Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. Specifically, Lemma 4 on page 307 states (without a proof) that

There exists a universal constant $M$ such that for each Banach space valued Gaussian random variable $X$ (having zero mean): $$ \mathsf{P}(\|X\|\ge u)\le \exp\left(-\frac{u^2}{M\mathsf{E}\|X\|^2}\right). $$

The authors refer to an older paper which is not available online. So I'm wondering how one proves this result.

As a first step, applying the generic Chernoff bound, one gets $$ \mathsf{P}(\|X\|\ge u)\le e^{-su}\mathsf{E}e^{s\|X\|} $$ for any $s>0$. Then the desired inequality holds if $\mathsf{E}e^{s\|X\|}$ is bounded by $e^{Cs^2\mathsf{E}\|X\|^2}$.

  • $\begingroup$ See Lemma 3.1 in the Ledoux-Talagrand book, available online. $\endgroup$ Jan 30, 2020 at 13:34
  • $\begingroup$ Indeed, but it’s a mean-zero random variable, so one should be able to control the median (using concentration)?.. $\endgroup$ Jan 30, 2020 at 14:56
  • $\begingroup$ Yes, I see now. I thought it would be absorbed into the universal constant M. $\endgroup$ Jan 30, 2020 at 15:15
  • $\begingroup$ I tried to derive this bound for Gaussian vectors in $\mathbb{R}^d$. For $X\sim N(0,\Sigma)$, the best I could obtain so far is $$ \mathsf{P}(\|X\|_2\ge u)\le 2\exp(-u^2/(2\operatorname{tr}(\Sigma))), $$ and I don't see a way to get rid of the factor of $2$ before the exponent. $\endgroup$
    – d.k.o.
    Jan 30, 2020 at 16:15
  • $\begingroup$ There's an unspecified universal constant $M$ in the OP -- doesn't that make any factor before the exponent meaningless? $\endgroup$ Jan 30, 2020 at 20:02

1 Answer 1


This inequality is false. E.g., consider the random vector $X_n:=(Z_1,\dots,Z_n)/\sqrt n$ in $\mathbb R^n$ with the Euclidean norm $\|\cdot\|$, where $Z_1,Z_2,\dots$ are independent standard normal random variables. Then $E\|X_n\|^2=1$ and, by the law of large numbers, $$\|X_n\|^2=\frac1n\,\sum_1^n Z_i^2\to1$$ in probability (as $n\to\infty$), so that $P(\|X_n\|\ge u)\to1$ for any $u\in(0,1)$. So, for any real constant $M>0$, any $u\in(0,1)$, and all large enough $n$ $$P(\|X_n\|\ge u)\not\le \exp\Big(-\frac{u^2}{M}\Big)=\exp\Big(-\frac{u^2}{M\,E\|X_n\|^2}\Big).$$

On the other hand, according to formula (3.5) in the Ledoux--Talagrand book, one has e.g. the inequality $$P(\|X\|\ge u)\le 4\exp\Big(-\frac{u^2}{8E\|X\|^2}\Big)$$ for $u\ge0$.

The constants here can be a bit improved by using Corollary 3, which states that $$P(\|X\|-E\|X\|\ge x)\le \exp\Big(-\frac{x^2}{2E\|X\|^2}\Big)\tag{1}$$ for $x\ge0$, which implies e.g. that $$P(\|X\|\ge u)\le\sqrt e\,\exp\Big(-\frac{u^2}{8E\|X\|^2}\Big)\tag{2}$$ for $u\ge0$.

Details on how to get (2) from (1): If $u^2<4E\|X\|^2$, then the upper bound on $P(\|X\|\ge u)$ in (2) is $>1$ and thus trivial. So, without loss of generality, $u^2\ge4E\|X\|^2\ge4(E\|X\|)^2$, so that $E\|X\|\le u/2$. Letting now $x:=u-E\|X\|\ge u/2$, we see that the left-hand of (1) becomes that of (2), and the right-hand of (1) is less than that of (2).

  • $\begingroup$ Interestingly, the same inequality is stated in another paper by Basu. $\endgroup$
    – d.k.o.
    Jan 31, 2020 at 17:54
  • $\begingroup$ @Iosif Can you kindly explain how you went from Corollary 3 of your paper to this form in terms of u ? $\endgroup$ Feb 23, 2020 at 23:37
  • $\begingroup$ @d.k.o. : By "the same inequality", do you mean the false one? Maybe, this is how false results propagate. Hopefully, this omission of a constant factor does not affect most of the applications. $\endgroup$ Feb 24, 2020 at 0:58
  • $\begingroup$ @gradstudent : I have added details on how to get (2) from (1). $\endgroup$ Feb 24, 2020 at 0:59

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