Let $X_1,...,X_n$ be points on $\mathbb S^1.$

We then define the expectation value $E(X)=\frac{1}{n}\sum_{i=1}^n X_i.$

Let $\frac{dS(X_1)}{2\pi}$ be the normalized surface measure of $\mathbb S^1,$ i.e. $X_i$ are uniformly distributed random variables on the circle.

I am curious to know:

How does

$$\int_{(\mathbb S^1)^n } \frac{1}{\vert E(X) \vert}\frac{dS(X_1)}{2\pi}...\frac{dS(X_n)}{2\pi}$$ scale with $n$?

  • 2
    $\begingroup$ A quick guess would be the Central limit theorem $\sim \sqrt{n}$ $\endgroup$
    – RaphaelB4
    Sep 3, 2020 at 10:04
  • $\begingroup$ @RaphaelB4 would you mind elaborating a bit on this point? $\endgroup$ Sep 4, 2020 at 14:15
  • $\begingroup$ See the distribution in Carlo.s answer : $e^{-R^2/n}$: the Gaussian with variance $n$. $\endgroup$
    – RaphaelB4
    Sep 4, 2020 at 14:46

1 Answer 1


The probability distribution $P(R)$ of $R=n|E(X)|$ was calculated by Kluyver (1906), it is given by $$P(R)=\frac{1}{2\pi}\int_0^\infty [J_0(x)]^n J_0(rx)x\,dx.$$ For $n\gg 1$ one has a Rayleigh distribution (here is derivation including higher order corrections): $$P(R)=\frac{2R}{n}e^{-R^2/n}.$$ The desired integral then becomes $$I=\int_0^{\infty}\frac{n}{R}P(R)\,dR\rightarrow \sqrt{\pi n}$$ in the limit $n\rightarrow\infty$.


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