Musings

Say we have a finite-dimensional family of probability distributions. We will call that family natural for the problem at hand, if the Bayes flow keeps us within that finite dimensional space.

A natural family of probability distributions for Problem 1 is the $\Gamma$-distribution. Let us choose a prior in that family $$p_0(\lambda) = f(\lambda; \kappa, a) = \frac{\kappa^a \lambda^{a-1} e^{-\kappa\lambda}}{\Gamma(a)}$$ where, for reasons that will be apparent in a moment, we require $\kappa\gt 0,\, a\gt 1$. For fixed $\lambda$, the expected wait time is $$\int_0^\infty d\tau e^{-\lambda \tau} \lambda\tau= \frac{1}{\lambda}$$ So, given our prior, we expect to wait $$\langle T\rangle = \int_0^\infty d\lambda \frac{p_0(\lambda)}{\lambda} = \frac{\kappa}{a-1}$$

Applying Bayes Theorem, our posterior distribution, after waiting a time $t$, is $$\begin{split} p(\lambda; t) &= \frac{e^{-\lambda t} p_0(\lambda)}{\int_0^\infty d\lambda' e^{-\lambda' t} p_0(\lambda')}\\ &= f(\lambda; \kappa+t, a) \end{split}$$ which, as announced, is just a shift of parameters of the $\Gamma$-distribution. We immediately conclude that our expected wait time $$\langle T\rangle = \frac{\kappa+t}{a-1}$$ has gone up. The longer we wait, the longer we expect to continue having to wait!

You should pause to convince yourself that’s the generic behaviour, whatever prior you assumed.

What about Problem 2? The first task is to compute, for fixed rates, $\lambda,\mu$, the probability that

• There are $n$ people waiting at the stop, when you arrive.
• You wait for a time, $t$, during which
  • no bus arrives, but
  • $N-n$ more riders arrive.

The answer is $$P(N,n; \lambda,\mu) = e^{-(\lambda+\mu)t} \frac{{(\mu t)}^{N-n}}{(N-n)!}\, \frac{\mu^n \lambda}{{(\lambda+\mu)}^{n+1}}$$

The second task is to find a natural family of probability distributions for this problem. I don’t know its name, but there is an obvious 5-parameter family, which is the natural generalization of the $\Gamma$-distribution, $$\begin{split} p_0(\lambda, \mu) &= g(\lambda,\mu;\kappa,\rho, a,b,c)\\ &= \frac{\kappa^{a+c}\rho^b}{\Gamma(a+c)\Gamma(b)\multiscripts{_2}{F}{_1}(b,-c,1-a-c;\kappa/\rho)}\, \lambda^{a-1} \mu^{b-1} {(\lambda+\mu)}^c e^{-(\kappa\lambda+\rho\mu)} \end{split}$$ Note that

• For $c=0$, this is just a product of independent $\Gamma$-distributions. $$g(\lambda,\mu;\kappa,\rho,a,b,0) = f(\lambda;\kappa,a)f(\mu;\rho,b)$$
• For positive integer $c$, the hypergeometric function is just a finite-order polynomial $$\multiscripts{_2}{F}{_1}(b,-c,1-a-c;x) = \frac{1}{{(a)}_c} \sum_{k=0}^c {(b-1)}_k {(a)}_{c-k}\begin{pmatrix}c\\ k\end{pmatrix} x^k$$
• There’s an obvious symmetry $$\begin{gathered} \lambda\leftrightarrow\mu \\ \kappa\leftrightarrow\rho \\ a\leftrightarrow b \end{gathered}$$

Applying Bayes Theorem, the posterior probability is $$p(\lambda,\mu; N,n,t) = g(\lambda,\mu;\kappa+t,\rho+t , a+1,b+N,c-n-1)$$ which, again, is just a shift of parameters of the distribution. The expected wait time $$\langle T\rangle = \frac{\kappa}{a+c-1} \frac{\multiscripts{_2}{F}{_1}(b,-c,2-a-c;\kappa/\rho)}{\multiscripts{_2}{F}{_1}(b,-c,1-a-c;\kappa/\rho)}$$ transforms accordingly. The dependence on $N,n,t$ is, alas, somewhat complicated. You can spend some hours convincing yourself that it is what you should expect.

Whether you think a Poisson process is a good model for a real public transit systems, probably depends on your political persuasion.

Update: 12/29/2013

In return, I should pose the followup problem:

3. Same setup as Problem 1 but, now, you’ve been to the bus stop $n$ times previously and had to wait for times $\{t_1,t_2,\dots,t_n\}$. What is your posterior probability distribution for $\lambda$ (hint: the $\Gamma$-distribution is natural for this problem, too), and what is your expected wait time?