这是一份 Imperial帝国理工大学 MATH97116作业代写的成功案例
which is an i.i.d. sequence of $\mathcal{U}\left([0,1]^{d}\right)$ distributed random variables. Then for any $M \in \mathbb{N}$ (large),
$$
I_{M}:=\frac{1}{M} \sum_{k=1}^{M} f\left(\vec{U}{k}\right) $$ is an unbiased estimator for $I$, that is, $$ \mathbb{E}\left[I{M}\right]:=\frac{1}{M} \sum_{k=1}^{M} \mathbb{E}\left[f\left(\vec{U}{k}\right)\right]=\mathbb{E}\left[f\left(\vec{U}{1}\right)\right]=I
$$
Since $f$ is integrable we have by the strong law of large numbers that
$$
I_{M} \rightarrow I, \quad \text { with probability } 1 \text { as } M \rightarrow \infty
$$
If $f^{2}$ is integrable too we may define
$$
\begin{aligned}
& \operatorname{Var}\left[f\left(\vec{U}{1}\right)\right]:=\mathbb{E}\left[\left(f\left(\vec{U}{1}\right)-\mathbb{E}\left[f\left(\vec{U}{1}\right)\right]\right)^{2}\right] \ =& \mathbb{E}\left[\left(f\left(\vec{U}{1}\right)-I\right)^{2}\right]=\int_{[0,1]^{d}}(f(x)-I)^{2} d x=: \sigma_{f}^{2}
\end{aligned}
$$
MATH97116 COURSE NOTES ：
Thus, in order to get a statistical accuracy of $\epsilon$, i.e.
$$
\sqrt{\mathbb{E}\left[\left(I_{M}-I\right)^{2}\right]} \sim \epsilon
$$
we need
$$
\sigma_{M}^{2}=\sigma_{f}^{2} / M \sim \epsilon^{2},
$$
at computation costs
$$
C(\epsilon) \sim M .
$$
Hence, the Monte Carlo complexity is
$$
C(\epsilon) \sim \frac{\sigma_{f}^{2}}{\epsilon^{2}},
$$