TITLE: Convergence of $\frac1{n-1} \sum_{k=1}^n (X_k- {\frac 1 n \sum_{k=1}^n X_k})^2$ to $\operatorname{Var}(X_i)$ with $X_i$ i.i.d
QUESTION [0 upvotes]: I'm trying to understand what is needed in the following problem and how to solve it [Ex. 22 of this]:

Let $(X_n)_{n\ge 1}$ be a sequence of i.i.d random variables with finite $E[X_1^2]$.
Show that $Z_n={\frac 1 {n-1}} \sum\limits_{k=1}^{n} \left(X_k- {{\frac 1 n} {\sum\limits_{k=1}^{n} X_k}}\right)^2$ for $n\ge 2$ converges a.s. to $\operatorname{Var}(X_i)$.

First of all, shall $E$ converge or $V$ converge (?) I assume that if it is "converging" then it's $E$ whom shall converge.
Now, how to solve it: Where is my mistake?
$E[{(X_k- {{\frac 1 n} {\sum X_t}})}^2]$=$E[X_1^2]-2E[X_1^2]+E[({{\frac 1 n} {\sum X_t}})^2]$
(since they are independent and identically disterbuted)
Now from the same reason:
$E[({{\frac 1 n} {\sum X_t}})^2]=E[X_1(X_1+...+X_n)+X_2(X_1+X_2...+X_n)...+X_n(X1+...+X_n)]$ = $\sum{E[{X_1}^2]}$
Which leads to total sum of
$...$
$E[{X_i}^2]$.
Can you help:/
Why doesn't it converge to $V(X_1)$?

REPLY [1 votes]: To be a little more precise, we are applying the following form of the strong law of large numbers:

Let $Y_1,Y_2,\ldots$ be iid random variables satisfying $\mathbb E|Y_k|<\infty$. Then $n^{-1}\Sigma_{k\leq n}Y_k$ converges a.s. to the mean.

Note that some textbooks prove a version of this that require more assumptions, like $\mathbb E|Y_k|^2<\infty$, which we will not be able to assume here - so it is important to use this stronger version. To give you a concrete reference, I am using Theorem 3.23 in Kallenberg's probability textbook.
We are going to apply the result with $Y_k=X_k^2$, which is valid due to your assumption that $\mathbb EX_k^2<\infty$ (this is why we can only assume $\mathbb E|Y_k|<\infty$, because $\mathbb EY_k^2=\mathbb EX_k^4$ might be infinite given your assumptions). Thus, we obtain that
$$
\frac{1}{n}\sum_{k=1}^nY_k\overset{a.s.}\to\mathbb EY_k,
$$
or in other words
$$
\frac{1}{n}\sum_{k=1}^nX_k^2\overset{a.s.}\to\mathbb EX_1^2.\qquad(\star)
$$
In addition, we will also use the fact (follows from Jensen's inequality) that $(\mathbb E|X_k|)^2\leq \mathbb EX_k^2$ to conclude that $\mathbb E|X_k|<\infty$ as well, and therefore
$$
\frac{1}{n}\sum_{k=1}^nX_k\overset{a.s.}\to\mathbb EX_1,
$$
which implies that
$$
\left(\frac{1}{n}\sum_{k=1}^nX_k\right)^2\overset{a.s.}\to(\mathbb EX_1)^2,\qquad (\star\star)
$$
Okay, now the analysis is out of the way and all that is left is the algebra. Specifically, we will expand out your formula for $Z_n$ to write it as a linear combination of $(\star)$ and $(\star\star)$, and then get the limit you desire.
{"set_name": "stack_exchange", "score": 0, "question_id": 3530456}