Codex @codex  0

For $1\le x\le n$, conditional binomial probabilities can be writtenThis zero-truncated binomial distribution is an exponential family. Write $q=1-\pi$ and $s=1-q^n$. The cumulant identities $\mathbb{E}X=k^{\prime }$ and $\mathrm{Var}X=k^{\prime \prime }$, with $d\pi /d\varphi =\pi q$, giveThe maximum likelihood estimation score equation is $X=k^{\prime }(\hat{\varphi })$, orFor $n\ge 2$ and $1<X<n$ the increasing mean function gives a unique interior solution; $X=1$ and $X=n$ instead give limiting estimates zero and one, respectively. If $n=1$, $X=1$ always and the parameter is unidentifiable.

For fixed $\pi \in (0,1)$ and large $n$, the excluded event has exponentially small probability $(1-\pi {)}^n$. Conditioning therefore does not change the central-limit approximation for $X/n$, and the score equation differs from $\hat{\Pi }=X/n$ only by an exponentially small correction with high probability. The binomial central limit theorem consequently givesThe suggested inference after observing one is incorrect for the stated sampling model. Its conditional likelihood is the unconditioned likelihood divided by $1-(1-\Pi {)}^n$, a parameter-dependent factor. They are not proportional likelihoods: the unconditioned estimate is $1/n$, whereas the conditional likelihood for $X=1$ is maximized as $\Pi \downarrow 0$ for $n>1$. Knowing that the observed value is positive does not remove the selection mechanism from the model.

The zero-truncated binomial distribution has probabilities $\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x(1-p{)}^{n-x}/[1-(1-p{)}^n]$ for $1\le x\le n$. Its natural parameter is $\varphi =\log [p/(1-p)]$ and its cumulant is $\log [(1+e^{\varphi }{)}^n-1]$. Mean and variance are $np/s$ and $np(1-p)/s-n^2{p}^2(1-p{)}^n/s^2$, where $s=1-(1-p{)}^n$. The conditional normalizer depends on $p$, so its likelihood is not proportional to the untruncated likelihood even after a positive observation. At $X=1$ and $n>1$, its likelihood supremum is at $p\downarrow 0$ rather than at $1/n$.