This is a follow-up of the previous post Law of the iterated logarithm in Hilbert space

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 $X_1 + \cdots + X_n$.

As indicated in the previous post by Iosif Pinelis, it is known (see the links in Law of the iterated logarithm in Hilbert space) that if $X, X_1, X_2, \ldots$ are iid random vectors in a separable Hilbert space $(H, \langle \cdot, \cdot \rangle, |\cdot |)$ with $E(X) = 0$ and $E(|X|^2) < \infty$, then $$ \limsup_{n \to \infty} \frac {|X_1 + \cdots + X_n|}{\sqrt {2n \ln \ln n}} = \sigma $$ almost surely where $$ \sigma = \sup \Big \{ \sqrt {E \big (\langle X, f \rangle ^2 \big) } : f \in H, |f| = 1 \Big\}. $$

Now the additional question is about the minimal assumption under which such a property holds. The real law of the iterated logarithm is known to be equivalent to $\sigma^2 = E(X^2) < \infty$. Does the law of the iterated logarithm in Hilbert space imply back that $E(|X|^2) < \infty$, or only $\sigma < \infty$ which seems weaker? Are there necessary and sufficient moment conditions for this law of the iterated logarithm in Hilbert space?


1 Answer 1


As a immediate corollary of the real-valued case, a necessary condition is that for all $f\in H$, $\langle X,f\rangle$ should be centered and have a finite moment of order two. For $n\geqslant 3$, denote $a_n:=\sqrt{2n\log\log n}$. If $\limsup_{n\to+\infty}\lVert S_n\rVert/a_n<+\infty$ almost surely, so is $\limsup_{n\to+\infty}\lVert X_n\rVert/a_n<+\infty$, since $a_n/a_{n-1}\to 1$. By the $0-1$ law, the latter $\limsup$ is almost surely constant (say equal to $C$) hence $\mathbb P\left(\limsup_{n\to+\infty}\left\{\lVert X_n\rVert/a_n>C+1\right\}\right)=0$. The second Borel-Cantelli lemma thus implies that $\sum_{n\geqslant 3}\mathbb P\left\{\lVert X_n\rVert/a_n>C+1\right\}$ converges. Using the fact that the random variables are identically distributed, we get that $\sum_{n\geqslant 3}\mathbb P\left\{\lVert X_1\rVert/a_n>C+1\right\}<\infty$. This implies that $\mathbb E\left[\lVert X_1\rVert^2/LL\left(\lVert X_1\rVert\right)\right]<+\infty$, where $L(x)=\max\{1,\log x\}$ and $LL(x)=L\circ L(x)$.

We thus have shown the following:

If $(X_i)_{i\geq 1}$ is i.i.d. and such that $\limsup_{n\to+\infty}\lVert S_n\rVert/\sqrt{2n\log\log n}<+\infty$, then

  1. For each $f\in H$, $\mathbb E\left[\langle X,f\rangle\right]=0$ and $\mathbb E\left[\langle X,f\rangle^2\right]$ is finite.
  2. $\mathbb E\left[\lVert X_1\rVert^2/LL\left(\lVert X_1\rVert\right)\right]<+\infty$.

Note that in the real-valued case, condition 2. is automatically satisfied as long as 1. is.

These conditions are actually also necessary for almost sure boundedness of $\limsup_{n\to+\infty}\lVert S_n\rVert/\sqrt{2n\log\log n}$. This is shown in Theorem 2 of the paper

de Acosta, A.; Kuelbs, J. Some Results on the Cluster Set ,$C\left(\left\{\frac{S_n}{a_n}\right\}\right)$ undefined and the LIL. Ann. Probab. 11 (1983), no. 1, 102--122. doi:10.1214/aop/1176993662. https://projecteuclid.org/euclid.aop/1176993662

  • $\begingroup$ Thank you very much, that is most helpful! This is an unexpected and striking equivalence. It is not so clear to me why $\sigma$ is involved rather than only $E(|X|^2)$? $\endgroup$
    – user166870
    Oct 12, 2020 at 14:02
  • $\begingroup$ At least from the one dimensional case one can see that the lower bound for the $\limsup_n$ of the norm of normalized partial sums is bigger than $\sigma$. I need to think more for an intuitive argument for the upper bound. $\endgroup$ Oct 15, 2020 at 10:00

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