If $X_1,\ldots X_n$ are iid random variables with $E(X_{i}) = 0$ and $E(X_{i}^2) = 1$, then $\frac{\sum_{i=1}^{n}X_i^2}{n} \to E(X_i^2)$?

Question

I am trying to see why it is that if $X_1,\ldots X_n$ are i.i.d. random variables with $E(X_{i}) = 0$ and $E(X_{i}^2) = 1$, then by the Law of Large Numbers we have that

$$ \frac{\sum_{i=1}^{n}X_i^2}{n} \to E(X_i^2). $$

Here, also would we have almost sure convergence or just convergence in probability? It seems obvious that we can do this, but then I assumed that the Law of Large Numbers will only hold for the sample average of the $X_i$'s but not necessarily so for $X_i^2$. Could anyone help me shed light on perhaps what I'm missing? Thanks!

Answer 1

The strong law of large numbers states that if $(Y_i)_{i\geqslant 1}$ is an i.i.d. sequence of integrable random variables, then for almost every $\omega\in\Omega$, $n^{-1}\sum_{i=1}^nY_i(\omega)\to \mathbb E\left[Y_1\right]$.

In your context, you apply this to the sequence $(Y_i)_{i\geqslant 1}$ where $Y_i=X_i^2$, which is still i.i.d. and $Y_i$ is integrable, to obtain the following convergence (in the almost sure sense) $$ \frac 1n\sum_{i=1}^nX_i^2\to \mathbb E \left[X_0^2\right].$$