Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 5

Let $D$ denote having the disease and $+$ a positive result. Assume I am sampled uniformly from the stated population, so $\mathbb{P}(D)=n/N$. By the law of total probability, $\mathbb{P}(+)=an/N+b(N-n)/N$. Thus Bayes' theorem givesThe denominator must be positive; otherwise the observed event has zero probability in this model and the conditional probability is undefined. The answer depends on disease prevalence as well as the test's two error rates.

An integrable nonnegative probability density of this form requires $k>0$. Normalization gives $1=A{\int }_0^{\infty }{e}^{-kt}dt=A/k$, so $\boxed{A=k}$. This is the exponential distribution with rate $k$.

The survival function of this exponential distribution is $\mathbb{P}(X\ge t)=e^{-kt}$ for $t\ge 0$. Since $\{X\ge 3\}\subseteq \{X\ge 1\}$, the conditional probability isThis is the memoryless property: after surviving the first unit, surviving two additional units has the same probability as it did initially.

The expectation is $\mathbb{E}[X]=k{\int }_0^{\infty }t{e}^{-kt}dt$. An integration by parts, with the boundary term $t{e}^{-kt}$ tending to zero, gives