I am trying to show that the sample variance converges to the population variance in using the Weak Law of Large Numbers

$$\begin{align} \\ \Rightarrow S_n= \frac{1}{n} \sum_{i=1}^n (X_i-\bar{X})^2 &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 + 2 (\mu-\bar{X}) \underbrace{\frac{1}{n} \cdot \sum_{i=1}^n (X_i-\mu)}_{\left(\frac{1}{n} \sum_{i=1}^n X_i\right)-\mu =(\bar{X}-\mu)} + (\mu-\bar{X})^2 \\ &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 - (\bar{X}-\mu)^2 \end{align}$$

By the Weak Law of large numbers we obtain

$$\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \to \mu \quad \text{WLN} $$

$$(\bar{X}-\mu)^2 \to 0 \quad\text{ Slutsky}$$

How do I show this term converges using only Weak Law $$ \\ \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 \to \mathbb{E} \left( (X_i-\mu)^2 \right)=\sigma^2 \quad \text{ How?} $$

  • $\begingroup$ I don't understand. Why doesn't weak law prove precisely what you want? $\endgroup$ – BCLC Dec 8 '15 at 0:50
  • $\begingroup$ Because I am not sure how to take the expression and use chebychev $\endgroup$ – jessica Dec 8 '15 at 1:16
  • $\begingroup$ To use Chebychev's inequality to complete the proof, you will need $E((X-\mu)^2) = \sigma^2$, and $Var((X-\mu)^2)$, 4th (and probably also the 3rd) moment of $X$ are required to exist. $\endgroup$ – kitman0804 Dec 8 '15 at 8:54

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