# Monte Carlo integration error

I have a question regarding the Monte Carlo integration method. This is probably a "noob" question, but I have searched the internet and haven't been able to find an answer...

Let's say I want to estimate an integral of some function over an area $D$, that I'll denote as $I$. I can choose $N$ points to randomly sample from $D$ in order to obtain the integral, the error should "decay" as $err\sim \frac{\sigma}{\sqrt{N}}$, so that $I\sim \overline{I}\pm\frac{\sigma}{\sqrt{N}}$. But what if I just ran the simulation $N$ times and would get different values like: $I_1,I_2,....I_N$ those should be distributed normally as well, right? So I can estimate the error from those values again as $I\sim\overline{I} \pm \frac{\sigma}{\sqrt{N}}$, would this give the same results or is my thinking wrong?

• Why do you think the integrals from your $N$ simulations would be Normally distributed? The rate of convergence of the Monte Carlo integral to the true value is not dependent upon any specific distributional assumption. You also appear to be confusing a simulation with $N$ points and $N$ simulations, unless you mean your $N$ simulations are each of one point... which of course is, in the case of independent simulations and points, exactly the same as one simulation of $N$ points. – jbowman Dec 20 '17 at 21:16
• So... what you’ saying is that if I make N simulations of N points and each simulation represents one points, as $\lim_{N\rightarrow\infty} I_N =I$ and all points are the same... And actually those different values for the repeated simulations I get for “different” $I_N$ are basically due to some random errors because $N$ is not infinite? But what would the distribution of those different values for $I$ be , why wouldn’t it be normal? – user3633438 Dec 20 '17 at 21:31
• Your initial sentence makes no sense: "if I make N simulations of N points and each simulation represents one points". If your simulation is of N points, it represents N points. What are you actually trying to say? And I cannot understand your last sentence either. Forget the Normal distribution, it has nothing to do with this problem! – jbowman Dec 20 '17 at 21:35
• I make a “simulation” of the integral $\int_D f(\textbf x) d\textbf x=I$ using N sampling points that lie in $D$, and then I calculate N such integrals and I lable them as $I_1,I_2,...I_N$, all of those values from $I_1$ to $I_N$ will be, in general, different. Maybe they are normally distributed and I can use that to estimate the error.... – user3633438 Dec 20 '17 at 21:45
• “if I make N simulations of N points....” means what if I calculate N simulations of the same integral that I compute using N sampling points from $D$... – user3633438 Dec 20 '17 at 21:47

1. $N$ simulations from a uniform distribution on the set $\mathfrak{D}$, $x_1,\ldots,x_N$, the estimator$$I_N=\frac{1}{N}\sum_{i=1}^N h(x_i)$$is an unbiased estimator of$$\mathfrak{I}=\int_\mathfrak{D} h(x)\text{d}x$$ that a.s. converges at the speed $\sqrt{N}$ in the sense of the Central Limit Theorem:$$\frac{1}{\sqrt{N}}\sum_{i=1}^N \{h(x_i)-\mathfrak{I}\}\stackrel{\mathcal{L}}{\longrightarrow}\mathcal{N}(0,\sigma^2)$$ which variance $\sigma^2$ can be estimated by the empirical variance of the $h(x_i)$'s. This means that $I_N$ is approximately Gaussian for large $N$'s, with variance $\sigma^2/N$
2. When considering an iid sequence $I_N^1,\ldots,I^N_N$ of such estimators, they all are unbiased estimators of $\mathfrak{I}$ with the same variance $\sigma^2/N$ that can be estimated by the empirical variance of the $I_N^i$'s, and again satisfy a CLT: $$\frac{\sqrt{N}}{\sqrt{N}}\sum_{i=1}^N \{I^i_N-\mathfrak{I}\}\stackrel{\mathcal{L}}{\longrightarrow}\mathcal{N}(0,\sigma^2)$$ This means that $I_N^1,\ldots,I^N_N$ is approximately a Gaussian sample for large $N$'s, again with variance $\sigma^2/N$
3. Letting $N$ grow to infinity in $I_N^1,\ldots,I^N_N$ does not serve a clear purpose, as opposed to computing a single $I_{N^2}$ based on the same number of simulations, $N^2$, which has a variance of $\sigma^2/N$