Question regarding the frequency a sequence takes specific values

Question

Consider a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n \in \{1,2,\dots,L\}$ for all $n$ $\in \mathbb{N}$. Assume that the following assumption holds for the sequence: \begin{equation} \frac{\sum_{k=1}^{n} \mathbb{1}_{x_k = i}}{n} \rightarrow b_i \end{equation} for all $i \in \{1,\dots,L\}$ where $b_i \in [0,1]$. Does it also hold that for any such $i$ and $m \geq 2$ \begin{equation} \frac{\sum_{k=(m-1)n+1}^{mn} \mathbb{1}_{x_k = i}}{n} \rightarrow b_i \end{equation} as $n \rightarrow \infty$. Any help would be greatly appreciated. Thanks!

Answer 1

Yes, that's true. To simplify notaion let $occ(k, n, i)$ be the number of $i$'s in $x_k\ldots x_n$ (here $i\in\{1, 2, \ldots L\}$). By assumption we know that $occ(1, n, i)/n\to b_i$ as $n \to \infty$ for any $i\in\{1, 2, \ldots, L\}$. We are asked to prove that for every $m\ge 2$ and $i\in\{1, 2, \ldots, L\}$ the following holds: $$occ((m - 1)n + 1, mn, i)/n \to b_i \mbox{ as } n\to\infty.$$

By definition we have that $$occ((m - 1)n + 1, mn, i) = occ(1, mn, i) - occ(1, (m - 1)n, i).$$

Using that we can write:

\begin{align*} \frac{occ((m - 1)n + 1, mn, i)}{n} &= \frac{occ(1, mn, i) - occ(1, (m - 1)n, i)}{n} \\ &= \frac{occ(1, mn, i)}{mn}\cdot m - \frac{occ(1, (m - 1)n, i)}{(m - 1) n} \cdot (m - 1). \end{align*}

This expression tends to $b_i \cdot m - b_i \cdot (m - 1) = b_i$ as $n\to\infty$.