I am stuck proving that this sequence $$\sigma^2_n:= \frac{1}{n-1} \sum_{i=1}^n (X_i - \frac{X_1+...+X_n}{n})^2$$ is converegent almost surely to $D^2X_i = \sigma^2$.

We assume that $X_1, X_2, X_3, ...$ are independent with identical distribution, $\mathbb{E}X_i =m, \ D^2X_i = \sigma^2 \ \ \ \forall i \in \mathbb{N_+}$.

I'll later write $\mathbb{E}X_1 =m, \ D^2X_1 = \sigma^2$.

What I need to prove is that for a fixed $\omega \in \Omega$, $ \ \forall \varepsilon>0 $ we can find a suitable $N \in \mathbb{N_+}$ such that $\forall n \ge N: \ \ |\sigma^2_n - \sigma^2| < \varepsilon$ .

Now, $\sigma^2 = \mathbb{E}((X_1 - \mathbb{E}X_1)^2)= \mathbb{E}(X_i^2) - (\mathbb{E}X_i)^2$.

We have $\mathbb{E}$ in the limit but no expected value in $\sigma^2_n$. So I thought I could maybe use Kolmogorow inequality:

$$X_1, X_2, ... \ \ \text{are iid}, \ \mathbb{E}X_i \in \mathbb{R}, \ S_k = X_1 + ... + X_k, \ \varepsilon>0 \ $$ then $$P(\max _{1 \le k \le n} |S_k - \mathbb{E}S_k| \ge \varepsilon) \le \frac{D^2(S_n)}{\varepsilon^2}$$

But the problem is that we don't have a sum $X_1 + ... + X_n$ here $X_i - \frac{X_1+...+X_n}{n}$

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    $\begingroup$ Please check your first equation for typos. As it is stated now, the claim is not correct. $\endgroup$ – saz Apr 19 '15 at 13:54
  • $\begingroup$ @saz Thank you. I've just squared the summands in the first equation. $\endgroup$ – Hagrid Apr 19 '15 at 14:13

Set $$S_n := \sum_{j=1}^n X_j.$$ Then $$\begin{align*} \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \frac{X_1+\ldots+X_n}{n} \right)^2 &= \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \frac{S_n}{n} \right)^2 \\ &= \frac{1}{n-1} \sum_{i=1}^n X_i^2 - 2 \frac{1}{n-1} \frac{S_n}{n} \sum_{i=1}^n X_i + \frac{1}{n-1} \frac{S_n^2}{n} \\ &= \frac{1}{n-1} \sum_{i=1}^n X_i^2 - \frac{n}{n-1} \left(\frac{S_n}{n} \right)^2. \end{align*}$$

Use the strong law of large numbers (twice) to show that the first term at the right-hand side converges to $\mathbb{E}(X_1^2)$ and that the second term converges to $(\mathbb{E}(X_1))^2$ as $n \to \infty$.

  • $\begingroup$ Ok. That's exactly what was missing! Thank you! $\endgroup$ – Hagrid Apr 19 '15 at 14:33
  • $\begingroup$ By the strong law of large numbers we have: $$\frac{X^2_1+...+X_n^2}{n} \to \mathbb{E}(X_1^2), $$ isn't that so? Whereas in the equalities we have $ \frac{X^2_1+..+X^2_n}{n-1}$. How can it be fixed? When it comes to the second summand, $\lim_{n \to \infty} \frac{n}{n-1}=1$, so it is clear to me that it converges to $(\mathbb{E}(X_1))^2$. $\endgroup$ – Hagrid Apr 19 '15 at 20:11
  • $\begingroup$ Ok, we can do the same for the first summand and get $$ \frac{n}{n-1} \frac{1}{n} \sum_{i=1}^n X_i^2.$$ Am I correct? $\endgroup$ – Hagrid Apr 19 '15 at 20:17
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    $\begingroup$ @Hagrid Yes, exactly. $\endgroup$ – saz Apr 20 '15 at 5:19

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