# Frequentist Accuracy of Bayesian Estimates

##### Posted on Mar 31, 2019

This note is for Efron’s slide: Frequentist Accuracy of Bayesian Estimates, which is recommended by Larry’s post: Shaking the Bayesian Machine.

In Bayesian Inference, we have

• Parameter: $\mu\in \Omega$
• Observed data: $x$
• Prior: $\pi(\mu)$
• Probability distributions: $\{f_\mu(x),\mu\in\Omega\}$
• Parameter of interest: $\theta = t(\mu)$

then the posterior estimate of $\theta$ is

$\E[\theta\mid x] = \frac{\int_\Omega t(\mu)f_\mu\pi(\mu)d\mu}{\int_\Omega f_\mu(x)\pi(\mu)d\mu}\,.$

This form recalls me a more generic form of the conditional expectation.

The usual uninformative prior is Jeffreys’ prior,

$\pi(\mu) = \vert I(\mu)\vert^{1/2}\,,$

where $I(\mu) = \Cov[\nabla_\mu\log f_\mu(x)]$.

But how accurate are the estimates? Then we will consider the frequentist variability of $\E[t(\mu)\mid x]$.

## General Accuracy Formula

• $\mu$ and $x\in\bbR^p$
• $x\sim (\mu, V_\mu)$ (here I guess $\mu$ is not necessarily the expectation of $x$ according to the following examples, this symbol might want to indicate $x$ is indexed by $\mu$)
• $\alpha_x(\mu)=\nabla_x\log f_\mu(x)$
$\hat E=E[t(\mu)\mid x]$ has gradient $$\nabla_xE =\Cov[t(\mu), \alpha_x(\mu)\mid x]\,.$$
The delta-method standard deviation of $E$ is $$\sd(\hat E) = \left\{\Cov[t(\mu),\alpha_x(\mu)\mid x]'V_x\Cov[t(\mu),\alpha_x(\mu)\mid x]\right\}^{1/2}\,.$$

## Implementation

Suppose we have the posterior sample $\mu_1^*,\ldots,\mu_B^*$, let $t_i^*=t(\mu_i^*)$ and $\alpha_i^*=\alpha_x(\mu_i^*)$, then

$\widehat{\cov} = \sum_{i=1}^B(\alpha_i^*-\bar \alpha)(t_i^*-\bar t)/B\,,$

it follows that

$\hat \sd(\hat E) = \left[\widehat\cov'V_x\widehat\cov\right]^{1/2}\,.$

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