= Zero-truncated binomial distribution
{title2=$X\sim\operatorname{Bin}(n,p)\mid X>0$}
The <zero-truncated binomial distribution> has probabilities $\binom nx p^x(1-p)^{n-x}/[1-(1-p)^n]$ for $1\leq x\leq n$. Its natural parameter is $\phi=\log[p/(1-p)]$ and its cumulant is $\log[(1+e^\phi)^n-1]$. Mean and variance are $np/s$ and $np(1-p)/s-n^2p^2(1-p)^n/s^2$, 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 $p\downarrow0$ rather than at $1/n$.
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