How to use a Bayesian parameter estimate?

This might be a naive question, but I still ask it since I couldn't easily find an answer to it.

Suppose we have an unknown parameter, $\theta$, and some function $\theta \mapsto f(\theta)$.

Suppose also that we have an a priori distribution for the value of $\theta$

Finally, we assume for simplicity that we have not obtained any information yet, so our posterior distribution equals the apriori distribution.

So our best estimate of $\theta$ is $\tilde{\theta} = E[\theta]$.

As I understand it, the typical thing to do in Bayesian statistics in order to compute $f$, and please correct me if I'm wrong, is now to simply evalute $f(E[\theta]) = f(\tilde{\theta})$.

However, wouldn't a more correct thing to do be to actually compute $E[f(\theta)]$?

If $f$ is convex for example, it follows that the latter is larger than the former, hence there will be a systematic difference between the two variants.

A reason why I'm asking this is that I have a model where computing $f(\theta)$ for given $\theta$ is expensive, so actually computing $E[f(\theta)]$ is far from desirable.

Edit: Ok, I just read about posterior predictive distributions and it seemed that I was probably wrong in my assumption that one merely used the expected value as the "true" parameter.

• This is an incorrect assumption: the Bayes estimate of $f(\theta)$ is $\mathbb{E}[f(\theta)]$ – Xi'an Mar 10 '16 at 10:11
• The main case when it makes no difference is $\hat{\theta}$ is the median and $f(\theta)$ is a a monotone function. I assume your situation is more complicated, since evaluating some simple function would presumably not be too hard. – Björn Mar 10 '16 at 12:00

This is a fairly general problem in parameter estimation. It is usually related to the use of loss functions which do not stay "coherent" under a change of parameters. It affects non-bayesian paparameter estimates as well. For example, if $\hat {\theta}$ is unbiased for $\theta$ then $f (\hat {\theta})$ is usually not unbiased for $f (\theta)$.