# How do you derive the Success-Run Theorem from the traditional form of Bayes Theorem?

In my industry it is common to test a sample of 20-30 and then use that data to draw conclusions about the reliability of the product with a certain confidence. We have tables for such things but it appears that for the case of 0 failures in the sample, the "Success Run Theorem" is used. In my references this appears as: $$R_c = (1-C)^{\frac{1}{(n+1)}}$$ where $C$ = confidence level, $R$ = reliability at confidence level $C$, and $n$ = sample size.

However, I cannot find an explanation of how to get to the above equation from Bayes' theorem:

Every attempt to talk myself through Bayes theorem to arrive at the Success Run Theorem gets me confused. Even more confusing is when I try to extend my understanding to cases where some failures are observed in the sampling. Then I know to use this formula:

But again I don't understand where it comes from (binomial?) or how it relates to the above two other formulas, if at all.

My specific question would be how you go from Bayes Theorem (written as probabilities) to the Success Run Theorem (written as confidence, reliability, sample size)?

Thank you for helping a poor engineer lost in the world of stats.

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First we need to state the problem, including its assumptions, precisely. Being clear about the assumptions is important not only for understanding the derivation, but also for understanding when the result is applicable.

We are studying the reliability of a product, by performing $n$ trials. In each trial, the product succeeds with probability $p$ and fails with probability $1-p$. Suppose that the trials are conditionally independent given $p$ (the distinction between conditional independence and unconditional independence is crucial in statistics).

Let $R$ be a desired reliability level, and $C$ be the corresponding confidence level, in the sense that, given the data, there is at least probability $C$ that the true reliability $p$ is at least $R$. For example, if $R=0.9, C=0.95$, we want to be able to say that there is at least a $95\%$ chance that $p$ is at least $0.9$. Given that the product succeeds all $n$ times, how are $R$ and $C$ related?

Another assumption is still needed though: we have not assigned a probability distribution to p, so what does $P(p \geq R)$ mean? If $p$ is a constant and does not have a distribution, then this probability is either 0 or 1. So to reflect our uncertainty about $p$, we will view it as a random variable (this is a Bayesian approach). Assume that the prior distribution is $p \sim \textrm{Unif}(0,1)$ (this is a simple way of quantifying being very, very uncertain about the value of $p$).

The posterior distribution of $p$, given the data, is $\textrm{Beta}(n+1,1)$. This follows from Bayes' theorem, and more specifically from the fact that Beta is the conjugate prior for the Binomial (I explain the Beta distribution and conjugacy in Lecture 23 of Statistics 110).

So we want to find $P(W \geq R)$ for $W \sim \textrm{Beta}(n+1,1)$. The CDF of $W$, evaluated at $w$ in $[0,1]$, is $F(w)=w^{n+1}.$ Thus, $$C=P(W \geq R) = 1 - F(R) = 1-R^{n+1},$$ as desired.

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Nice answer -- welcome to CrossValidated, Joe. Your other answer on Quora are excellent and I am glad to see your contribute here. – Cam.Davidson.Pilon Dec 30 '13 at 4:18