Let $\theta_i\sim\mathcal{U}(0,2\pi)$, $i = 1,\dots,n$ be $n$ i.i.d. uniformly distributed random variables. Let: $$ \frac{D_n^2}{n} = 1 + \frac{2}{n}\sum_{i<j}^{n}\cos(\theta_i-\theta_j) $$ I have the evidence (by Monte Carlo simulation) that $\frac{D_n^2}{n}\to Z\sim Exp(1)$ in distribution as $n\to\infty$. I really have tried all the strategies I know, but none of them seem to work in this case. Note that the random variables $\cos(\theta_i-\theta_j)$ in general are not indipendent (although uncorrelated).

Note how $\frac{D_n^2}{n}$ has mean 1 and variance equal to $1-\frac{1}{n}$ (so at least asymptotically they certainly have the first two moments equal to each other).

I have proved that $\frac{D_n^2}{n}$ has all the moment bounded, in particular: $$ \mathbb{E}\Bigl(\Bigl(\frac{D_n^2}{n}\Bigr)^k\Bigr)\leq\frac{(2k)!}{2^k k!} \mbox{ $\forall k\in\mathbb{N}$} $$ although I really think, since I have evidence of the convergence in distribution of $\frac{D_n^2}{n}$ to an exponential distribution, that: $$ \lim_{n\to\infty}\mathbb{E}\Bigl(\Bigl(\frac{D_n^2}{n}\Bigr)^k\Bigr) = k! $$ which is indeed the $k$-th moment of an exponential of parameter 1.

I was also trying to understand if $\frac{D_n^2}{n}$ was sub-exponential or sub-gaussian $\forall n\in\mathbb{N}$, and I think the answer is yes, although again I don't know how to use this information to prove this convergence. I have also tried to use the Portmanteau theorem, without success (probably I didn't find the right trick to prove it).

To simulate this process using the software $R$, here is a code I made for computing $\frac{D_n^2}{n}$:

it = 10000
n = 1000
thetamat = matrix(rep(0,n*it),n,it)
  D1 = rep(0,it)
  for(i in 1:n) {
    thetamat[i,] = runif(it,0,2*pi)
  for(k in 1:it) {
    for(i in 1:(n-1)) {
      for(j in (i+1):n) {
        D1[k] = D1[k] + cos(thetamat[i,k]-thetamat[j,k])
  D = 1 + 2/n*D1

Any help will be so much appreciated. Thank you in advance.


1 Answer 1


Let $Y_n=n^{-1/2}\sum_{i=1}^n \cos(\theta_i)$ and $Y'_n=n^{-1/2}\sum_{i=1}^n \sin(\theta_i)$. For each constants $a$ and $b$, the sequence $(aY_n+bY'_n)$ converges to a Gaussian random variable having variance $a^2\mathbb E[\cos^2(\theta_1)]+b^2\mathbb E[\sin^2(\theta_1)]$. We derive from this that $(Y_n,Y'_n)$ converges to a Gaussian vector, say $(U,V)$, whose covariance matrix is $I_2/2$. The continuous mapping theorem guarantees that if $(Y_n,Y'_n)$ converges in distribution to $(U,V)$, then $Y_n^2+(Y'_n)^2$ converges in distribution to $U^2+V^2$.

  • $\begingroup$ Thank you for your reply! But.. from this, how can we say that $Y_n^2+Y_n^2'$ converges to an exponential distribution? For all n, these variables are not indipendent, but (I think) asymptotically indipendent. $\endgroup$ Commented Nov 19, 2023 at 21:10
  • 1
    $\begingroup$ Consider $(\cos \theta, \sin \theta)$ as a random vector . Its covariance matrix is $I_2/2.$ Apply the central limit theorem to it. $\endgroup$ Commented Nov 20, 2023 at 7:01
  • $\begingroup$ @LucaOnnis Yes you are right. $\endgroup$ Commented Nov 20, 2023 at 10:21
  • $\begingroup$ Thank you Davide. $\endgroup$ Commented Nov 20, 2023 at 12:47

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