If $(X_n)_n$ is i.i.d. and $\frac{S_n}{n} \to a$ almost surely then $a=\Bbb E[X_1]$

Question

$(X_n)_{n \in \Bbb N}$ independent and identical. $S_n = \sum_{i=1}^n X_i$ Now I want to show: $$\frac{S_n}{n} \xrightarrow{\text{almost surely}} a, \mathrm{with}\: a \in \Bbb R\Rightarrow a=\Bbb E[X_1]$$ In our lecture we showed: $(X_n)_{n \in \Bbb N}$ independent and identical, with $\Bbb E[X_1] \lt \infty$ , then $\frac{S_n}{n} \xrightarrow{\text{almost surely}} \Bbb E[X_1]$

In other words, either there's a problem with the uniqueness of the limit or the fact that we don't know from the start if the expectation values are finite. I'm stuck on this problem.
Does someone have any ideas or tipps on how to solve this? Thanks in advance!

Answer 1

I'll prove that $\frac{S_n}{n}\to c$ ($c\in \mathbb{R}$) only if $\mathbb{E}[|X|]<\infty$. The if statement, namely, $c = \mathbb{E}[X]$ is a consequence of SLLN.

Notice that given condition is equivalent to having $$\frac{1}{n}\underbrace{\sum_{k=1}^n (X_k-c)}_{:=T_n} \to 0.$$

Now, observe that $$\frac{(X_n - c)}{n} = \frac{T_n}{n}-\frac{n-1}{n}\frac{T_{n-1}}{n-1}\to 0$$ as $n\to\infty$. Now, recall that the condition $Y_n\to 0$ almost surely is equivalent to having for every $\epsilon>0$, $\mathbb{P}(|Y_n|>\epsilon, \text{ infinitely often})=0$. Taking $\epsilon=1$, and $Y_n = \frac{X_n-c}{n}$, we notice that the condition implies $$ \sum_{n=1}^\infty \mathbb{P}(|X-c|>n)=\sum_{n=1}^\infty \mathbb{P}(|X_n-c|>n)<\infty, $$ otherwise if the RHS were $\infty$, we would, by Borel-Cantelli lemma, get that the aforementioned probability is $1$, which is a contradiction.

Now, we'll use a final trick. For a non-negative random variable $Y$, $\mathbb{E}[Y]=\int_0^\infty \mathbb{P}(Y>x)dx$. Using a partition $\{0,1,2,\dots\}$ for $\mathbb{R}$, and bounding the integral via upper and lower Riemann sums, we get $$ \sum_{n=1}^\infty \mathbb{P}(Y>n)\leq \mathbb{E}[Y] = \int_0^\infty \mathbb{P}(Y>x)dx \leq \sum_{n=0}^\infty \mathbb{P}(Y>n). $$ Taking $Y=|X-c|$, we get $\mathbb{E}[|X-c|]<\infty$, hence (by Triangle inequality), $\mathbb{E}[|X|]\leq \mathbb{E}[|X-c|]+|c|<\infty$.

Hence $\frac{S_n}{n}$ converges to a finite constant only when $X\in L^1$, that is, $\mathbb{E}[|X|]<\infty$. Under this, it is a consequence of SLLN (or repeating it, if you like) that this constant must be equal to $\mathbb{E}[X]$.