Standard error of importance sampling estimator

Standard error of importance sampling estimator

pI am currently studying Monte Carlo simulation and more specifically
independent, importance sampling. Consider the following context:/p
pAssume we have some posterior pdf $p(\boldsymbol{\theta}|\mathbf{y})$
which we do not know in its entirety, in other words, assume we only know
$p$ up to its integrating constant, hence write
$p(\boldsymbol{\theta}|\mathbf{y})=c \times
p^*(\boldsymbol{\theta}|\mathbf{y})$ where we know
$p^*(\boldsymbol{\theta}|\mathbf{y})$. Our aim is to estimate the
expectation, $E_p[g(\boldsymbol{\theta}|\mathbf{y})]$, with respect to
$p(\boldsymbol{\theta}|\mathbf{y})$. Thus we can write:
\begin{align*}E_p[g(\boldsymbol{\theta}|\mathbf{y})] amp; =
\int_{\boldsymbol{\theta}} g(\boldsymbol{\theta})
p(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta} \\ amp; =
\int_{\boldsymbol{\theta}} g(\boldsymbol{\theta}) c
p^*(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta} \\ amp; = c
\int_{\boldsymbol{\theta}}g(\boldsymbol{\theta})p^*(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta}
\\ amp; =
\frac{\int_{\boldsymbol{\theta}}g(\boldsymbol{\theta})p^*(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta}}{\int_{\boldsymbol{\theta}}p^*(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta}}
\\ amp; = \frac{\int_{\boldsymbol{\theta}} g(\boldsymbol{\theta})
\frac{p^*(\boldsymbol{\theta}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}}|\mathbf{y})}f(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta}}{\int_{\boldsymbol{\theta}}
\frac{p^*(\boldsymbol{\theta}|\mathbf{y})}{f(\boldsymbol{\theta}|\mathbf{y})}f(\boldsymbol{\theta}|\mathbf{y})d\boldsymbol{\theta}}
\\ amp; =E_f\left[g(\boldsymbol{\theta})
\frac{p^*(\boldsymbol{\theta}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}}|\mathbf{y})}\right]
/
E_f\left[\frac{p^*(\boldsymbol{\theta}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}}|\mathbf{y})}\right]
\end{align*}/p pwhere $f(\boldsymbol{\theta}|\mathbf{y})$ is a known pdf
from which we sample $\boldsymbol{\theta}$ from, assume we have a sample
of $M$, ie, $\boldsymbol{\theta}^{(1)}, \boldsymbol{\theta}^{(2)}, \cdots,
\boldsymbol{\theta}^{(M)}$ then the numerator can be estimated as
$$\frac{1}{M} \sum_{j=1}^M g(\boldsymbol{\theta}^{(j)})
\frac{p^*(\boldsymbol{\theta}^{(j)}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}^{(j)}}|\mathbf{y})}
\rightarrow^{a.s} E_f\left[g(\boldsymbol{\theta})
\frac{p^*(\boldsymbol{\theta}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}}|\mathbf{y})}\right]
$$ /p pand the denominator can be estimated as $$\frac{1}{M} \sum_{j=1}^M
\frac{p^*(\boldsymbol{\theta}^{(j)}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}^{(j)}}|\mathbf{y})}
\rightarrow^{a.s}
E_f\left[\frac{p^*(\boldsymbol{\theta}|\mathbf{y})}{f(\mathbf{\boldsymbol{\theta}}|\mathbf{y})}\right]$$
/p pusing the strong law of large numbers./p pHence defining the
importance weights as $w(\boldsymbol{\theta}^{(j)}) =
\frac{p^*(\boldsymbol{\theta}^{(j)}|\mathbf{y})}{f(\boldsymbol{\theta}^{(j)}|\mathbf{y})}$,
we have the estimator for $E_p[g(\boldsymbol{\theta}|\mathbf{y})]$:/p
p$$\overline{g(\boldsymbol{\theta})}^{IS} =
\sum_{j=1}^M\left(g(\boldsymbol{\theta}^{(j)})w(\boldsymbol{\theta}^{(j)})\right)
/ \sum_{j=1}^M w(\boldsymbol{\theta}^{(j)}) \rightarrow^{a.s}
E_p[g(\boldsymbol{\theta}|\mathbf{y})]$$/p pNow I was told the
strongstandard error/strong of $\overline{g(\boldsymbol{\theta})}^{IS}$
is:/p p$$ \displaystyle
\frac{\sqrt{\sum_{j=1}^M\left(w(\boldsymbol{\theta}^{(j)})\left[g(\boldsymbol{\theta}^{(j)})-\overline{g(\boldsymbol{\theta})}^{IS}\right]\right)^2}}{\sum_{j=1}^M
w(\boldsymbol{\theta}^{(j)})}$$/p pI've tried to derive this expression
but to no avail, can anyone please shed some light as to how it's
derived?/p