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?


1 Answer 1


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, 2020 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. $\endgroup$ Oct 7, 2020 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, 2020 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. $\endgroup$ Oct 7, 2020 at 20:25

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