Show that $ \frac{Y_{1}+\cdots+Y_{n}}{n} \stackrel{\mathrm{P}}{\longrightarrow} \mu \text { as } n \rightarrow \infty $

Question

Let $Y_{1}, Y_{2}, \ldots$ be a sequence of identically distributed random variables with $\mathbb{E} Y_{1}=\mu$ and $\operatorname{Var} Y_{1}=\sigma^{2}<\infty$. Furthermore, let $Y_{j}$ be independent of $Y_{j-k}$ and $Y_{j+k}$ for all $k \geq 2$ and for all $j \in \mathbb{N}$. Using the Chebyshev's inequality, we will show that

$$ \frac{Y_{1}+\cdots+Y_{n}}{n} \stackrel{\mathrm{P}}{\longrightarrow} \mu \text { as } n \rightarrow \infty $$

Attempt/Idea:

Chebyshev's inequality states that for any random variable $Z$ with finite variance, and for any $a > 0$:

$$ \Pr(|Z-\mathbb{E}Z| \geq a) \leq \frac{{\operatorname{Var}Z}}{{a^{2}}} $$

Let $Z_n = \frac{Y_{1}+\cdots+Y_{n}}{n}$ be the sample mean of the first $n$ random variables. We want to show that $\Pr(|Z_n - \mu| \geq \epsilon) \rightarrow 0$ as $n \rightarrow \infty$, for any $\epsilon > 0$.

We know that $\operatorname{Var}(Z_n) = \frac{\operatorname{Var}(Y_1)}{n}$ since the variables $Y_{1}, Y_{2}, \ldots$ are independent and identically distributed. (*)

Now, applying the Chebyshev's inequality with $Z = Z_n$, $\mathbb{E}Z = \mu$, and $a = \epsilon$, we have:

$$ \Pr(|Z_n - \mu| \geq \epsilon) \leq \frac{{\operatorname{Var}(Z_n)}}{{\epsilon^{2}}} = \frac{{\operatorname{Var}(Y_1)}}{{n \epsilon^{2}}} = \frac{{\sigma^{2}}}{{n \epsilon^{2}}} $$

Since $\sigma^{2}$ is a constant (finite variance) and $n \epsilon^{2}$ grows unboundedly as $n$ approaches infinity, we can conclude that $\Pr(|Z_n - \mu| \geq \epsilon) \rightarrow 0$ as $n \rightarrow \infty$.

Therefore, we have shown that $\frac{Y_{1}+\cdots+Y_{n}}{n} \stackrel{\mathrm{P}}{\longrightarrow} \mu$ as $n \rightarrow \infty$ using the Chebyshev's inequality.

I'm stuck at (*) and don't know if Y_n is actually completely independent (see prerequisite) or not. Can someone help me with this? Otherwise, it would be quite easy to solve as it is already given here.

The idea would be to use the following form and check all covariances: For the variance of any sum of random variables$ X=a_{1} X_{1}+\cdots+a_{n} X_{n} $ holds in general: $$ \begin{aligned} \operatorname{Var}(X) & =a_{1}^{2} \operatorname{Var}\left(X_{1}\right)+\cdots+a_{n}^{2} \operatorname{Var}\left(X_{n}\right)+2 a_{1} a_{2} \operatorname{Cov}\left(X_{1}, X_{2}\right)+2 a_{1} a_{3} \operatorname{Cov}\left(X_{1}, X_{3}\right)+\cdots \\ & =\sum \limits_{i=1}^{n} a_{i}^{2} \operatorname{Var}\left(X_{i}\right)+2 \sum \limits_{i<j} a_{i} a_{j} \operatorname{Cov}\left(X_{i}, X_{j}\right) \end{aligned} $$

Answer 1

You have $Cov(Y_{i}Y_{j})=Cov(Y_{1}Y_{2})\bigg(\delta_{j,i+1}+\delta_{j,i-1}\bigg)$ where $\delta_{i,j}$ is the Kronecker Delta.

Now, if $S_{n}=\sum_{k=1}^{n}Y_{k}$ , then $$Var(S_{n})=\sum_{k=1}^{n}Var(Y_{k})+\sum_{i\neq j}Cov(Y_{i}Y_{j})=n\cdot Var(Y_{1})+(2n-2)Cov(Y_{1}Y_{2})$$

Now $Cov(Y_{1}Y_{2})=E(Y_{1}Y_{2})-E(Y_{1})E(Y_{2})$

Now $E(|Y_{1}Y_{2}|)\leq \bigg(E(Y_{1}^{2})\bigg)^{\frac{1}{2}}\bigg(E(Y_{1}^{2})\bigg)^{\frac{1}{2}}$ due to Cauchy-Schwartz. Which gives that $Cov(Y_{1}Y_{2})$ is finite.

Thus $Var(S_{n})=O(n)$

Which is basically what is required. This directly gives that $Var(Z_{n})= \frac{O(n)}{n^{2}}\to 0$