# Convergence in distribution of a trig sequence of random variables

Let $$\theta_i\sim\mathcal{U}(0,2\pi)$$, $$i = 1,\dots,n$$ be $$n$$ i.i.d. uniformly distributed random variables. Let: $$\frac{D_n^2}{n} = 1 + \frac{2}{n}\sum_{i I have the evidence (by Monte Carlo simulation) that $$\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.

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 $$U^2+V^2$$.
• 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. Commented Nov 19, 2023 at 21:10
• Consider $(\cos \theta, \sin \theta)$ as a random vector . Its covariance matrix is $I_2/2.$ Apply the central limit theorem to it. Commented Nov 20, 2023 at 7:01