Modifications of Weak Law of Large Numbers Let $X_1,...,X_n$ be random variables with finite variances $\text{Var}[X_i]=\sigma_i^2$ and expectations $\mathbb{E}[X_i]=\mu_i$ for all $i=1,...,n$.
The weak law of large numbers that I know states, that for
$\overline {X}_{n}={\tfrac  1n}\textstyle \sum \limits _{{i=1}}^{{n}}(X_{i}-E({X}_{i}))$
we have 
$\lim _{{n\rightarrow \infty }}\operatorname {P}\left(\left|\overline {X}_{n}\right|>\varepsilon \right)=0$
If we have iid. random variables with variance $\sigma^2$ and expectation $\mu$ we get as a consequence
${\displaystyle \overline {X}_{n}\ {\xrightarrow {p}}\ \mu}$ 
Is it true that for iid. random variables it holds that
${\tfrac  1n}\textstyle \sum \limits _{{i=1}}^{{n}}(X_{i}-\mu)^2{\xrightarrow {p}}\ \sigma^2$ ?
How can it be derived from the version stated in the beginning?
Is there an analogue statement if the random variables are not iid?
 A: The statement that you cite as weak law of large numbers is wrong. Without additional assumptions it is not true that $\lim_{n\to\infty}\operatorname {P}\bigl(\bigl|\overline {X}_{n}\bigr|>\varepsilon \bigr)=0$ for any random variables $X_1,\ldots,X_n$ with finite variances $\text{Var}[X_i]=\sigma_i^2$ and expectations $\mathbb{E}[X_i]=\mu_i$ for all $i=1,...,n$.
Say, you can take $X_1=\ldots=X_n=X$ with $\mathbb E[X]=\mu$, $\operatorname{Var}[X]=\sigma^2>0$ and get $\overline X_n=X-\mu$. This value does not depend on $n$ and does not converge in probability to $0$.
When r.v.'s are i.i.d., the statement is true. There can exists a number of other sufficient conditions for convergence ${\frac  1n}\textstyle \sum \limits _{{i=1}}^{{n}}(X_{i}-E({X}_{i}))\xrightarrow{p} 0$. Say, if $X_i$, $i=1,2,\ldots$ are pairwise independent and $\frac{\sigma_1^2+\ldots+\sigma_n^2}{n^2}\to 0$, it is valid. This is a consequence of Chebyshev inequality. 
For i.i.d. case, it is sufficient that the first moment of $X_1$ exists. So for $${\tfrac  1n}\textstyle \sum \limits _{{i=1}}^{{n}}(X_{i}-\mu)^2{\xrightarrow {p}}\ \sigma^2$$
in i.i.d. case it is sufficient that $\mathbb E(X_1-\mu)^2=\sigma^2<\infty$.
