We have a stationary sequence of random variables $X_{j}:j\geq 0$ and let $D$ be a Borel subset of $\mathbb{R}^{d}$. For each n, let $Y_{n}$ be the number of indices $i \in \{0,1, \ldots, n-d\}$ such that $(X_{i+1}, X_{i+2}, \ldots X_{i+d}) \in D$. Show that $\frac{Y_{n}}{n}$ converges $as$.
My attempt:
Attempt 1: $Y_{n}=i$ has equal probability to be anything from the set ${0,1, \ldots, n-d}$ so we can say that $Y_{n}$ follows a uniform distribution with $P(Y_{n}=i) = \frac{1}{n-d+1}$.
Now, I know that I can if I can show that $Y_{n}$ is a subadditive process, stationary, and integrable, then I can use the sub-additive ergodic theorem to prove that it converges $as$. I don't know to prove that $Y_{n}$ is integrable because nothing has been given about the random variables $X_{j}$ except that they form a stationary sequence.
Attempt 2:
Since nothing's given about the integrability, if I can prove that the transformation is ergodic then I can use the Birkhoff ergodic theorem to show that $Y_{n}/n$ converges $as$.
For a simple example, if I fix d=1, then for different values of n, the different values of $Y_{n}$ correspond to shifted set of random variables belong to $A$. For $n=1$, $i = {0,1}$ which corresponds to $(X_{1}) \in D$ or $(X_{2}) \in D$ but both of them have the same distribution because $X_{j}$ is a stationary sequence.
But I cannot go further. Any help would be great.
The Ergodic theorem gives that if $W_n$ is stationary and integrable, then $\sum_{i=1}^n W_i/n$ converges a.s. (Note that the random variable it converges to will only be a constant when $W$ is ergodic. )
Let $Z_i$ be an indicator function on whether $(X_{i+1},...,X_{i+d})\in D$. $Z_i$ is measurable (Since the X are measurable and $D$ is Borel) and bounded (it’s 0 or 1) so it’s integrable. Since $X_i$ are stationary, for any nonnegative integer $m$, $X_i$ through $X_{i+m+d}$ have the same distribution over all $i$. Therefore, $Z_i$ through $Z_{i+m}$ have the same distribution over all $i$, so $Z$ is stationary. Applying the Ergodic theorem to $Z$ gives your result.