Let $\xi_{1},\xi_{2},\dots$ be random variables (i.e measurable functions) such that $\mathbb{P}(\xi_{n}=-3^{n})=2^{-n}$ and $\mathbb{P}(\xi_{n}=1)=1-2^{-n}$ Let $S_{n}=\displaystyle\sum_{j=1}^{n}\xi_{j}$. Prove that $\mathbb{E}[S_{n}]\to -\infty$ whereas $S_{n}\to \infty$ a.s
Solution:
so by the definition of expectation
$\mathbb{E}[S_{n}]=-3^{n}2^{-n}+1(1-2^{-n})=-3^{n}2^{-n}+1-2^{-n}<1-(\frac{3}{2})^{n}$
but $\displaystyle\sum_{n=1}^{\infty}(\frac{3}{2})^{n}=\infty$ by d'Alembert convergence test so ${E}[S_{n}]=\to-\infty$
Just one question to this part: $S_{n}$ is defined as a finite sum of random variables whereas when we test convergence of the series we consider infinite sum....
Now for the second part I am unsure which Borel- Cantelli lemma has been used.
By Borel-Cantelli $\xi_{n}=1$ a.s for all but for finitely many $n$. Hence a.s $S_{n}\to \infty$
How do we get the last result?
Borel-Cantelli I: if $\displaystyle\sum_{n=1}^{\infty}\mu(E_{n})<\infty$ then $\mu(\displaystyle\limsup_{n}(E_{n}))=0$ where $E_{1},E_{2}....\in\mathcal{M}$
Borel-Cantelli II:Let $E_{1},E_{2}....\in\mathcal{M}$ be independent events, if $\displaystyle\sum_{n=1}^{\infty}\mathbb{P}(E_{n})=\infty$ then $\mathbb{P}(\displaystyle\limsup_{n}(E_{n}))=1$ i. e infinitely many $E_{1},E_{2},,,$ will occur with prob =1.
It's the first version (we don't have necessarily independence). Take $E_n=\{\xi_n=-3^n\}$. We have $\mu(\limsup_n(E_n))=0$, hence there is $\Omega'\subset\Omega$ such that $\mu(\Omega')=1$ and for all $\omega\in\Omega'$, there is $N(\omega)$ such that if $n\geqslant N(\omega)$, then $\omega\notin E_n$. Since $\xi_n$ takes only two values, this means that $\xi_n(\omega)=1$ for these $n$.