# Intuition Behind $O_p(\frac{1}{n})$

So let's say $$X_n = O_p(\frac{1}{n})$$.

According to Wikipedia's definition this means that $$\forall \; \epsilon>0, \; \exists$$ finite $$M>0,N>0$$ such that $$P(|nX_n|>M)=P(|X_n|>\frac{M}{n})<\epsilon \; \forall \; n>N$$.

I am trying to build an intuition and visualize what this means.

Is it fair to say that as $$n\to \infty, \;X_n$$ becomes arbitrarily small "most of the time" (i.e. except with a very small probability)? And that we are guaranteed for any $$y$$ and $$\epsilon$$ (and perhaps a very small $$y$$ I wish to use to bound $$X_n$$) there will eventually be some $$N$$ such that $$P(|X_n|>y)<\epsilon \; \forall \; n>N$$?

• Be careful, it is $O_p(1/n)$. $O(1/n)$ means something nonrandom. Feb 5, 2020 at 17:49
• @Zhanxiong I have edited the question accordingly Feb 5, 2020 at 18:23

$$X_n = O_p(\frac{1}{n})$$ means it's not terrible to think of $$X_n$$ as something like $$Y / n.$$ This is a single random variable over a changing nonrandom constant.
Indeed $$Y/n = O_p(\frac{1}{n})$$ because $$P(|Y/n|n > M) = P(|Y|>M)$$
can be made arbitrarily small by increasing $$M$$.
Then there's one more thing to consider. The probability inequality in this definition only needs to hold for $$n$$ greater than some chosen large $$N$$. This means that eventually your sequence of random variables "feels like" a single random variable over a constant.
• This is true, but what the OP appears to have meant was $O_p(\frac{1}{n})$, not $O(\frac{1}{n})$. Feb 5, 2020 at 17:52
• @Taylor along the same line of your reasoning is it fair to say if $Y$ a r.v. then $\frac{Y}{n}=O_p(\frac{1}{n})$ and in general $\text{Var}[O_p(\frac{1}{n})]$ decreases at a rate of $\frac{1}{n}$? Feb 5, 2020 at 20:16