# Bayesian surprise

For reasons not entirely unconnected with NZ election polling, I’ve been thinking about surprise in Bayesian inference again: what happens when you get a result that’s a long way from what you expected in advance? Yes, your prior is badly calibrated and you should feel bad, but what should you believe?

A toy version of the problem is inference for a location parameter. We have a prior $$p_\theta(\theta)$$ for the parameter, and a model $$p_X(x|\theta)$$. Consider two extremes

• $$\theta\sim N(0,1)$$ and $$X\sim\textrm{Cauchy}(\theta)$$
• $$\theta\sim\textrm{Cauchy}(0)$$ and $$X\sim N(\theta, 1)$$

Suppose we take a single observation $$x$$ of $$X$$ and it’s very large. What do we end up believing about $$\theta$$ in each case?

Heuristically, the first case says the data can sometimes be a long way from $$\theta$$, but $$\theta$$ has to be not that far from 0. The second case says $$\theta$$ can sometimes be a long way from 0 but $$X$$ can’t be that far from $$\theta$$. So in the the first case the posterior for $$\theta$$ should be concentrated fairly near zero and in the second it should be concentrated fairly near $$X$$. That’s exactly what happens when you do the maths.

Under the first model, the posterior density is proportional to
$e^{-\theta^2/2}\frac{1}{1+(x-\theta)^2}$ and the posterior mode solves
$\tilde\theta =\frac{(x-\tilde\theta)}{1+(x-\tilde\theta)^2}.$
For $$x\to\infty$$ we can’t have  $$x-\theta$$ bounded, which in turn means $$\tilde\theta=O((x-\tilde\theta)^{-1})$$, giving $$\theta\to 0$$.

Under the second model, the posterior is proportional to $e^{(x-\theta)^2/2}\frac{1}{1+\theta^2}$ and the posterior mode solves
$x-\tilde\theta=\frac{2\tilde\theta}{1+\tilde\theta^2}.$
If $$x\to\infty$$, the solution to this equation must have $$x-\tilde\theta$$ bounded, which implies $$\tilde\theta\to\infty$$, which implies $$x-\tilde\theta\to 0$$.

If the two distributions are both Normal the posterior mode will be about halfway between $$x$$ and 0. If they’re both Cauchy, the posterior will be bimodal, with one mode near $$x$$ and another near 0.

The basic observation here goes back a long way, with a relatively recent summary by O’Hagan in JASA, 1990: given a surprising observation, Bayesian inference can (sensibly) end up just ‘rejecting’ which ever of the prior and model have heavier tails.

Working it out for simple cases makes a nice straightforward stats theory question.  It’s also a good low-dimensional example of the problem common in high-dimensional problems that it’s quite hard to be sure what features of your model and prior are going to matter for inference.