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 $P(|X_n|>y)<\epsilon \; \forall \; n>N$?

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

1 Answer 1


$X_n = O_p(\frac{1}{n})$ 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.

  • $\begingroup$ This is true, but what the OP appears to have meant was $O_p(\frac{1}{n})$, not $O(\frac{1}{n})$. $\endgroup$
    – jbowman
    Commented Feb 5, 2020 at 17:52
  • $\begingroup$ @jbowman yes sorry about that I have edited the original question $\endgroup$ Commented Feb 5, 2020 at 18:25
  • $\begingroup$ @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 $\frac{1}{n}$? $\endgroup$ Commented Feb 5, 2020 at 20:16
  • $\begingroup$ This statement could still apply in situations where there are nonexistent/infinite variances. $\endgroup$
    – Taylor
    Commented Feb 6, 2020 at 5:04

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