A Poisson-Binomial variable $X\sim PB(p_1, \dots, p_n)$ is the sum of $n$ independent, not necessarily identically distributed, Bernoulli variables $X_1, \dots, X_n$: $$ X=\sum_{i=1}^n X_i, $$ with $X_i\sim Ber(p_i)$.

The Poisson-Binomial distribution is unimodal and $\mathbb E[X]=\sum_{i=1}^np_i=\mu.$

Question: is it always the case that the mode is either $\lfloor \mu \rfloor$ or $\lceil \mu \rceil$?

It seems to me that this is the case. The PB distribution is a generalization of the Binomial distribution, and has lower or equal entropy. The probability mass is thus more concentrated around the mean in some sense, which suggests that the answer is yes.

I have done some numerical experiments and have not found a counterexample, which reinforces my suspicion.


Darroch, J. N. "On the distribution of the number of successes in independent trials." The Annals of Mathematical Statistics 35.3 (1964): 1317-1321,

proved that the mode of a Poisson binomial variable satisfies the following:

\begin{equation} mode= \begin{cases} k \hspace{3mm}if\hspace{3mm}k \leq \mu \leq k+\frac{1}{k+2},\\ k \hspace{3mm}or\hspace{3mm} k+1\hspace{3mm}if\hspace{3mm} k+\frac{1}{k+2} \leq \mu \leq k+1 - \frac{1}{n-k+1},\\ k+1\hspace{3mm}if\hspace{3mm}k+1 - \frac{1}{n-k+1} \leq \mu \leq k+1. \end{cases} \end{equation}

Consequently, the mode differs from the mean by at most $1$. Note that a Poisson binomial distribution can have either one or two consecutive modes.

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  • $\begingroup$ +1. I found the answer elsewhere myself, but it is always nice to have different statements and proofs of the same result. $\endgroup$ – cangrejo Jan 15 at 10:41
  • $\begingroup$ Thanks for your comment and +1. I am pretty sure Darroch was the first to establish this property. $\endgroup$ – Mickybo Yakari Jan 15 at 10:49

I found the answer in a paper by Samuels:

Samuels, Stephen M. "On the number of successes in independent trials." The Annals of Mathematical Statistics 36.4 (1965): 1272-1278.

As a consequence of Theorem 1 we have that if there is an integer $k$ satisfying $$ k\leq \mu \leq k+1 $$ then $$ Pr(X=k-1)<Pr(X=k) \text{ and } Pr(X=k+1)>Pr(X=k+2). $$ So the answer is yes.

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