Exercise about the Strong Law of Large Numbers

Question

This is Exercise 5.3.2 from Achim Klenke: »Probability Theory — A Comprehensive Course«.

Exercise: Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of independent identically distributed random variables with $\frac{1}{n}(X_1 + \cdots + X_n) \overset{n\rightarrow\infty}{\longrightarrow} Y$ almost surely for some random variable $Y$. Show that $X_1 \in \mathcal{L}^1(\mathbf{P})$ and $Y = \mathbf{E}[X_1]$ almost surely.

Hint: First show that \begin{align*} \mathbf{P}\bigl[ |X_n| > n \text{ for infinitely many } n \bigr] = 0 \Longleftrightarrow X_1 \in \mathcal{L}^1 (\mathbf{P}). \end{align*}

My solution so far:

We follow the hint...

Assume that $\mathbf{P}\bigl[|X_n|>n \text{ for infinitely many } n\bigr] = 0$. Because $X_n$ is independent, the events $\{|X_n| > n\}$ must also be and it follows by the second Borel-Cantelli-Lemma, that \begin{equation*} \infty > \sum_{n=1}^\infty \mathbf{P}\bigl[|X_n| > n\bigr] = \sum_{n=1}^\infty \mathbf{P}\bigl[|X_1| > n\bigr] \, . \end{equation*}

We can add $\mathbf{P}\bigl[|X_1| > 0 \bigr]$ and the sum is still finite, of course, so $\infty > \sum_{n=0}^\infty \mathbf{P}\bigl[|X_1| > n\bigr] \geq \int |X_1| \, d\mathbf{P}$ and we conclude that $X_1 \in \mathcal{L}^1(\mathbf{P})$.

Now assume that $X_1 \in \mathcal{L}^1(\mathbf{P})$, then \begin{equation*} \infty > \int |X_1| \, d\mathbf{P} \geq \sum_{n=1}^\infty \mathbf{P}[|X_1| > n] = \sum_{n=1}^\infty \mathbf{P}[|X_n| > n] \end{equation*} and by the first Borel-Cantelli-Lemma we conclude that $\mathbf{P}\bigl[|X_n| > n \text{ for infinitely many }n\bigr] = 0$.

But what shall I do now?

Somehow we must end up using this theorem:

Theorem 5.17 (Etemadi’s strong law of large numbers (1981)) Let $X_1, X_2, \ldots \in \mathcal{L}^1 (\mathbf{P})$ be pairwise independent and identically distributed. Then $(X_n)_{n\in \mathbb{N}}$ fulfills the strong law of large numbers.

Can somebody help me, please?

Answer 1

Note that $$ \frac{S_{n+1}}{n+1}-\frac{S_n}n=-\frac{S_n}{n(n+1)}+\frac{X_{n+1}}{n+1},$$ hence the sequence $(X_n/n)_{n\geqslant 1}$ converges to $0$ almost surely. The series $\sum_n \mathbb P(|X_n|\geqslant n) $ is convergent; otherwise, by the second Borel-Cantelli's lemma, we would have $\mathbb P( \limsup_n\{|X_n|\geqslant n\})=1$, which contradicts the almost sure convergence to $0$ of the sequence $(X_n/n)_{n\geqslant 1}$.