De Finetti's representation theorem, which is important in mathematical statistics, shows that something like this is true for exchangeable random variables. It says that for any sequence of exchangeable random variables, there exists another random variable conditional on which the former are IID.
But now I'll attack your problem more directly. Let's formalize it like this:
Conjecture: 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 $ω ∈ Ω$, $f(Y_1(ω), …, Y_m(ω)) = (X_1(ω), …, X_n(ω))$.
This statement is false. For a counterexample, consider the case when:
- $n = 2$
- $Ω = \{1, 2, 3\}$
- $μ(1) = \tfrac{1}{2}$, $μ(2) = μ(3) = \tfrac{1}{4}$
- $X_1(1) = 0$, $X_1(2) = X_1(3) = 1$
- $X_2(1) = 0$, $X_2(2) = 0$, $X_1(3) = 1$
Notice that $X_1$ and $X_2$ are dependent, since
$$P(X_1 = 0, X_2 = 0) = \tfrac{1}{4} ≠ \tfrac{3}{8} = \tfrac{1}{2} \cdot \tfrac{3}{4} = P(X_1 = 0)P(X_2 = 0).$$
Obviously, $m$ will need to be greater than 1. However, on this probability space, there is no pair of independent Bernoulli random variables. There are only 8 distinct Bernoulli random variables (because $2^{|Ω|} = 2^3 = 8$), and all $\binom{8}{2} = 28$ pairs of these are dependent. The following Python program proves this by checking every pair.
import numpy as np
from itertools import combinations
probs = np.array([2, 1, 1]) # Multiplied by 4 so we can do integer arithmetic.
ys = [np.array([a, b, c])
for a in (False, True) for b in (False, True) for c in (False, True)]
for y1, y2 in combinations(ys, 2):
def f(a, b):
v1 = y1 if a else ~y1
v2 = y2 if b else ~y2
return np.sum(probs[v1 & v2]) == np.sum(probs[v1]) * np.sum(probs[v2])
if f(0, 0) and f(0, 1) and f(1, 0) and f(1, 1):
print "Independent"
else:
print "Dependent"
