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

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

• Funny how it was not your "idea" to consider covariances at all before I posted my answer and someone downvoted it after it already had been upvoted. Essentially the last 4 lines of your question were non-existent before I posted my answer an hour ago. Commented Jun 9, 2023 at 15:11
• Not saying you had anything to do with that btw. Commented Jun 9, 2023 at 15:13
• The downvote I don´t know, but I just changed the idea thing, because it is included in your answer. Commented Jun 9, 2023 at 15:14
• But I tried your idea and the longer version above this and it worked out. Thank you! Commented Jun 9, 2023 at 15:20
• There was nothing left to "try" anyway. I provided a complete answer which explicitly gives you the variance and shows that it is of the order of $n$. Which directly gives you that $Var(Z_{n})$ is of the order of $\frac{1}{n}$ which goes to $0$ and this allows you to conclude convergence in Probability using Chebycheff's Inequality. Commented Jun 9, 2023 at 20:26

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 $$Var(Z_{n})= \frac{O(n)}{n^{2}}\to 0$$