Show that $\frac{1}{n(n-1)} \sum_{i\neq j} \sin\left(X_i X_j\right)$ converges almost surely to a constant

Question

Let $X_i$ be iid random variables. How does one show that

$$ \frac{1}{n(n-1)}\sum_{i\neq j}^n \sin\left(X_i X_j\right) $$ converges almost surely to a constant?

Answer 1

I think it is worth elaborating @StubbornAtom's sharp comment as applying the U-statistics theory is perhaps the most direct way to prove this proposition. To this end, identify the random variable of interest as the U-statistic \begin{align} U_n := U(X_1, \ldots, X_n) := \frac{1}{\binom{n}{2}}\sum_{1 \leq i < j \leq n}h(X_i, X_j), \end{align} where $h(x, y) = \sin(xy)$ is the "kernel" function of this U-statistic. Now the result immediately follows by the following theorem (cf. Approximation Theorems of Mathematical Statistics by R.Serfling, Theorem 5.4.A) because $|h| \leq 1$:

If $E[|h|] < \infty$, then $U_n$ converges to $\theta = E[h(X_1, X_2)]$ almost surely.

For more details (e.g., which references provided the proof to this theorem), refer to the same reference, Section 5.4. At high-level, the key of the proof is the Hoeffding representation of the quantity $U_n - \theta$.

Answer 2

In the case of the weak LLN

We can prove that the variance of the sum decreases for increasing $n$ (then applying Chebyshev's inequality).

For simplicity of writing the equations, I will define $$Z_{\mathscr{ij}} = \sin(X_\mathscr{i}X_\mathscr{j})$$ and $$S_n = \frac{1}{n(n-1)} \sum_{\mathscr{i}\neq \mathscr{j}} Z_{\mathscr{ij}}$$

The variance of the sum can be computed as

$$\operatorname{Var}(S_n) = \left(\frac{1}{n(n-1)} \right)^2 \sum_{\mathscr{i}\neq \mathscr{j}} \sum_{\mathscr{k}\neq\mathscr{l}} \operatorname{Cov}(Z_{\mathscr{ij}} Z_{\mathscr{kl}}).$$

We have three situations of the covariance terms

$$\operatorname{Cov}(Z_{\mathscr{ij}} Z_{\mathscr{kl}}) = \begin{cases} V & \quad \text{if $\mathscr{ij} = \mathscr{kl}$ or if $\mathscr{ij} = \mathscr{lk}$} \\ C & \quad \text{if ($\mathscr{i} = \mathscr{k}$ and $\mathscr{j}\neq \mathscr{l}$) or ($\mathscr{i} = \mathscr{l}$ and $\mathscr{j} \neq \mathscr{k}$)} \\ 0 & \quad \text{if $\mathscr{i} \neq \mathscr{j} \neq \mathscr{k} \neq \mathscr{l}$} \\ \end{cases}$$

For each $\mathscr{ij}$ pair, there are 2 cases of $V$ and $2 (n-2)$ cases of $C.$

Then the variance can be computed as

$$\begin{align} \operatorname{Var}(S_n) &= \left(\frac{1}{n(n-1)} \right)^2 \sum_{\mathscr{i}\neq \mathscr{j}} (2V+ 2(n-2)C) \\[6pt] &= \left(\frac{1}{n(n-1)} \right)^2 (n(n-1)) (2V+ 2(n-2) C) \\[6pt] &= 2 \frac{V+(n-2)C}{n(n-1)} \end{align} $$ and this approaches zero for increasing $n$.

Answer 3

Acting hastily, I first thought that the boundeness of $Y_{ij}\mathrel{:=\sin(X_iX_j)}$ and $\mathop{\mathbb E}\left[Y_{ij}^2\right] \leq \mathbb E[1] = 1$ would be enough to solve the problem with a version of the strong law of large numbers for pairwise uncorrelated random variables.
However, clearly, in general we don't have pairwise uncorrelatedness of $Y_{ij}$ and $Y_{ik}$ for $j \neq k$ if $X_i \overset{\mathrm{i.i.d.}}{\sim} \mathbb P$ for an arbitrary distribution $\mathbb P$ on $\mathcal B(\mathbb R).$

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We can exploit the following result (e.g., Brown, 1992, Proposition 1) to show almost sure convergence:

Suppose that $\left(X_k\right)_{k \in \mathbb N_{\geq 1}}$ is a sequence of complex-valued random variables of bounded modulus and that $$ \sum_{n=1}^\infty\frac{1}{n}\mathop{\mathbb E}\left[\left|\frac{1}{n} \sum_{k=1}^nX_k\right|^2\right] < \infty. $$ Then $\frac{1}{n} \sum_{k=1}^nX_k \overset{n \to \infty}{\longrightarrow} 0$ almost surely.

This result is a slight generalization of Theorem 1 in Lyons (1988), in which $|X_k|\leq 1$ almost surely $\forall \,k \in \mathbb N_{\geq 1}$ is assumed, and can be proved analogously.

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Let $\mu \mathrel{:=} \mathbb E[Y_{ij}].$
Applying the above result to the sequence $\left(Y_{ij}-\mu\right)_{i \neq j}$ with $$ \mathop{\mathbb E}\left[\left|\frac{1}{n\left(n-1\right)} \sum_{i \neq j}^n \left(Y_{ij}-\mu\right)\right|^2\right] = \mathop{\mathbb V}\left[\frac{1}{n\left(n-1\right)} \sum_{i \neq j}^n Y_{ij}\right] $$ and $\mathop{\mathbb V}\left[\frac{1}{n\left(n-1\right)} \sum_{i \neq j}^n Y_{ij}\right] = 2 \frac{V+(n-2)C}{n(n-1)} \leq c/n$ (for all $n \in \mathbb N_{\geq 2}$ and a constant $c \in \mathbb R_{>0}$) from Sextus Empiricus' answer yields $$ \frac{1}{n\left(n-1\right)} \sum_{i \neq j}^n \left(Y_{ij}-\mu\right)\underset{n \to \infty}{\overset{\mathrm{a.s.}}{\longrightarrow}} 0, $$ and hence $$ \frac{1}{n\left(n-1\right)} \sum_{i \neq j}^n Y_{ij} \underset{n \to \infty}{\overset{\mathrm{a.s.}}{\longrightarrow}} \mu. $$

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References

• Brown, G. (1992). Some new applications of Riesz products. Miniconference on Probability and Analysis, 29, 1–13.
• Lyons, R. (1988). Strong laws of large numbers for weakly correlated random variables. Michigan Mathematical Journal, 35(3), 353–359.