The formal definition of stochastic boundedness $O_p(1)$ of a sequence of random variables $\{X_n\}$ goes

$$\{X_n\} = O_p(1) \implies \forall \varepsilon >0, \quad \exists\, N_{\varepsilon}, \delta_{\varepsilon} \quad \text{ such that } \Pr(|X_n| \geq \delta_{\varepsilon}) \leq \varepsilon \quad \forall n> N_{\varepsilon}.$$

Suppose now that somebody says to us: "$X_n$ is NOT stochastically bounded."

Since the definition of $O_p(1)$ contains more than one condition, it would appear that the above verbal statement could translate to more than one situation.

For example, one could think of the following:

There exists some $\varepsilon_0$, say $\varepsilon_0=0.001$, for which we cannot find $N_{\varepsilon}, \delta_{\varepsilon}$ such that...

This would imply, I guess,

$$\forall\, \delta: \quad \Pr(|X_n| \geq \delta) > \varepsilon_0=0.001 \quad \forall\, n.$$

This appears to say that the sequence maintains strictly positive probability to obtain an arbitrary large value.

But consider also

There is no $\varepsilon >0 $ such that...

or even better

$$... {\rm only}\; \forall\, n \leq N_{\varepsilon}.$$

Q: Do the above, or any other, describe useful ways in which the definition of stochastic boundedness does not hold? And what is the verbal-intuitive description of each such situation?


2 Answers 2


If you understand that "$\{X_n\}$ is stochastically bounded" is just a synonym for "the probability measures $\mu_{X_n}$ induced by $X_n$ (I will call it $\mu_n$ for simplicity) is tight", and recall the following definition (Probability and Measure (3rd edition) by Patrick Billingsley, p. 336):

A sequence of probability measures $\mu_n$ on $(R^1, \mathscr{R}^1)$ is said to be tight if for each $\epsilon$ there exists a finite interval $(a, b]$ such that $\mu_n(a, b] > 1 - \epsilon$ for all $n$.

then an easy-to-interpret negation of "$\{\mu_n\}$ is not tight" is clearly "there exists an $\epsilon_0 > 0$ such that for every finite interval $(a, b]$, there exists some $n_0$ such that $\mu_{n_0}(a, b] \leq 1 - \epsilon_0$."

Billingsley also made the following important analogy which may help you reinforce the understanding of this concept (I think this also explains how the terminology "stochastically bounded" was coined this way):

Tightness of sequence of probability measures is analogous to boundedness of sequences of real numbers.

and the remark:

Tightness is a condition preventing the escape of (probability) mass into infinity.

With this interpretation, it might be better to rephrase my previous statement to "there exists an $\epsilon_0 > 0$ such that for every finite interval $(a, b]$, there exists some $n_0$ such that $\mu_{n_0}((a, b]^c) \geq \epsilon_0$."


Hi Alecos: It's not necessarily an answer but I needed space.

According to the link below, it means that, it is NOT the case that, for every $n$, we can find an $\epsilon$, such that $P(||X_{n} || < M) > (1-\epsilon)$.


  • $\begingroup$ Indeed. But the interest here is to arrive at a positive statement: from what does not happen, to what does happen as a consequence, and what sequence of r.v. s are we looking at then. $\endgroup$ Commented Oct 23, 2023 at 15:09
  • $\begingroup$ Hi Alecos: Okay. I understand. That's interesting and I was way off. I'll think about it but I doubt that I'll arrive at anything. My best guess is that the positive statement will have something to do with the limsup since that has to do with the event happening infinitely often. So, my guess is that the sequence becomes greater than M infinitely often but that's just a guess. If no one answers, I would send it to math.stackexchange. There are some serious probabilists over there also. $\endgroup$
    – mlofton
    Commented Oct 24, 2023 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.