Showing that $ S_{n}^{2}$ converges to $ \sigma ^{2} $ in probability Let $x_{1},...,x_{n} \sim F$ where the expected value of $F$ is $\mu$ and the variance is $\sigma^{2}$.
$ S_{n}^{2}=\frac{1}{n-1}\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2} $
converges in probability to the variance. What I have tried so far is to use Chebyshev's in equality - since I know that $S_{n}$ is unbiased, given an $\epsilon >0$  we get:
$P\left(\left|S_{n}^{2}-\sigma^{2}\right|\geq\epsilon\right)\leq\frac{Var\left(S_{n}^{2}\right)}{\epsilon^{2}}$
I thought somehow with manipulation I can bound it with something that can converge to zero. However finding the variance of $S_{n}^{2}$ got really complicated, and I think that there should be an easier way. Should I proceed with finding the variance of $S_{n}^{2}$ ? Can somebody give me a clue? I would really appreciate it. Thanks!
 A: $$S_{n}^{2}=\frac{1}{n-1}\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}=\dfrac{n}{n-1}\left({\overline{x^2}-\left(\overline x\right)^2}\right),$$
where $\overline{x^2} = \dfrac{\sum_{i=1}^n x_i^2}{n}$.
Law of Large Numbers implies that
$$
\overline{x^2} = \dfrac{\sum_{i=1}^n x_i^2}{n} \xrightarrow{p} \mathbb Ex_1^2=\sigma^2+\mu^2 \text{ as } n\to\infty
$$
and
$$
\overline{x} = \dfrac{\sum_{i=1}^n x_i}{n} \xrightarrow{p} \mathbb Ex_1=\mu \text{ as } n\to\infty.
$$
And $\tfrac{n}{n-1}\to 1$ as $n\to\infty$. By the properties of convergence in probability, 
$$
S_{n}^{2}= \underbrace{\dfrac{n}{n-1}}_{\begin{matrix}\downarrow\\ 1\end{matrix}}\biggl({\underbrace{\overline{x^2}}_{\begin{matrix}\downarrow_p \\ \sigma^2+\mu^2\end{matrix}}-\underbrace{\left(\overline x\right)^2}_{\begin{matrix}\downarrow_p \\ \mu^2\end{matrix}}}\biggr) \xrightarrow{p} 1\cdot (\sigma^2+\mu^2 - \mu^2)=\sigma^2.
$$
You cannot use Chebyshev inequality without any knowledge on higher moments of samples. The variance $Var(S_n^2)$ exists if 4th moment exists: $$\mathbb Ex_1^4<\infty.$$
And we are given only that first two moments exist, and this is sufficient for convergence in probability of $S_n^2$.
