Solution
Two Quiet Days at the Help Desk
Show the problem again
A help desk receives urgent tickets as a Poisson process with an unknown rate $\lambda$ per day. Your prior for $\lambda$ is Gamma with shape 2 and rate 1 (mean 2). You then observe 4 tickets in total over 2 days. What is the posterior predictive probability that the next 2 days together produce no urgent tickets at all? Give your answer as a reduced fraction.
Worked solution
The Gamma prior is conjugate: after observing $k$ events in $t$ days the posterior is Gamma(shape 2+4, rate 1+2) = Gamma(6, 3). No tickets in a further stretch of 2 days has probability $e^{-2\lambda}$ given $\lambda$. For $\lambda \sim$ Gamma(shape $a$, rate $b$), $E[e^{-s\lambda}] = (b/(b+s))^a$ (the Gamma moment generating function). With $a=6$, $b=3$, $s=2$: $$\left(\frac{3}{5}\right)^6 = \frac{729}{15625}.$$ The answer is **729/15625**, about 0.0467. Plugging in the posterior mean $\lambda=2$ would give $e^{-4}\approx 0.018$, a very different number: parameter uncertainty fattens the predictive tail.
Source: Original problem written for AxiomIQ (standard Gamma-Poisson posterior predictive, i.e. negative binomial). Statement written for AxiomIQ.