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?


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 $(X_n/n)_{n\geqslant 1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.