What is the probability limit and limit distribution of the estimators given that$ X_i$ are iid

Question

This is more of a practice question but I'm not sure how to really proceed.

Say that $E(X)=0$ and $Var(X)=\sigma^2$.

Firstly I am required to find the probability limit of the estimator as $n$ go to infinity.

I tried to break it up into

I know the average of $X_i$ converges in probability to $\mu$ according to Khinchine but I'm not sure what to do the the average of $X_{i+1}$.

Secondly the limiting distribution

I was thinking of squaring the term so I'll get $1/n$ but then I'm not sure how to proceed with a squared summation.

Answer 1

For the first part, note that $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{i=1}^n X_{i+1}=\lim_{n\to\infty}\frac{n-1}{n}\frac{1}{n-1}\sum_{i=1}^{n-1} Y_i$$ where $Y_i=X_{i+1},\ i\ge 1$. So $Y_i$ are i.i.d and hence again by WLLN (Khinchin's law) it will converge to $\mu$ in probability.

Now, for the second part, use Central Limit Theorem to get $$\frac{1}{\sqrt{n}}\sum_{X_i}X_i\overset{d}{\to} X\sim \mathcal{N}(0,\sigma^2)$$ Same thing goes for $\frac{1}{\sqrt{n}}\sum_{i=1}^n X_{i+1}$