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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!

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    What is $O_P(1)$? – 2017-02-12
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    This makes little sense as is. Do you mean let $X_n = O_p(1)$ and does $X_n/n\to 0$ a.s.? (or $X_t = O_p(1) $and $X_t/t\to 0$ for a continuous s.p.) – 2017-02-12
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    @Crostul I assume they mean https://en.wikipedia.org/wiki/Big_O_in_probability_notation – 2017-02-12
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    @ spaceisdarkgreen thanks for your comment。 I mean $X_t/t\to 0$ – 2017-02-12

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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 $n>M/\delta.$

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    Assuming independence. – 2017-02-12
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    @Did that's right (edited) – 2017-02-12
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    @spaceisdarkgreen Thanks all your guys. So $\frac{X_{T}}{T}\overset{a.s.}{\rightarrow } 0$ is right? – 2017-02-12
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    @weilinhy No, it is wrong. I produced a counterexample. – 2017-02-12
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    @spaceisdarkgreen Thanks. If $X\sim \mathbb{N} (0,1)$, then $\frac{X}{T}\overset{a.s.}{\rightarrow } 0$ is right? – 2017-02-12
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    @weilinhy Yes, if $X_n$ are independent $N(0,1)$ then $X_n/n\to 0$ a.s. $\sum_n P(|X_n/n|>\delta) < \infty$ This is true for i.i.d for any finite variance RV by the chebyshev inequality – 2017-02-12
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    @spaceisdarkgreen Thanks for your useful and elegent response. – 2017-02-12
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    @weilinhy notice the counterexample I used wasn't identically distributed. – 2017-02-12
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    @spaceisdarkgreen Thanks for kindly help. – 2017-02-12