One way to find the volume of a sphere would be to imagine the sphere in a box, randomly select points in the box, and count how many of these points fall inside the sphere. In principle this would work in any dimension.

## The problem with naive Monte Carlo

We could write a program to estimate the volume of a high-dimensional sphere this way. But there’s a problem: very few random samples will fall in the sphere. The ratio of the volume of a sphere to the volume of a box it fits in goes to zero as the dimension increases. We might take a large number of samples and none of them fall inside the sphere. In this case we’d estimate the volume as zero. This estimate would have small absolute error, but 100% relative error.

## A more clever approach

So instead of actually writing a program to randomly sample a high dimensional cube, let’s **imagine** that we did. Instead of doing a big Monte Carlo study, we could **be clever and use theory**.

Let *n* be our dimension. We want to draw uniform random samples from [−1, 1]^{n} and see whether they land inside the unit sphere. So we’d draw *n* random samples from [−1, 1] and see whether the sum of their squares is less than or equal to 1.

Let *X*_{i} be a uniform random variable on [−1, 1]. We want to know the probability that

*X*_{1}² + *X*_{2}² + *X*_{3}² + … + *X*_{n}² ≤ 1.

This would be an ugly calculation, but since we’re primarily interested in the case of large *n*, we can approximate the sum using the central limit theorem (CLT). We can show, using the transformation theorem, that each *X*_{i}² has mean 1/3 and variance 4/45. The CLT says that the sum has approximately the distribution of a normal random variable with mean *n*/3 and variance 4*n*/45.

## Too clever by half

The approach above turns out to be a bad idea, though it’s not obvious why.

The CLT does provide a good approximation of the sum above, **near the mean**. But we have a sum with mean *n*/3, with *n* large, and we’re asking for the probability that the sum is less than 1. In other words, we’re asking for the probability **in the tail** where the CLT approximation error is a bad (relative) fit. More on this here.

This post turned out to not be about what I thought it would be about. I thought this post would lead to a asymptotic approximation for the volume of an *n*-dimensional sphere. I would compare the approximation to the exact value and see how well it did. Except it did terribly. So instead, this post a cautionary tale about remembering how convergence works in the CLT.

While plain Monte Carlo and the central limit theorem don’t help in this case, importance sampling can give a Monte Carlo estimate with low relative error. I wrote up the details here: https://mathstoshare.com/2024/08/07/monte-carlo-integration-in-high-dimensions/

Thanks for the thought provoking post!