Proof. Let \(M(\theta , t) = E\left (e^{\theta X(t)}\right ) = \sum _ne^{n\theta }P_n(t)\), so that \(M(\theta , 0) = e^{a\theta }\), \(\frac {\partial M}{\partial \theta } = \sum _nne^{n\theta }P_n(t)\) and \(\frac {\partial M}{\partial t} = \sum _ne^{n\theta }P_n'(t)\). Multiplying (5.1) by \(e^{n\theta }\) and summing over \(n\), \[\frac {\partial M}{\partial t} = -\lambda \frac {\partial M}{\partial \theta } + \lambda e^\theta \frac {\partial M}{\partial \theta }, \qquad \text {that is}\qquad \frac {\partial M}{\partial t} + \lambda (1 - e^\theta )\frac {\partial M}{\partial \theta } = 0 .\] This first-order equation says that \(M\) is constant along the characteristic curves \(\frac {dt}{1} = \frac {d\theta }{\lambda (1 - e^\theta )}\). Writing \(\lambda \,dt = \frac {d\theta }{1 - e^\theta } = \frac {e^{-\theta }\,d\theta }{e^{-\theta } - 1}\) and integrating, \(e^{\lambda t}(e^{-\theta } - 1)\) is constant along each curve, so \(M(\theta , t) = \Phi \left (e^{\lambda t}(e^{-\theta } - 1)\right )\) for some function \(\Phi \). At \(t = 0\), \(\Phi (e^{-\theta } - 1) = e^{a\theta }\); putting \(y = e^{-\theta } - 1\) gives \(\Phi (y) = (1 + y)^{-a}\). Hence \[M(\theta , t) = \left [1 - e^{\lambda t}\left (1 - e^{-\theta }\right )\right ]^{-a} .\] A negative binomial variable counting the trials needed for \(r\) successes with success probability \(p\) has m.g.f. \(\left (\frac {pe^\theta }{1 - (1 - p)e^\theta }\right )^r = \left [1 - \frac 1p(1 - e^{-\theta })\right ]^{-r}\), which is \(M(\theta , t)\) with \(r = a\) and \(p = e^{-\lambda t}\). The mean and variance are those of this distribution, \(\frac rp = ae^{\lambda t}\) and \(\frac {r(1 - p)}{p^2} = ae^{\lambda t}(e^{\lambda t} - 1)\). □