$\frac{1}{n} \sum_{k=1}^{n-s} X_{k+s}X_{k}$ the same as $\frac{1}{n} \sum_{k=1}^{n} X_{k+s}X_{k}$ for $n \rightarrow \infty$?

Question

I need to show that $$ \frac{1}{n} \sum_{k=1}^{n-s} X_{k+s}X_{k}$$ for some number $s$ is essentially the same (asymptotically negligible) as $$ \frac{1}{n} \sum_{k=1}^{n} X_{k+s}X_{k}$$ as $n \rightarrow \infty$, for a zero-mean time series $(X_t)$.

First, I know that $\frac{n-s}{n}$ goes to $1$ very rapidly (sublinearly, even logarithmically ). If I write them as a fraction then $$ \frac{ \sum_{k=1}^{n-s} X_{k+s}X_{k}}{\sum_{k=1}^{n} X_{k+s}X_{k}}=\frac{ \sum_{k=1}^{n-s} X_{k+s}X_{k}}{\sum_{k=1}^{n-s} X_{k+s}X_{k} + \sum_{k=n+1-s}^{n} X_{k+s}X_{k}}$$ Then $\sum_{k=n+1-s}^{n} X_{k+s}X_{k}$ goes to zero az $n \rightarrow \infty$.

Is this enough of an argument? It seems to me that is very simple, but I can say the same for the question? Any ideas?

Thank you!

EDIT: $(X_t)$ are (wlog) with zero mean and $\mathbb{E} [X_{t+s}X_t]$ exists and it is finite for all $s$, and does not depend on $t$.

Answer 1

Fix $s \in \mathbb{N}$, and set

$$Y_n := \frac{1}{n} \sum_{k=1}^{n-s} X_{k+s} X_k \qquad \qquad Z_n := \frac{1}{n} \sum_{k=1}^n X_{k+s} X_k.$$

Obviously, we have

$$Y_n = \frac{n-s}{n} Z_{n-s}. \tag{1}$$

We want to show that $$Y_n \stackrel{d}{\to} Y \quad \Leftrightarrow \quad Z_n \stackrel{d}{\to} Y \tag{2}$$ where $d$ denotes convergence in distribution and $Y$ is an abritrary random variable. To see this we note the following two facts.

1. $Z_{n-s} \stackrel{d}{\to} Y \Leftrightarrow Z_n \stackrel{d}{\to} Y$; this follows straight from the definition of convergence in distribution (recall that $s$ is fixed).
2. If $X_n \stackrel{d}{\to} \mu$ and $(a_n)_n$ is a sequence of real numbers such that $a_n \to a \in \mathbb{R}$, then $a_n \cdot X_n \stackrel{d}{\to} a \cdot X$.

Combining these two, we get $(2)$ from $(1)$.