The standard law of the iterated logarithm expresses that if $X_1, X_2, \ldots$ are iid real random variables with mean zero and variance $\sigma^2$, $$ \limsup_{n \to \infty} \frac {X_1 + \cdots + X_n}{\sqrt {2n \ln \ln n}} = \sigma $$ almost surely. Together with the same result for $-X_1, -X_2, \ldots$, the same limit holds true with absolute values around the sum $X_1 + \cdots + X_n$.

Is there a corresponding result when the $X_i$'s take values in a Hilbert space, and the absolute values are replaced by the Hilbert norm?


There are such versions of the law of the iterated logarithm even for independent random vectors in an arbitrary separable Banach space. See e.g. Theorems 4.1 and 4.2.

In the case when the Banach space is a Hilbert one, Theorem 4.1 implies the following:

Theorem 1: Let $X,X_1,X_2,\dots$ be iid random vectors in a separable Hilbert space $(H,\langle\cdot,\cdot\rangle,|\cdot|)$ with $EX=0$ and $E|X|^2<\infty$. Let $S_n:=X_1+\cdots+X_n$. Then $$\limsup_n\frac{|S_n|}{\sqrt{2n\ln\ln n}}=\sigma$$ almost surely, where $$\sigma:=\sup\big\{\sqrt{E\langle X,f\rangle^2}\colon f\in H,|f|=1|\big\}.$$

To deduce Theorem 1 from Theorem 4.1, one only needs to note the following two points:

(i) In view of formula (2.5), $\sup_{x\in K}|x|=\sigma$.

(ii) $E|S_n|\le\sqrt{E|S_n|^2}=\sqrt{nE|X|^2}=o(\sqrt{2n\ln\ln n})$, so that condition (ii) of Theorem 4.1 holds.

In the case when $H=\mathbb R$, Theorem 1 becomes the law of the iterated logarithm cited in your post.

  • $\begingroup$ Thank you for the reference. I would be also interested in the minimal assumptions for such a property to hold. $\endgroup$ – Wolfgang Oct 7 '20 at 8:46
  • $\begingroup$ @Wolfgang : I have added details showing that your one-dimensional law of the iterated logarithm extends almost literally to Hilbert spaces. $\endgroup$ – Iosif Pinelis Oct 7 '20 at 15:24
  • $\begingroup$ Thank you again. About minimal moment assumptions, on the real line, the second moment is necessary and sufficient for the law of the iterated logarithm to hold. Do I understand correctly that it is the same in Hilbert space with the condition E(|X|^2) finite? $\endgroup$ – Wolfgang Oct 7 '20 at 16:56
  • $\begingroup$ @Wolfgang : I don't know if the condition $E|X|^2<\infty$ is necessary here. I see that your posted question "Is there a corresponding result" did not say anything about minimal assumptions. $\endgroup$ – Iosif Pinelis Oct 7 '20 at 20:25

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.