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My understanding is that this is equivalent to looking for values of $p$ such that the pmf is strictly increasing. The pmf of a binomial function is not easily differentiable though, so I doubt that's the right way to think about it. For $p$ close to 1 this will be the case, but I'm having a hard time putting an actual range.

Edit: after some more thought and playing around with numbers, it occurred to me that the 'boring' answer - that this only happen when $p=1$ - might actually be the good one. My rationale is that since the distribution is discrete, $n$ will not be the maximum value unless $E[X]=n$, from which it follows that $p=1$. It is my intuition that the maximum and mean are the same in the binomial distribution, but I can't really find a convincing argument for it.

Hortense
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2 Answers2

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Let's say the maximum of the binomial PMF, hereby denoted by $B(n,p)$ occurs at $k \in \{0,1,...,n\}$. Then $k$ must satisfy: $B(k-1, p) \leq B(k,p) \leq B(k+1,p)$.

Given the general expression for the PMF: $B(n,p) = \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$, it is fairly easy to show that the previous inequality reduces to: $(n+1)p-1 \leq k \leq (n+1)p$.

If we set $k=n$, we get $\frac{n}{n+1} \leq p \leq 1$. The right side of the inequality has to be true by definition. The left side gives you the smallest value of $p$ for which the PMF will peak at $n$.

Intuitively, it makes sense that the values of $p$ at which the maximum of the PMF occurs at $n$ are close to 1. As $p \rightarrow 1$, $n$ independent Bernoulli trials are increasingly likely to result in $n$ successes.

You can also easily check that at $p = \frac{n}{n+1}$, the maximum value of the PMF is $\left( \frac{n}{n+1} \right) ^n$, which approaches $1/e$ as $n \rightarrow \infty$.

Godfather
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It can be proved that $P(X=k)$ as a function on $\{0,1,\dots,n\}$ by ascending $k$ first increases, then reaches a maximum and then decreases.

This by observing factor $P(X=k)/P(X=k-1)$ (for small $k$ it exceeds $1$ and larger $k$ it does not exceed $1$).

In that light it can be concluded that there will be a maximum at $n$ iff: $$P(X=n)\geq P(X=n-1)$$ which can be translated into: $$p^n\geq np^{n-1}(1-p)$$

Working that out we find the condition:$$p\geq\frac{n}{n+1}$$

drhab
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