Index: The Book of Statistical Proofs ▷ Statistical Models ▷ Count data ▷ Binomial observations ▷ Maximum likelihood estimation (n)

Theorem: Let $y$ be the number of successes resulting from an unknown number $n$ of independent trials with known success probability $p$, such that $y$ follows a binomial distribution:

\[\label{eq:Bin} y \sim \mathrm{Bin}(n,p) \; .\]

Suppose $0 < p < 1$. Then, the maximum likelihood estimator of $n$ is

\[\label{eq:Bin-MLE-Trials} \hat{n} = \begin{cases} \lfloor \frac{y}{p} \rfloor \; , & \text{if } p \nmid y \\ \frac{y}{p} \text{ and } \frac{y}{p}-1, \; , & \text{otherwise} \; . \end{cases}\]

Proof: With the probability mass function of the binomial distribution, equation \eqref{eq:Bin} implies the following likelihood function:

\[\label{eq:Bin-LF} \begin{split} \mathrm{p}(y|p) &= \mathrm{Bin}(y; n, p) \\ &= {n \choose y} \, p^y \, (1-p)^{n-y} \; . \end{split}\]

Thus, the log-likelihood function is given by

\[\label{eq:Bin-LL} \begin{split} \mathrm{LL}(p) &= \log \mathrm{p}(y|p) \\ &= \log {n \choose y} + y \log p + (n-y) \log (1-p) \; . \end{split}\]

Note that ${n \choose y} = \frac{n!}{y!(n-y)!} = \frac{\Gamma(n+1)}{\Gamma(y+1)\Gamma(n+1-y)}$ and that $\frac{\mathrm{d}}{\mathrm{d}n} \log \Gamma(n) = \psi(n)$ by definition. Now, since $\Gamma(n + 1) = n \Gamma(n)$, we have $\psi(n+1) = \psi(n) + \frac{1}{n}$, so $\psi(n+1)$ differs from the $n$-th harmonic number $H_n = \sum_{k=1}^{n} \frac{1}{k}$ only by a constant.

Thus, the derivative of the log-likelihood function \eqref{eq:Bin-LL} with respect to $n$ is

\[\label{eq:dLL-dn} \begin{split} \frac{\mathrm{d}\mathrm{LL}(n)}{\mathrm{d}n} &= \psi(n+1) - \psi(n+1-y) + \log (1-p) \\ &= H_n - H_{n-y} + \log (1-p) \; . \end{split}\]

The log-likelihood derivative \eqref{eq:dLL-dn} can be bounded below by $\log\left( \frac{n+1}{n+1-y} \right) + \log(1 - p)$ and above by $\log\left( \frac{n}{n-y} \right) + \log(1 - p)$ using the Hermite-Hadamard inequality.

Since these bounding functions are continuous and monotone for $n > y$, setting them to zero and solving gives bounds for the maximum likelihood estimate of $n$:

\[\label{eq:n-MLE} \begin{split} \log\left( \frac{\hat{n}_\mathrm{lower}+1}{\hat{n}_\mathrm{lower}+1-y} \right) + \log (1 - p) &= 0 \\ \frac{(1 - p)\hat{n}_\mathrm{lower}+1}{\hat{n}_\mathrm{lower}+1-y} &= 1 \\ (1 - p)\hat{n}_\mathrm{lower}+1 &= \hat{n}_\mathrm{lower}+1-y \\ \hat{n}_\mathrm{lower} &= \frac{y}{p} - 1 \; . \end{split}\]

Likewise, $\hat{n}_\mathrm{upper} = \frac{y}{p}$.

Thus, we have $\frac{y}{p} - 1 \leq \hat{n} \leq \frac{y}{p}$. If $p \nmid y$, then $\lfloor \frac{y}{p} \rfloor$ is the only integer in this interval. Otherwise, both bounds are integers yielding equal values for the likelihood function.

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Metadata: ID: P556 | shortcut: bin-mlen | author: jdonland | date: 2026-09-28, 23:42.