In the book Bayesian Data Analysis, the authors state on page 128:

The concept of calibration of the posterior mean [is] the Bayesian analogue to the classical notion of bias.

They define the miscalibration of the posterior mean of some parameter $\theta$ as $$ \text{miscalibration} = E(\theta|\hat{\theta}) - \hat{\theta} $$ where $$ \hat{\theta} = E(\theta|y) $$ and $y$ is the observed data.

Can someone clarify what this means? How does this $E(\theta|\hat{\theta})$ term work? It's a conditional expectation that's conditional on something that is itself a conditional expectation. How does that work? Can you show that the miscalibration is 0 if the prior distribution is true, i.e. if the data are constructed by first drawing $\theta$ from $p(\theta)$ and then drawing $y$ from $p(y|\theta)$?


1 Answer 1


It may be confusing that the $\hat \theta = E[\theta \mid y]$ seems to use the chosen prior and the posterior distribution then implied by the observation, while the $E\left[\theta \mid \hat \theta\right]-\hat \theta$ in the miscalibration calculation seems to use the actual (though presumably unknown) distribution for $\theta$. It is probably easier with an example.

Suppose we have $\theta \in [0,1]$ as the parameter of a Bernoulli random variable $Y$ which is $1$ with probability $\theta$ and $0$ otherwise

If we use a prior $p(\theta)=2\theta$, then

  • observing $Y=1$ will suggest to us a posterior distribution $p(\theta \mid Y=1)=3\theta^2$ with mean $\hat \theta =\frac34$,
  • observing $Y=0$ will suggest to us a posterior distribution $p(\theta \mid Y=0)=6\theta(1-\theta)$ with mean $\hat \theta =\frac12$,

and there is a $1-1$ relationship here between the value of $Y$ and $\hat \theta$.

If the prior is the correct distribution of $\theta$ then

  • when $\hat \theta =\frac34$, we have $E\left[\theta \mid \hat \theta=\frac34\right] - \hat\theta =\frac34-\frac34=0$

  • when $\hat \theta =\frac12$, we have $E\left[\theta \mid \hat \theta=\frac12\right] - \hat\theta =\frac12-\frac12=0$

automatically as asserted, and also

  • the probability of observing $Y=1$ and estimating $\hat \theta=\frac34$ is $\frac23$
  • the probability of observing $Y=0$ and estimating $\hat \theta=\frac12$ is $\frac13$

so $E\left[\hat\theta\right]=\frac23 \times \frac34 +\frac13\times \frac12 =\frac23 = E[\theta]$ and all is well with the world.

But if the prior were wrong and in fact $\theta$ in reality had a density of $2(1-\theta)$ i.e. with $\theta$ and $Y$ more likely to be smaller than the assumed prior suggested, then the actual conditional expectations for $\theta$ would be smaller

  • when $\hat \theta =\frac34$, we would have $E\left[\theta \mid \hat \theta=\frac34\right] - \hat\theta =\frac12-\frac34=-\frac14$
  • when $\hat \theta =\frac12$, we would have $E\left[\theta \mid \hat \theta=\frac12\right] - \hat\theta =\frac14-\frac12=-\frac14$

so a miscalibration of $- \frac14$ (in more complicated cases it does not have to be constant) and

  • the probability of observing $Y=1$ and estimating $\hat \theta=\frac34$ would have been $\frac13$
  • the probability of observing $Y=0$ and estimating $\hat \theta=\frac12$ would have been $\frac23$

making $E[\hat\theta]=\frac13 \times \frac34 +\frac23\times \frac12 =\frac7{12} \not = \frac13 = E[\theta]$, which may be undesirable.

  • $\begingroup$ I still don't understand how you calculate $E(\theta | \hat{\theta})$. For example above you simply state that in your first example $E(\theta | \hat{\theta} = \frac{3}{4}) = \frac{3}{4}$. But I just fundamentally don't understand how such an expectation conditional on another expectation works. Could you perhaps elaborate this in baby steps? I don't see where that conditional comes in, or what probability distribution is used exactly to calculate the expectation. $\endgroup$
    – Willem
    Dec 20, 2021 at 13:54
  • $\begingroup$ @Willem If your prior is $p(\theta)=2\theta$ then, having observed $Y=1$, your posterior would be $p(\theta \mid Y=1) = 3\theta^2$, leading to $\hat \theta = \mathbb E[\theta \mid Y=1]= \frac34$. But if instead your observe $Y=0$ you get $\hat \theta = \frac12$. So with this prior, the event $Y=1$ is equivalent to the event $\hat \theta = \frac34$ while the event $Y=0$ is equivalent to the event $\hat \theta = \frac12$. If the prior is correct then you get $E[\theta \mid \hat \theta =\frac34] =\frac34$ $\endgroup$
    – Henry
    Dec 29, 2021 at 22:58
  • $\begingroup$ Thank you for your reply. Am I correct in understanding that this means conditioning on $\hat{\theta}$ is equivalent to conditioning on $Y$? The way I'm understanding it now is that really miscalibration = $E_{\text{true prior}}(\theta | y) - E_{\text{chosen prior}}(\theta | y)$. Is this interpretation correct? $\endgroup$
    – Willem
    Jan 10, 2022 at 20:29
  • $\begingroup$ @Willem - yes - I think that is a way of looking at it $\endgroup$
    – Henry
    Jan 10, 2022 at 20:54

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