I am studying the Law of large numbers for independent and identically distributed (i.i.d) random variables.

Assume there are i.i.d variables $(\xi_k)_{k\ge 1}$ taking values in $(0,1)$. From the law of large numbers, we know $\frac{\sum_{k\le n} \xi_k}{n} \to \mathbb{E}(\xi_1)$ almost surely (a.s.).

Then we know, if $\mathbb{E}(\log\xi_1^{-1})<\infty$:

$$\sup_{n}\sum_{k\le n} \xi_1\cdot \xi_2....\xi_k=\sup_{n}\sum_{k\le n} e^{-(\log\xi_1^{-1} + \log \xi_2^{-1}+....+\log \xi_k^{-1})}\precsim\sup_n\sum_{k\le n}e^{-i\mathbb{E}(\log\xi_1^{-1})} < \infty$$ a.s. (geometric series).

Let's consider $\sup_{n}\sum_{k\le n} \xi_n\cdot \xi_{n-1}....\xi_k$.

Can we prove this is also finite?


1 Answer 1


If the $\xi$ take values arbitrarily close to 1 (e.g. the $\xi_i$ are Unif[0,1]), this quantity is infinite: for any $\epsilon>0$, there exist arbitrarily long strings where all of the $\xi_j$’s are at least $1-\epsilon$, which means that the sup is larger than$1/\epsilon$.

  • $\begingroup$ oh, yeah, it is infinity. nice construction! Many thanks! $\endgroup$
    – jason
    May 10, 2018 at 2:36

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