Suppose we want to calculate the expectation of some function of a normal random variable $E(f(X))$. Instead of using the estimator

$\frac{1}{N}\sum_{i=1}^N f(X_i)$

(1)

it is well-known that we can use the estimator

$\frac{1}{2N} \sum_{i=1}^{N} f(X_i)+f(-X_i)$

(2)

It can be trivially verified that the second estimator has smaller variance than that of the first estimator. What if $X$ is not normal when it could be non-symmetric or discrete ?

Note that all random variables have something to do with uniform-zero-one distribution: let $A(x)$ be the accumulated density, i.e. and $A(x) ≡ \Pr(X \leq xU$ be a uniform-zero-one random variable, then the number $X \equiv \min\{x |U \leq A(x)\}$ has $A(x)$ as its accumulated density.

Because $1 - U$ is also a uniform-zero-one random variable, therefore we can have this estimator:

$\frac{1}{2N} \sum_{i=1}^N f(\min\{x |U_i \leq A(x)\}) + f(\min\{x |1 - U_i \leq A(x)\})$

(3)

It can also be trivially verified that this estimator has (1) as its special case and smaller variance.