I am a bit up over my head here, I will present an argument and then I hope you guys will say if my reasoning is correct or what should be changed, ultimately I am hoping to say something qualified about the size of $\varepsilon$ as to be defined.
Let $X$ be some quite unmanageable random variable (it is an integral of a stochastic process) and assume of which I would like to approximate $E[X]$. Using $M$ operations (approximating it with trapezoidal discrete integral on M points) I can get $Y_M$ such that $X-Y_M$ is asymptotically $O(1/M^2)$. I can get $N$ of such $Y_M$'s (by simulation), i.e. $Y_M^1,\dots Y_M ^N$ and approximate $E[X]\approx \tilde{S}_N^M = 1/N \sum_{i=1}^{N} Y_M ^i$ by the central limit theorem the error from doing this is asymptotically $O(1/\sqrt{N})$. So if I choose $M= \sqrt[3]{N}$ my error will be asymptotically $O(1/\sqrt{N})$ and if I consider the error $$ \varepsilon =\frac{\lvert E[X] -\tilde{S}_N^M \rvert}{E[X]} $$ it will be in the order of magnitude $1/\sqrt{N}$. In particular if I get a value of say $0.05$ and have $N=10^6$ I can expect that $\varepsilon \approx 0.1 \%$ or that I can be certain of my result up to $1\%$?