In Bayesian estimation, integral expressions such as
$$
f_{\pi}(x)=\int_{\Theta} f(x \mid \theta) \pi(\theta) d \theta
$$
or the mean cost function
$$
\int_{E_{1} \Theta} C(\theta, T(x)) f(x \mid \theta) \pi(\theta) d x d \theta
$$
cannot generally be computed analytically. In order to solve this kind of problem, we consider the general expression
$$
I=\int_{E} h(x) f(x) d x
$$
where $f$ is a Probability Density Function . We can compute a numerical approximation of $I$ by simulating $n$ realisations of independent random variables $X_{1}, \ldots, X_{n}$, with the same $f(x)$. We then consider the estimator of $I$ defined by
$$
\hat{I}=\frac{1}{n} \sum_{k=1, n} h\left(X_{k}\right)
$$
This is an unbiased estimator, and it is clear that its variance is given by
$$
\operatorname{var}[\bar{I}]=\frac{1}{n}\left(\int_{E} h^{2}(x) f(x) d x-\left(\int_{E} h(x) f(x) d x\right)^{2}\right)
$$
STAT4027 COURSE NOTES ：
$$
\operatorname{var}[\mathbb{E} h(X) \mid Z]] \leq \operatorname{var}[h(X)]
$$
which suggests that estimators of $\int_{E} h(x) f(x) d x$ of the form
could have a weaker variance than
$$
\frac{1}{n} \sum_{k=1, n} h\left(X_{k}\right) \text {. }
$$
This is evident in the particular case where the random variables $X_{k}$ are uncorrelated. The conditions under which using this technique, called $R a o$ Blackwellisation, is justified, can be found.