# Gaussian concentration inequality

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 $$e^{Cs^2\mathsf{E}\|X\|^2}$$.

• See Lemma 3.1 in the Ledoux-Talagrand book, available online. – Aryeh Kontorovich Jan 30 at 13:34
• Indeed, but it’s a mean-zero random variable, so one should be able to control the median (using concentration)?.. – Aryeh Kontorovich Jan 30 at 14:56
• Yes, I see now. I thought it would be absorbed into the universal constant M. – Aryeh Kontorovich Jan 30 at 15:15
• 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. – d.k.o. Jan 30 at 16:15
• There's an unspecified universal constant $M$ in the OP -- doesn't that make any factor before the exponent meaningless? – Aryeh Kontorovich Jan 30 at 20:02

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).

• Interestingly, the same inequality is stated in another paper by Basu. – d.k.o. Jan 31 at 17:54
• @Iosif Can you kindly explain how you went from Corollary 3 of your paper to this form in terms of u ? – gradstudent Feb 23 at 23:37
• @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. – Iosif Pinelis Feb 24 at 0:58
• @gradstudent : I have added details on how to get (2) from (1). – Iosif Pinelis Feb 24 at 0:59