How to prove $\frac{X_{T}}{T}\overset{a.s.}{\rightarrow } 0$ as $T$ tends to infinite with $X_{T}=O_{P}(1)$

Question

I've encountered a problem, which might not be difficult: Let $X_{T}=O_{P}(1)$. Can we obtain $\frac{X_{T}}{T}\overset{a.s.}{\rightarrow } 0$ as $T$ tends to infinite? Thanks all!

Answer 1

Let $(X_n)$ be independent with $X_n = n$ with probability $1/n$ and $X_n=0$ with probability $1-1/n.$ Then $X_n=O_p(1)$ since for any $\epsilon\in(0,1)$, $P(X_n>1/\epsilon)<\epsilon$ for all $n.$

Yet $X_n/n$ is $1$ with probability $1/n$ and $0$ with probability $1-1/n.$ Since $\sum_n 1/n=\infty,$ $X_n/n=1$ infinitely often by Borel Cantelli and thus $X_n/n$ does not converge with probability one.

However, it is the case that if $X_n = O_p(1)$ then $X_n/n\to 0$ in probability. Let $\epsilon>0$ and pick $M$ such that $P(|X_n|>M) < \epsilon$ for all $n$. Let $\delta>0.$ Then $$ P(|X_n|/n >\delta) < \epsilon$$ for all $n>M/\delta.$