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## 26.1 Bayes is calibrated by construction

Suppose a Bayesian model is given in the form of a prior density $$p(\theta)$$ and sampling density $$p(y \mid \theta).$$ Now consider a process that first simulates parameters from the prior, $\theta^{\textrm{sim}} \sim p(\theta),$ and then simulates data given the parameters, $y^{\textrm{sim}} \sim p(y \mid \theta^{\textrm{sim}}).$ By the definition of conditional densities, the simulated data and parameters constitute an independent draw from the model’s joint distribution, $(y^{\textrm{sim}}, \theta^{\textrm{sim}}) \sim p(y, \theta).$ From Bayes’s rule, it follows that for any observed (fixed) data $$y$$, $p(\theta \mid y) \propto p(y, \theta).$ Therefore, the simulated parameters constitute a draw from the posterior for the simulated data, $\theta^{\textrm{sim}} \sim p(\theta \mid y^{\textrm{sim}}).$ Now consider an algorithm that produces a sequence of draws from the posterior given this simulated data, $\theta^{(1)}, \ldots, \theta^{(M)} \sim p(\theta \mid y^{\textrm{sim}}).$ Because $$\theta^{\textrm{sim}}$$ is also distributed as a draw from the posterior, the rank statistics of $$\theta^{\textrm{sim}}$$ with respect to $$\theta^{(1)}, \ldots \theta^{(M)}$$ should be uniform.

This is one way to define calibration, because it follows that posterior intervals will have appropriate coverage . If the rank of $$\theta^{\textrm{sim}}$$ is uniform among the draws $$\theta^{(1)}, \ldots, \theta^{(M)},$$ then for any 90% interval selected, the probability the true value $$\theta^{\textrm{sim}}$$ falls in it will also be 90%. The same goes for any other posterior interval.

### References

Dawid, A Philip. 1982. “The Well-Calibrated Bayesian.” Journal of the American Statistical Association 77 (379): 605–10.
Gneiting, Tilmann, Fadoua Balabdaoui, and Adrian E Raftery. 2007. “Probabilistic Forecasts, Calibration and Sharpness.” Journal of the Royal Statistical Society: Series B (Statistical Methodology) 69 (2): 243–68.