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Let $(Y_n)$ be a collection of i.i.d nonnegative random variables with $E[\log^+Y_1]$ finite, where $\log^+ X=\max(0,\log(X))$. How does one show that this is equivalent to the fact that $ \sum\limits_{k=0}^\infty \frac{Y_k}{c^k} $ is finite a.s. for any $c>1$?

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up vote 1 down vote accepted

I will assume $Y_n$ are positive a.s. such that $\log Y_1$ are well-defined. Clearly, we have

$$ (\log a) \mathbb{P}(a < Y_n \leq b) \leq \int_{\{a < Y_n \leq b\}} \log Y_n \; d\mathbb{P} \leq (\log b) \mathbb{P}(a < Y_n\leq b).$$

In particular for $(a, b) = (c^k, c^{k+1})$ for $k = 0, 1, 2, \cdots$,

$$ k (\log c) \mathbb{P}\left(c^{k} < Y_n \leq c^{k+1}\right) \leq \int_{\{c^k < Y_n \leq c^{k+1}\}} \log^{+} Y_n \; d\mathbb{P} \leq (k+1)(\log c) \mathbb{P}\left(c^{k} < Y_n \leq c^{k+1}\right). $$

Thus if we let

$$ S(c) = (\log c) \sum_{k=1}^{\infty} k \mathbb{P}\left(c^{k} < Y_n \leq c^{k+1}\right),$$

then summing the inequality above gives

$$ S(c) \leq \mathbb{E}[\log^{+} Y_n] \leq S(c) + \log c. $$

Since $Y_n$ are i.i.d., $S$ does not depend on $n$. Thus this inequality implies that $\mathbb{E}[\log^{+} Y_1] < \infty$ if and only if $S(c) < \infty$. Here, we observe that the finiteness of $S(c)$ also does not depends on the choice of $c > 1$.

Now rearranging,

$$\begin{align*} S(c) &= (\log c) \sum_{k=1}^{\infty} \sum_{j=1}^{k} \mathbb{P}\left(c^{k} < Y_1 \leq c^{k+1}\right) \\ &= (\log c) \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} \mathbb{P}\left(c^{k} < Y_1 \leq c^{k+1}\right) \\ &= (\log c) \sum_{j=1}^{\infty} \mathbb{P}\left( c^j < Y_1 \right) \\ &= (\log c) \sum_{j=1}^{\infty} \mathbb{P}\left( c^j < Y_j \right) \end{align*}$$

Now we prove the prescribed equivalence.

$(\Longrightarrow)$ : By our previous observation, $\mathbb{E}[\log^{+} Y_n] < \infty$ if and only if $S(b) < \infty$ for $1<b<c$. The first Borel-Cantelli lemma says that $S(b) < \infty $ implies

$$ \mathbb{P} \left( b^{j} < Y_j \text{ for infinitely many } j \right) = 0. $$

Thus $Y_j / c^j < (b/c)^{j}$ except for finitely many exceptions a.s., and hence the finiteness of $\sum_{j} Y_j / c^j$ a.s..

$(\Longleftarrow)$ : We prove the contraposition. Assume $\mathbb{E}[\log^{+} Y_n] = \infty$. This is equivalent to say that $S(c) = \infty$. Since $Y_j$ are i.i.d., the events $\{ c^{j} < Y_j \}$ are mutually independent, hence we can apply the second Borel-Cantelli lemma to obtain

$$ \mathbb{P} \left( c^{j} < Y_j \text{ for infinitely many } j \right) = 1. $$

This proves that $\sum_{j} Y_j / c^j$ is not a.s. finite (in fact, infinite a.s.). ////

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+1. For the direct implication, $S(2c)$ does not work and should be replaced by $S(b)$ for some $1\lt b\lt c$. – Did May 23 '12 at 7:41
@Didier, you,re right. I fixed it. – Sangchul Lee May 23 '12 at 14:10
Thanks, that is gonna help me solve my problem! – Jean-Sébastien May 23 '12 at 15:33

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