# Can the sum of identically distributed dependent Bernoulli trials be binomially distributed?

If you have $$n$$ identically distributed Bernoulli trials whose sum is binomially distributed random variable, does it then follow that the $$n$$ Bernoulli trials are independent?

• I think this question would be more appropriate for math.stackexchange.com. The answer is no because you could take $X_1,\ldots,X_n$ to be iid Bernoulli and define $Y_1,\ldots,Y_n$ to be the increasing rearrangement of those random variables. The sum has the same distribution. Feb 22, 2019 at 15:25
• @AnthonyQuas: I might misunderstand your reasoning, but these $Y_i$ are, I think, not identically distributed? Feb 22, 2019 at 15:31
• @Steve: Ah yes... Feb 22, 2019 at 15:38

That's false even for $$n=3$$.

Denote by $$B(p)$$ the Bernoulli distribution which has probability $$p$$ of being 1 and $$(1-p)$$ to be 0.

Define the three Bernoulli variables $$(X_1, X_2, X_3)$$ by $$X_1 \sim B(0.5) \\ X_2|(X_1=1) \sim B(0.25);~ X_2|(X_1=0) \sim B(0.75) \\ X_3|(X_1+X_2=2) \sim B(1);~ X_3|(X_1+X_2=0) \sim B(0);~ X_3|(X_1+X_2=1) \sim B(0.5)$$

Quick calculation shows they're all $$B(0.5)$$ distributed (basically by symmetry) and that the sum has the same distribution as the sum of three i.i.d. trials.

• Okay, that was actually a quite easy counterexample. Thanks. Feb 22, 2019 at 16:19
• Another way of writing the same counterexample: Let $Y_1, Y_2, Y_3$ be independent Bernouli($1/2$). Let $X_i = Y_i$, except in the following cases: If $(Y_1, Y_2, Y_3) = (0,0,1)$ then take $X_3 = 0$, and either $(X_1,X_2) = (0,1)$ or $(1,0)$, with equal probabilities. If $(Y_1,Y_2,Y_3) = (1,1,0)$ then take $X_3 = 1$, and either $(X_1,X_2) = (0,1)$ or $(1,0)$, with equal probabilities. Feb 22, 2019 at 18:23

The answer is no for large enough $$n$$, because the independence of the $$n$$ Bernoulli random variables (r.v.'s) is given by about $$2^n$$ equations, whereas to describe the individual distributions of the Bernoulli r.v.'s and their sum one needs only $$O(n)$$ equations.

• It's good intuition but not exactly a proof, since there are a further $2^n$ inequalities to be satisfied: that the probability of every outcome must be nonnegative. Feb 22, 2019 at 15:43
• @NateEldredge : Of course, you are right. However, with this intuition, one immediately knows there should be a counterexample, which is then easy to find, by using e.g. the Mathematica command FindInstance[]. Feb 22, 2019 at 18:02