# Monte Carlo Integration for non-square integrable functions

I hope this is the right place to ask, if not feel free to move it to a more appropriate forum.

I've been wondering for quite a while now how to treat non-square integrable functions with Monte Carlo Integration. I know that MC still gives a proper estimate but the error is unrealiable (divergent?) for those kind of functions.

Let's restrict us to one dimension. Monte Carlo integration means that we approximate the integral

$$I = \int_0^1 \mathrm{d}x \, f(x)$$

using the estimate

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

with $x_i \in [0,1]$ uniformly distributed random points. The law of large numbers makes sure that $E \approx I$. The sample variance

$$S^2 = \frac{1}{N-1} \sum_{i=1}^N (f (x_i) - E)^2$$

approximates the variance $\sigma^2$ of the distribution induced by $f$. However, if $f$ is not square-integrable, i.e. the integral of the squared function diverges, this implies

$$\sigma^2 = \int_0^1 \mathrm{d} x \, \left( f(x) - I \right)^2 = \int_0^1 \mathrm{d} x \, f^2(x) - I^2 \longrightarrow \infty$$

meaning that also the variance diverges.

A simple example is the function

$$f(x) = \frac{1}{\sqrt{x}}$$

for which $I = \int_0^1 \mathrm{d}x \, \frac{1}{\sqrt{x}} = 2$ and $\sigma^2 = \int_0^1 \mathrm{d}x \, \left( \frac{1}{x} - 2 \right) = \left[ \ln x - 2x \right]_0^1 \rightarrow \infty$.

If $\sigma^2$ is finite one can approximate the error of the mean $E$ by $\frac{S}{\sqrt{N}} \approx \frac{\sigma}{\sqrt{N}}$, but what if $f(x)$ is not square-integrable?

• I don't get it: you start out by noting that none of the $E_i$ has a variance and then ask whether the variance of their average would be a reasonable estimator of--that nonexistent variance! Or do I misread this question: perhaps by "statistically independent estimations" you have some different (perhaps robust) estimator of the integral in mind? – whuber Jun 21 '13 at 18:14
• I didn't say $E$ doesn't have a variance, only that I cannot define a variance for it by $S^2$. So the question is whether I can define an error at all and if $\bar{S}^2$ is a reasonable candidate. By statistically independent I mean that the $E_i$ are obtained using different random numbers, e.g. by using differently seeded random number generators (I hope thats the right term then). – cschwan Jun 22 '13 at 6:38
• Please explain what you mean by not being able to "define a variance for it by $S^2$." I cannot make sense of this using the standard definitions of variance and $S^2$. – whuber Jun 24 '13 at 14:45
• Well, the function is not square-integrable so, if I am not mistaken, $S^2$ should diverge. If this is the case the definition for $S^2$ makes no sense in the first place, right? By means of the central limit theorem, however, $E$ will still converge to the true value of the integral, but without an error this value alone makes no sense (how 'good' is this result?). – cschwan Jun 26 '13 at 13:55
• Sorry, I meant to say "law of large numbers" of course, not CLT. – cschwan Jun 26 '13 at 14:04