# $\lim_{n \to \infty}P\left(\left|{1 \over n}\sum_{i=1}^n\left(X_i-{{X_1+X_2+\cdots + X_n} \over n }\right)^2 - \sigma^2 \right| >\epsilon\right)=0$

Let $(X_n)_n$ be independent and identically distributed random variables with expected value $\mu$, variance $\sigma^2$ and $E(X_n^4)<\infty$. How can I show that

$$\lim_{n \to \infty}P\left(\left|{1 \over n}\sum_{i=1}^n\left(X_i-{{X_1+X_2+\cdot \cdot\cdot + X_n} \over n }\right)^2 - \sigma^2 \right| >\epsilon\right)=0$$ for all $\epsilon > 0$.

Some help is much appreciated.

## 1 Answer

Using the Strong Law of Large Numbers, we can actually say that $$\frac1n \sum_{i = 1}^n \bigg( X_i - \frac{X_1 + \cdots + X_n}{n} \bigg)^2$$ converges almost surely to $\sigma^2$. This is stronger than convergence in probability. To see this, for the $i$-th summand we can expand to get $$X_i^2 - 2 \bar S_n X_i + \bar S_n^2$$ Here $\bar S_n = \frac1n (X_1 + \cdots + X_n)$. For the first term, $$\frac1n \sum_{i = 1}^n X_i^2$$ converges almost surely to $E(X_1^2)$ by the Strong law. For the second term, $$- 2 \frac1n \sum_{i =1}^n X_i \bar S_n = - 2 (\bar S_n)^2$$ converges to $- 2 \mu^2$ as $n \to \infty$, again by the strong law. Similarly, the last term $\frac1n \cdot n \bar S_n^2 = \bar S_n^2$ converges to $\mu^2$. Putting it all together, the almost-sure limit is $$E(X_1^2) - \mu^2 = \sigma^2 \, .$$

I would point out that all we used was the fact that $E(X_1^2) < \infty$ to invoke the Strong Law for $\frac1n \sum_{i = 1}^n X_i^2$.