5
$\begingroup$

I have read in a paper about the following result:

Let $V$ be a separable Hilbert space and $(\Omega,A_{\Omega},P)$ a probability space. Suppose that $Y_1,Y_2,...$ is a sequence of independent $V$-valued random variables. If $E\left(\Vert Y_i\Vert_{V}^m\right)\leq \frac{1}{2} m! B^2 L^{m-2}$ $\forall m\geq 2$, then, $\forall n\in N$ and $\epsilon>0$

$$ P\big(\big\Vert \frac{1}{n} \sum_{i=1}^n Y_i\big\Vert_{V}>\epsilon\big)\leq 2 \exp\Big(-\frac{n\epsilon^2}{B^2+L\varepsilon+B\sqrt{B^2+2L\epsilon}}\Big).$$

There was no proof given but it was mentioned that the result is well known. I think I found some literature that shows the result if we would assume that $E(Y_i)=0$ for all $i\in N$ holds (see Theorem 3.3 of Pinelis). But I don't know much about martingales so maybe it wouldn't follow. Does anyone know if the result is actually true and if $E(Y_i)=0$ is a necessary condition?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The condition $EY_i=0$ cannot be dropped.

Indeed, if e.g. the $Y_i$'s are iid with $\mu=EY_i\ne0$ and $n\to\infty$, then, by the law of large numbers, the left-hand side of the inequality in question will go to $1$ for each $\epsilon\in(0,\|\mu\|_V)$, whereas the right-hand side of the inequality will go to $0$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Okay thank you! But if $E X_i=0$ holds, is it right that the result follows from Theorem 3.3 and/or Theorem 8.6 as a special case? If yes, does it make a difference in using Theorem 3.3 or Theorem 8.6 to get the result above? $\endgroup$
    – Hugo10T
    Commented Mar 3, 2022 at 1:22
  • $\begingroup$ @Hugo10T : Theorem 8.6 is only for a one-dimensional case (even though it does not follow from the "multi-/infinite-dimensional" Theorem 3.3). So, use indeed Theorem 3.3, from which the inequality in question does follow if $EY_i=0$ for all $i$. There is also a published version of this paper, at projecteuclid.org/journals/annals-of-probability/volume-22/… . $\endgroup$
    – Iosif Pinelis
    Commented Mar 3, 2022 at 2:40
  • $\begingroup$ Previous comment continued: As you apparently know, there was a typo in the statement of this theorem: a term was missing -- see the correction at projecteuclid.org/journals/annals-of-probability/volume-27/… . $\endgroup$
    – Iosif Pinelis
    Commented Mar 3, 2022 at 2:40
  • $\begingroup$ Thank you very much! $\endgroup$
    – Hugo10T
    Commented Mar 3, 2022 at 10:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .