5
$\begingroup$

Let be $(\Omega, \mathcal{F}, \mathcal{P})$ a probability space, $(X_{n})$ a sequence of randiom variables and $X$ a random variable. Let be $\Omega$ countable and $\mathcal{F}$ a power set of $\Omega$.

Show that $X_n\xrightarrow{p} X, n \rightarrow \infty \ \ \implies X_n \xrightarrow{a.s.} X, n \rightarrow \infty$

I have done some research and I have found out that the statement is not true in general but I have no idea how to prove it for this special case. I appreciate your help in advance a lot.

$\endgroup$

1 Answer 1

8
$\begingroup$

Let $F=\{\lim_{n\to\infty}X_n=X\}$ and suppose that $\mathsf{P}(F)<1$. Since $F^c$ is countable, there exists $\omega'\in F^c$ such that $\mathsf{P}(\{\omega'\})=p>0$. Thus, there exists a subsequence $n_k$ such that $|X_{n_k}(\omega')-X(\omega')|>\epsilon>0$. However, this implies $$ \mathsf{P}(|X_{n_k}-X|>\epsilon)\ge \mathsf{P}(\{\omega'\})=p\not\to 0. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .