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?
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2$\begingroup$ 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. $\endgroup$– Anthony QuasFeb 22, 2019 at 15:25
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$\begingroup$ @AnthonyQuas: I might misunderstand your reasoning, but these $Y_i$ are, I think, not identically distributed? $\endgroup$– SteveFeb 22, 2019 at 15:31
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$\begingroup$ @Steve: Ah yes... $\endgroup$– Anthony QuasFeb 22, 2019 at 15:38
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2 Answers
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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.
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$\begingroup$ Okay, that was actually a quite easy counterexample. Thanks. $\endgroup$– KristianFeb 22, 2019 at 16:19
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$\begingroup$ 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. $\endgroup$ Feb 22, 2019 at 18:23
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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.
answered Feb 22, 2019 at 15:40
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$\begingroup$ 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. $\endgroup$ Feb 22, 2019 at 15:43
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1$\begingroup$ @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[]. $\endgroup$ Feb 22, 2019 at 18:02