1
$\begingroup$

Tail bounds on random series in Hilbert space

Let $X_n$, $n \in \mathbb {N}$, be independent $\pm 1$ symmetric random variables, and $a_n$, $n \in \mathbb {N}$, be a sequence in a Hilbert space $H$ such that $\sum_n \|a_n\|^2 < \infty$, say $= 1$. Set $Z = \| \sum_n a_n X_n \|$ so that $E(Z^2) = \sum_n \|a_n\|^2 = 1$. Since $E(Z) \leq 1$, it is of interest to bound the probability $ P( Z > 1 + t)$, $t >0$. What would be a (sharp, exponential?) tail estimate in terms of the coefficients $a_n$, and in which range of $t >0$?

$\endgroup$
1
$\begingroup$

You can use e.g. Theorem 3.5, whereby $$P(Z\ge t)\le2e^{-t^2/2}\tag{1}$$ for all real $t\ge0$.

The coefficient $1/2$ in the exponent is of course sharp, in view of the standard central limit theorem (in $\mathbb R$), whereby $\sum_{j=1}^nX_j/\sqrt n$ converges to a standard normal random variable in distribution.


$\newcommand\si\sigma$Write $$Z=\Big\|\sum_n a_n X_n\Big\|=\sup_{\|x\|\le1}\sum_n X_n\langle a_n,x\rangle $$ and let $$\si^2:=4\sup_{\|x\|\le1}\sum_n \langle a_n,x\rangle^2 \le4\sum_n \|a_n\|^2=4. $$ Then, by Talagrand's concentration inequality for product measures (see e.g. Section 2.2), $$P(|Z-m_Z|\ge t)\le4\exp\Big\{-\frac{t^2}{4\si^2}\Big\}$$ for all real $t\ge0$, where $m_Z$ is a median of $Z$ and hence $$P(|Z-EZ|\ge t)\le P(|Z-m_Z|\ge t-\sqrt{8\pi}\,\si) \le4\exp\Big\{-\frac{(t-\sqrt{8\pi}\,\si)^2}{4\si^2}\Big\}\tag{2}$$ for all real $t\ge\sqrt{8\pi}\,\si$. The bound (2) will be better than (1) if $\si^2<1/8$ and $t$ is large enough.

$\endgroup$
4
  • $\begingroup$ I was told that there is somewhere an inequality involving the coefficients a_n by the (smaller) quantity sup \sum_n \langle y, a_n \rangle^2, where the sup is over all y in H with norm one? $\endgroup$ – Yilmis Jan 23 at 15:13
  • $\begingroup$ @Yilmis : They apparently meant something like the now added bound (2). $\endgroup$ – Iosif Pinelis Jan 24 at 2:30
  • $\begingroup$ Thank you very much for your help! $\endgroup$ – Yilmis Jan 25 at 7:25
  • $\begingroup$ @Yilmis : So, are you now satisfied with the answer? $\endgroup$ – Iosif Pinelis Jan 25 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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