This seems like a typical Bayes' Rule question, but I'm having trouble formulating it properly.

Say I'm a speaker on a conference tour which moves from city to city. I'm trying to estimate how many people will attend my talk in the next city, given how many attended in previous cities.

For example, say the total number of attendees in each city is 100, and that there are five simultaneous, equally likely tracks, so I start in the first city assuming that 20 people will attend (i.e., my first prior is 0.2). In actuality, 30 people attend. How do I use that evidence to predict how many will attend in the next city? How about the one after that?

It feels like I ought to be able to write a simple Bayes' Rule expression for this, but everything I try seems wrong. Any help would be appreciated.


if $x$ is number of attendees and $n$ is the available attendees, then we can assume $X$ has a binomial distribution with Bernoulli parameter $p$. In frequentist statistics, we would say that $p$ should be 0.3 now because we saw 30 people arrive. In a Bayesian framework, we are going to incorporate our belief that $p$ is really 0.2 with our new data with this formula $$ \begin{aligned} P(p|x) &= \frac{P(x|p)*P(p)}{P(x)} \\ &= \frac{P(x=30|p=0.2) * P(p=0.2)}{P(x=30)} \end{aligned} $$ Calculating the right hand side will give you an updated estimate of $p$. The tough parts are:

  1. what should the pdf of $p$ be?
  2. Calculating the marginal probability (denominator)

Depending on how you frame the 2 above, this could be very difficult or very simple to solve. Commonly, $p$ is taken to be uniform so you end up with a beta distribution posterior if you work it out.

| cite | improve this answer | |
  • 1
    $\begingroup$ And then that beta posterior becomes the prior for the next meeting and you go through the same steps to get another posterior, which becomes the prior for the city after that, ... $\endgroup$ – Greg Snow Mar 8 '13 at 0:05
  • $\begingroup$ Thanks for the answer, and I do appreciate it, but is it obvious what the "very simple" solution would look like? I'm quite happy to assume that p is uniform, and using a binomial distribution is quite reasonable. I'm just looking for a back-of-the-envelope type of calculation. $\endgroup$ – kousen Mar 8 '13 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.