Inference in binomial with zero successes and fixed number of trials

Question

Let $X \sim \mathsf{Bin}(n, p)$ where $n$ is known and $p$ is to be inferred from the data. Suppose further that $X = 0,$ so that we had no successes.

We can reason in the following way. We want to reject any value of $p$ that makes the observed data very unlikely. Let's say we want to mistakenly reject a true $p$ at most $100\alpha$% of the time; in other words, we want to make Type I error with level $\alpha.$ For any given $p,$ $\mathbf{P}(X = 0) = (1-p)^n;$ thus, we reject this $p$ if $(1 - p)^n \leq \alpha.$ This leads to reject $p$ if $1-\alpha^{\frac{1}{n}} \leq p \leq 1$. Therefore, the value this way of thinking will not reject are in the interval $(0, 1-\alpha^{\frac{1}{n}}).$ Note that when $n \to \infty,$ this interval shrinks, which is good (the more observations we have and if we still have zero successes the more certain we are $p \approx 0$).

Is this method used in practice?

To compare: I ask this because I was checking some materials and these appeal to asymptotic statistics like the Likelihood Ratio, or Rao's score statistic, or Bayesian methods where they choose either a uniform prior or a beta prior. Using Rao's score with $n = 25,$ we would get $(0, 0.132)$ at level $\alpha = 0.05$ while the Bayesian method with uniform prior will give $(0.001, 0.132)$ for a 95% equal-tail posterior interval. The method I described above give $(0, 0.113),$ and the interpretation is simple, any value outside this interval makes what we saw to happen only $5$% of the time or worse (much worse as $p$ grows, actually).

Answer 1

There may be a philosophical issue whether a $0$ success probability makes sense. For such a problem, one could use a Bayesian credible interval that gives an upper bound on the success probability $\theta,$ based on a sequence of say $n = 50$ or $n = 500$ failures in a row.

Suppose your prior distribution is that $\theta$ is small: $\theta \sim \mathsf{Beta}(1, 10)$ and density $f(\theta) \propto (1-\theta)^9.$

Then after $n=50$ trials with $x=0$ successes, the binomial likelihood is $g(x|\theta) \propto (1-\theta)^{50}.$

Thus, the posterior density is $(1-\theta)^9\times (1-\theta)^{50}$ and the posterior distribution on $\theta$ is $\mathsf{Beta}(1,60),$ which gives a one-sided 95% posterior interval estimate of $(0, 0.049).$ [Using R.]

qbeta(.95, 1, 60)
[1] 0.04870291

Similarly, with $n = 500$ consecutive failures, the 95% Bayesian posterior interval estimate would be that $\theta < 0.0059.$

qbeta(.95, 1, 510)
[1] 0.005856767

Note: If you approach this problem without any opinion as to the value of $\theta,$ then you might use the non-informative Jeffries prior $\mathsf{Beta}(.5,.5).$ Then 95% Bayesian credible intervals for $50$ and $500$ consecutive failures, respectively, results would be as shown in R below (rounded to five places).

round(qbeta(c(.025,.975), .5, 50.5),6)
[1] 0.000010 0.048758
round(qbeta(c(.025,.975), .5, 500.5),6)
[1] 0.000001 0.005009