The weakest concentration inequality I know of is markov's inequality:

$$\mathbb P (X \geq t) \leq \frac{\mathbb E X}{t}$$

where $X$ is a nonnegative random variable with first moment $\mathbb E X$ and $t > 0$. Of course, this bound assumes that the first moment exists.

Is there a weaker and still useful concentration inequality than the markov inequality? Does it even make sense to talk about concentration when the first moment does not exist?

For example in the specific case of a cauchy, I know that the measure concentrates around $0$, but I don't know of a useful bound on the measure far away from this concentration point that applies to general random variables without finite first moments.

EDIT: naively, I could imagine the existence of finite "partial moments" implying weaker inequalities. By partial moments I mean: $\mathbb E X^p$ for $p < 1$ (although I dont think negative $p$ could give us concentration type properties).


1 Answer 1


On $\mathbb{R}$ (and Polish spaces in general) all probability measures are tight, so some kind concentration inequality always exists and can be derived directly by the tightness property. (See https://en.wikipedia.org/wiki/Tightness_of_measures)

I would also add that for any bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$ and any ranom variable $X$ on $\mathbb{R}$ it always holds that $\mathbb{E}f(X) < \infty $. If $f$ is also strictly increasing then we can always use Markov's inequality to derive the concentration bound $$ P(X>f^{-1}(t))\leq \frac{\mathbb{E} f(X)}{t}.$$

  • $\begingroup$ I'm familiar with the fact that all probability measures are tight (but did not know it was limited to Polish spaces, thanks for specifying), but it's not exactly the type of result I'm looking for. As for bounded implying finite first moment, in that case we also have Hoeffdings inequality, which is certainly stronger than Markov. The comment about $f$ strictly increasing is interesting: perhaps in some cases we can find an $f$ such that $\mathbb E f(X)$ exists and the bound in your answer is not vacuous. I wonder if there are conditions we can impose to ensure that! $\endgroup$
    – dmh
    May 17, 2021 at 22:23
  • $\begingroup$ Do you mean a case where $f$ is not bounded but it still holds that $\mathbb{E} f(X) < \infty$? $\endgroup$
    – Giacomo
    May 18, 2021 at 8:49
  • $\begingroup$ I was looking for a general condition that would ensure the existence of an $f$ such that the bound above is useful (RHS is less than 1) for some values of $t$. Although on second thought I wonder if such an $f$ always exists (the boundedness condition on $f$ is probably enough) $\endgroup$
    – dmh
    May 18, 2021 at 12:57
  • 1
    $\begingroup$ Yes, for any $f:\mathbb{R_+} \rightarrow [0, 1]$ it holds that $\mathbb{E}f(X) \leq 1$. For example one could use the the hyperbolic tangent or any other sigmod function. $\endgroup$
    – Giacomo
    May 18, 2021 at 13:59

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