Prove that $M_{n , k} = \frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X_{n}})^{k} \overset{p}\longrightarrow \mu_{k}$ I want to prove that $M_{n , k} = \frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X_{n}})^{k} \overset{p}\longrightarrow \mu_{k}$
Where $X_{1} , X_{2} , ... , X_{n}$ are i.i.d with expected value $E(X_{i}) = \mu$
Also let the k-th central moment be
$$\mu_k = E(X - \mu)^{k}$$
A natural estimator is the k-th sample moment,given by
$$M_{n , k} = \frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X_{n}})^{k}$$
I know that for $k = 2$ due to the law of large numbers and the fact that $g(x) = x^{2}$ is a continuous function. That
$$M_{n , 2} = \frac{1}{n} \sum_{i = 1}^{n}(X_{i} - \bar{X_{n}})^{2} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}^{2} - \bar{X_{n}}^{2} \overset{p}\longrightarrow \mu_{2}$$
But how do I prove this for any k since.
$$M_{n , K} = \frac{1}{n} \sum_{i = 1}^{n}(X_{i} - \bar{X_{n}})^{K} \ne \frac{1}{n} \sum_{i = 1}^{n} X_{i}^{K} - X_{n}^{k}$$
 A: Without loss of generality, we can take $E(X_i)=\mu=0$ (why?).
Simply expand the sum using binomial theorem:
$$\frac1n\sum_{i=1}^n (X_{i} - \overline X_n)^k =\frac1n\sum_{i=1}^n X_i^k-\binom{k}{1}\overline X_n\cdot\frac1n\sum_{i=1}^n X_i^{k-1}+\binom{k}{2}\overline X_n^2\cdot \frac1n\sum_{i=1}^n X_i^{k-2}+\cdots+(-1)^{k}\overline X_n^{k} $$
By law of large numbers, $\overline X_n\stackrel{P}\to 0$ and more generally, $\frac1n\sum\limits_{i=1}^n X_i^j\stackrel{P}\to  E(X_i^j)$ for $j=1,\ldots,k$.
Now use the following theorem:

If $(U_n)$ and $(V_n)$ are two sequences of random variables such that $U_n\stackrel{P}\to u$ and $V_n\stackrel{P}\to v$, then $U_n+V_n\stackrel{P}\to u+v$ and $U_nV_n\stackrel{P}\to uv$.

So the first term in $\frac1n\sum\limits_{i=1}^n (X_{i} - \overline X_n)^k$ converges in probability to $\mu_k$, the remaining terms converge in probability to $0$.
A: Expanding $\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X_{n}})^{k}$ using the binomial theorem.
$ \frac{1}{n} \sum_{i = 1}^{n} \left(X_{i} - \bar{X_{n}}^{k}\right) = \frac{1}{n}\left(-\sum^{n}_{i = 1}\bar{X_{i}}^{k} + \binom{k}{1}  \bar{X_{n}}\sum^{n}_{i = 1}\bar{X_{i}^{k - 1}} - \binom{k}{2}  \bar{X_{n}}^{2}\sum^{n}_{i = 1}X_{i}^{k - 2} \dots + (-1)^{k -1 }\binom{k}{k-1}  \bar{X_{n}}^{k-1}\sum^{n}_{i = 1}X_{i} + (-1)^{k} n\bar{X}^{k}\right) $
We can now say that $m_{n , k}$ converges in probability to the expression below which is equal to $\mu_{k}$. This is because of the law of large numbers and the fact that if $Y_{n}$ converges in probability to $c$ and $g$ is a function which is continuous at $c$, then $g(Y_{n}) \overset{p}\longrightarrow g(c)$. This is true for all the functions that encompasses the $X_{n}$.
Therefore $\frac{1}{n}\sum_{i = 1}^{n} X_{i}^{k} \overset{p}\longrightarrow E(X^{k})$ and $(\bar{X_{n}})^{k} \overset{p} \longrightarrow{\mu^{k}}$ etc...
We can now conclude that:
$\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X_{n}})^{k} \overset{p} \longrightarrow  -E(X^{k}) + \binom{k}{1} E(X^{k-1})\mu - \binom{k}{2} E(X^{k-1})\mu^{2} ....\: + (-1)^{k-1}\binom{k}{k-1} E(X^{k-1})\mu^{k-1} + (-1)^{k} \mu^{k} = \mu_{k}$
