5
$\begingroup$

A simultaneous confidence band denotes the probability

$$p \big(\hat{f}(x) - w(\hat{f}(x)) \le f(x) \le \hat{f}(x) + w(\hat{f}(x)) \ \ \forall x \big)=1-\alpha$$

where $f$ a function of $x$, $\hat{f}$ and estimator for the value of $f$ at $x$, and $w(\hat{f}(x))$ an estimated interval length at $x$, see also Wikipedia entry.

Is there an existing analogue definition in Bayesian statistics and how is the resulting band estimated? (Using the posterior or posterior predictive distribution.)

$\endgroup$
0

1 Answer 1

6
$\begingroup$

Given a prior distribution $\pi$ on a functional space and observations about the function values at some points, or noisy observations of the function itself, the posterior distribution $\pi(\cdot|\mathcal{D})$ can be used to derive an HPD region, $$\left\{f\in\mathcal{F}\,,\ \pi(f |\mathcal{D})\ge k_\alpha\right\}$$ at least in principle since the derivation may prove too complex in a general situation.

For instance,

Breth, M. (1978) Bayesian confidence bands for estimating a function. Annals of Statistics Vol. 6, No. 3, pp. 649-657

seems to address this problem.

$\endgroup$
5
  • $\begingroup$ Is it possible to simulate this region using the samples from a posterior? $\endgroup$
    – tomka
    Commented Nov 1, 2017 at 9:05
  • $\begingroup$ Not that I know of, because the samples are functions. Except by considering the convex envelope of these functions, providing a conservative HPD region. $\endgroup$
    – Xi'an
    Commented Nov 1, 2017 at 9:11
  • $\begingroup$ This paper seems relevant in this regard. I wonder if this has been extended to the multivariate case. onlinelibrary.wiley.com/doi/10.1111/1467-9876.00053/pdf $\endgroup$
    – tomka
    Commented Nov 1, 2017 at 11:07
  • $\begingroup$ I am not sure; can't we consider $f(x)$ a bivariate parameter if $x$ is binary? $\endgroup$
    – tomka
    Commented Nov 1, 2017 at 11:14
  • $\begingroup$ Yes, but this changes everything in your question! $\endgroup$
    – Xi'an
    Commented Nov 1, 2017 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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