Suppose we have a sequence of positive random variables $X_1,X_2,…,X$. I am trying to prove a characterization of almost sure convergence.
It states that $X_n \rightarrow X$ almost surely iff for every $\epsilon >0$, $\lim_{n \rightarrow \infty} P[\sup_{k \ge n} \frac{X_k}{X} > 1+ \epsilon]=0$ and $\lim_{n \rightarrow \infty} P[\sup_{k \ge n} \frac{X}{X_k} > 1+ \epsilon]=0$.
If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.
Best Answer
Since all the random variables are positive, \begin{align} &\{\omega:X_n(\omega)\not\to X(\omega)\}=\{\omega:X_n(\omega)/X(\omega)\not\to 1\} \\ &\qquad=\{\limsup X_n/X > 1\} \cup \{\liminf X_n/X < 1\}. \end{align} Thus, $$ \mathsf{P}(X_n\not\to X)\le\mathsf{P}(\limsup X_n/X>1)+\mathsf{P}(\liminf X_n/X<1). $$ But \begin{align} \mathsf{P}(\limsup X_n/X>1)&=\lim_{m\to\infty}\mathsf{P}(\limsup X_n/X\ge 1+ m^{-1}) \\ &=\lim_{m\to\infty}\lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n} X_k/X\ge 1+ m^{-1}\right). \end{align} That is, $\mathsf{P}(\limsup X_n/X>1)=0$ is equivalent to $$ \lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n} X_k/X\ge 1+\epsilon\right)=0 \quad\forall \epsilon>0, $$ and, similarly, $\mathsf{P}(\liminf X_n/X<1)=0$ is equivalent to $$ \lim_{n\to\infty}\mathsf{P}\!\left(\inf_{k\ge n} X_k/X\le 1-\epsilon\right)=0 \quad\forall \epsilon>0. $$
The last condition is equivalent to $\lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n} X/X_k\ge 1+\epsilon\right)=0 \quad\forall \epsilon>0$.