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?



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


