B Mean of the Exponential Distribution

The average (mean) waiting time for an exponential process with rate constant \(k\) is given by the expectation value of \(t\) over the waiting time distribution \(p_\mathrm{wait}(t) = k\exp(-kt)\):

\[ \langle t \rangle = \int_0^\infty t\, p_\mathrm{wait}(t)\,\mathrm{d}t = \int_0^\infty t\, k\exp(-kt)\,\mathrm{d}t \]

We evaluate this by integration by parts, with \(u = t\) and \(\mathrm{d}v = k\exp(-kt)\,\mathrm{d}t\), giving \(\mathrm{d}u = \mathrm{d}t\) and \(v = -\exp(-kt)\):

\[ \langle t \rangle = \left[-t\exp(-kt)\right]_0^\infty + \int_0^\infty \exp(-kt)\,\mathrm{d}t \]

The boundary term vanishes at both limits: at \(t = 0\) it is zero, and as \(t \to \infty\) the exponential decays faster than \(t\) grows. The remaining integral gives \(1/k\), so:

\[ \langle t \rangle = \frac{1}{k} \]

A process with a large rate constant \(k\) has a short average waiting time, and vice versa.