Variance reduction

Remarks

Example: $q = \int_0^1 e^u\, du = e - 1$ and $Y = e^U$ with $U$ uniform on $(0, 1)$, so $\operatorname{Var} Y = \frac{e^2 - 1}{2} - (e - 1)^2 \approx 0.2420$. Take $Z = \frac12 \bigl(e^U + e^{1 - U}\bigr)$. Clause 1 holds because $1 - U$ is also uniform, clause 3 is clear, and a direct computation gives $\operatorname{Var} Z = \frac{e^2 - 1 + 2e}{4} - (e - 1)^2 \approx 0.0039$, so clause 2 holds. Non-example: $Z = 1 + U + \frac12 U^2$, the Taylor polynomial of $e^U$, has variance about $0.189 < 0.2420$, but $\mathbb E Z = \frac53 \ne e - 1$, so clause 1 fails and the estimate is biased. A fair comparison also counts the cost. If one copy of $Z$ costs $c$ times as much as one copy of $Y$, the reduction pays off when $c \operatorname{Var} Z < \operatorname{Var} Y$. In the example $c = 2$ and $2 \cdot 0.0039 < 0.2420$. Some techniques, such as stratified sampling, change the whole estimator rather than one sample. For them one compares the variances of two estimators of the same cost.