ Ω$, $f(Y_1(ω), …, Y_m(ω)) = (X_1(ω), …, X_n(ω))$.

(Intuitively, the conjecture says that the values of the independent $Y_i$s suffice to determine the values of the not necessarily independent $X_i$s. Notice that $m$ need not equal $n$, and indeed may need to be much larger.)

This statement can be readily disproved by choosing a small finite probability space $(Ω, Σ, μ)$ that has no pair of independent Bernoulli random variables. But suppose we require $(Ω, Σ, μ)$ to be [0, 1] with the Borel σ-algebra and Lebesgue measure. Now we have a roomier and more familiar underlying probability space. Is the conjecture true now?

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If you decompose the joint distribution of $X_1,\ldots,X_n$ as $$p(x_1)p(x_2|x_1)\cdots p(x_n|x_1,\ldots,x_{n-1})$$ then each term in the product is a Bernoulli distribution, which probability depends on the earlier realisations of the $X_i$'s.

Hence, if I draw

  1. $Y_1$ as a Bernoulli variate with probability $p=\mathbb{P}(X_1=1)$
  2. two independent Bernoulli variates $Y_2^1$ and $Y_2^0$ with probabilities $\mathbb{P}(X_2=1|X_1=1$ and $\mathbb{P}(X_2=1|X_1=0)$
  3. four independent Bernoulli variates $Y_3^{1,1}$, ... and $Y_3^{0,0}$ with probabilities $\mathbb{P}(X_3=1|X_1=1,X_2=1$, ... and $\mathbb{P}(X_3=1|X_1=0,X_2=0)$
  4. ...
  5. $2^{n-1}$ independent Bernoulli variates $Y_n^{x_1,x_2,...,x_{n-1}}$ with probabilities $\mathbb{P}(X_n=1|X_1=x_1,X_2=x_2,...,X_{n-1}=x_{n-1})$

I have a large enough collection of independent Bernoulli variates to reconstitute the $X_i$'s.

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