MMP.13:5.1 - Zero recorded failures: two different uncertainty claims

A device is tested for a fixed four operations. All failures are recorded, and the supplied model gives independent Bernoulli outcomes with one unchanged failure probability p. No failures occur, so K=0 and the likelihood is proportional to (1-p)^4.

Suppose the requested target is the probability of at least one failure in two further operations under the same condition:

q=1-(1-p)^2.

The future operations are assumed independent conditional on the same p. The question concerns q, rather than every detail of an operating model.

Frequentist construction. The likelihood estimate is p_hat=K/4=0. Inserting it into the familiar normal standard-error expression gives zero estimated standard error and the interval [0,0]. At p=0.2, zero failures occur with probability 0.8^4=0.4096, and [0,0] excludes the true p on every such occurrence. This alone rules out 95% coverage.

For a one-sided 95% binomial upper confidence procedure, invert the lower-tail probability: for k<4 choose U(k) satisfying

P_(p=U(k))(K<=k)=0.05,

and set U(4)=1. Binomial test inversion gives coverage at least 95%, allowing conservatism from discreteness. At k=0,

(1-U)^4=0.05; U=1-0.05^(1/4)=0.5271.

Since q increases with p, its upper confidence limit is

1-(1-U)^2=1-sqrt(0.05)=0.7764.

This is the result of a covering procedure, not a statement that q has a 95% probability of being below 0.7764 after these records.

Bayesian construction. Choose a uniform prior for p on [0,1]. Multiplying and normalizing gives posterior density

pi(p|K=0)=5*(1-p)^4; 0<=p<=1.

Thus E[p|K=0]=1/6, and the posterior 95% upper quantile of p is 1-0.05^(1/5)=0.4507. Transforming that quantile gives a posterior 95% upper quantile of q of 1-0.05^(2/5)=0.6983. Its smaller value does not make it a uniformly better confidence limit; it expresses a different conditional claim with a prior.

To obtain the probability of a failure in the next pair, average q itself:

E[q|K=0] = 1 - integral_0^1 (1-p)^2*5*(1-p)^4 dp = 2/7.

Using the posterior mean of p first would give 1-(5/6)^2=11/36, a different value. The posterior predictive probability is 2/7; the event of a failure in that pair remains a binary future outcome.

Changed assumption. Hold the four records fixed but replace the uniform prior with density 9*(1-p)^8, favoring lower failure probabilities. The posterior becomes 13*(1-p)^12. The posterior probability of p<0.2 changes from 1-0.8^5=0.67232 to 1-0.8^13=0.94502; the predictive probability for a failure in the next pair becomes 2/15.

These changes come entirely from the prior. They neither add operations to the observed test nor establish that the new prior is appropriate. If its relevance is unresolved, return the conditional results and their difference. The frequentist bound remains available without that prior under the original fixed-sample observation model.