As suggested in this discussion, I would like to inquire about the following question:
Consider a matrix B of size $n\times n$ defined as:
$$B_{ij}(\pmb{\theta})=(\theta_i-\theta_j)\sin(\theta_i-\theta_j), 1\leq i<j\leq n$$ where other entries are equal to 0. The vector $b(\pmb{\theta})$ is defined as follows:
\begin{equation}
b(\pmb{\theta}) = \left[(\theta_1-\theta_2)\sin(\theta_1-\theta_2), \ldots, (\theta_1-\theta_n)\sin(\theta_1-\theta_n), (\theta_2-\theta_3)\sin(\theta_2-\theta_3), \ldots, (\theta_2-\theta_n)\sin(\theta_2-\theta_n), \ldots, (\theta_{n-1}-\theta_n)\sin(\theta_{n-1}-\theta_n)\right]^T,
\end{equation}
where $\pmb{\theta} = [\theta_1,\ldots,\theta_n]$ satisfies $|\theta_i-\theta_j|\leq \pi/2$ for all $i,j$ and $\theta_i-\theta_j\neq 0$ for all $i,j$.
The ratio of interest is defined as:
$$\frac{\|b\|_1^2}{\|b\|_2^2}=\frac{\Big(\sum_{i<j}(\theta_i-\theta_j)\sin(\theta_i-\theta_j)\Big)^2}{\sum_{i<j}(\theta_i-\theta_j)^2\sin^2(\theta_i-\theta_j)}$$
I am interested in studying the asymptotic behavior of this ratio. In my numerical experiments, I sample $1000$ sets of $\pmb{\theta}$ uniformly from the set ${\theta \mid |\theta_i-\theta_j|\leq \pi/2, \forall i,j}$ and compute the ratio $\frac{|b|_1^2}{|b|_2^2}$ for each set. I then take the average of these ratios. The resulting figure displays the average ratios as blue dots, corresponding to different values of $n$ ranging from $4$ to $500$.
From the figure, it appears that the ratio grows at least as fast as $n^{1.7}$ for the uniformly sampled $\pmb{\theta}$. However, we can also observe that when $\pmb{\theta}=[\pi/2,0,\ldots,0]$, the ratio is equal to $n-1$. This implies that the ratio grows faster than $n^{1.7}$ is not universally true. Given the numerical experiments with uniformly sampled $\pmb{\theta}$ and the specific case where the ratio is $n-1$, my question is:
How does the ratio behave for 'most' points $\pmb{\theta}$ in the set ${\theta \mid |\theta_i-\theta_j|\leq \pi/2, \forall i,j}$? (I am very unsure how to define 'most points'. This statement is the intuition obtained from numerical experiments, it seems that although the ratio can take small value such as $n-1$, most of ratios are actually much larger than that.
I would appreciate any hints and thank you in advance!
Best Answer
$\newcommand\th\theta$Suppose that the $\th_i$'s are i.i.d. random variables each with a non-degenerate distribution. Then \begin{equation*} \frac{\|b\|_1}{\binom n2}=U_n:=\frac1{\binom n2}\sum_{1\le i<j\le n}h(\th_i,\th_j), \end{equation*} \begin{equation*} \frac{\|b\|_2^2}{\binom n2}=W_n:=\frac1{\binom n2}\sum_{1\le i<j\le n}h(\th_i,\th_j)^2, \end{equation*} where $h(\th_i,\th_j):=(\th_i-\th_j)\sin(\th_i-\th_j)$.
So, $U_n$ and $W_n$ are U-statistics. So, \begin{equation*} EU_n=c_1:=Eh(\th_1,\th_2), \quad EV_n=c_2:=Eh(\th_1,\th_2)^2, \end{equation*} and, in view of a formula by Hoeffding (see e.g. Theorem 3.4), $Var\,U_n=O(1/n)$ and $Var\,V_n=O(1/n)$. So, by Chebyshev's inequality, \begin{equation*} \frac{\|b\|_1}{\binom n2}=U_n\to c_1 \end{equation*} and \begin{equation*} \frac{\|b\|_2^2}{\binom n2}=W_n\to c_2 \end{equation*} in probability (as $n\to\infty$).
Thus, for the ratio in question we have \begin{equation*} \frac{\|b\|_1^2}{\|b\|_2^2}=\frac{U_n^2}{V_n}\binom n2\sim\frac{c_1^2}{c_2}\binom n2 \tag{1}\label{1} \sim cn^2 \end{equation*} in probability, where $c:=\frac{c_1^2}{2c_2}$.
Moreover, by (say) Theorem 2.3., asymptotic equivalence \eqref{1} holds almost surely.
Furthermore, by (say) the Cauchy--Schwarz inequality, $U_n^2\le V_n$. So, by dominated convergence, \begin{equation*} E\frac{\|b\|_1^2}{\|b\|_2^2}\sim cn^2. \end{equation*}
E.g., if each of the $\th_i$'s is uniformly distributed in the interval $[0,\pi/2]$, then \begin{equation*} c_1=\frac{4 (4-\pi)}{\pi ^2},\quad c_2=\frac{3}{\pi ^2}+\frac{\pi ^2}{48}-\frac{1}{4}, \end{equation*} and hence $c=0.233\dots$.
Shown below are 8 realizations of the (connected) random graph $\Big\{\dfrac{U_n^2}{2cV_n}\colon n\in\{1,\dots,20\}\Big\}$ (note that, by \eqref{1}, $\dfrac{U_n^2}{2cV_n}=\dfrac{\|b\|_1^2}{2c\|b\|_2^2}\big/\displaystyle{\binom n2} \sim\dfrac{\|b\|_1^2}{\|b\|_2^2}\big/\displaystyle{(cn^2)}\to1$):
We see that, in accordance with asymptotic equivalence \eqref{1} holding almost surely, all the 8 realizations are rather close to $1$ even for $n$ as small as $10$.