# Proof with probability inequalities and infinite sequences

I am reading Berkes et al. (2003) about the GARCH model. Could someone help me figure out the proof of one lemma in the paper?

$\mathbf{Lemma.}$ If $\{\xi_k, 0\leq\xi_k<\infty\}$ is a sequence of identically distributed random variables satisfying $$\mathrm{E}\log^{+}|\xi_0|<\infty, \tag{1}$$ then $\sum_{k=0}^{\infty}z^{k}\xi_k$ converges with probability one for any $|z|<1$. Note, $\log ^{+} x=\log x$ if $x>1$, and $0$ otherwise.

$\mathbf{Proof.}$ By the Borel-Cantelli lemma it is enough to prove that, for any $\zeta>1$, $$\sum_{k=1}^{\infty}P\{|\xi_{k}|>\zeta^{k}\}<\infty. \tag{2}$$

The distribution of $\xi_k$ does not depend on $k$, so \begin{align} \sum_{k=1}^{\infty} P\{|\xi_{k}|>\zeta^{k}\} &=P\{\log^{+}|\xi_k|>k\log\zeta\}\nonumber \\ &=\sum_{k=1}^{\infty} P\{\log^{+}|\xi_0|>k\log\zeta\} \tag{3} \\[10pt] &\leq\mathrm{E} \log^{+} |\xi_0|/\log \zeta, \tag{4} \end{align} and thus $(1)$ implies $(2)$.

$\Box$

$\mathbf{Question.}$ It seems to me that $(3)$ does not imply $(4)$. It is natural to apply the Markov inequality to $(3)$ and we have $\sum_{k=0}^{\infty} P\{\log^{+}|\xi_0|>k\log\zeta\}\leq \sum_{k=1}^{\infty}\frac{\mathrm{E}\log^{+}|\xi_0|}{k\log\zeta}$. Since the harmonic sequence, $\sum_{k=1}^{\infty}\frac{1}{k}$, does not converge, we cannot get $(4)$ by using the Markov inequality. Did I miss something here?

References:

• You should add a reference to the paper. Commented Aug 5, 2016 at 3:53
• I think I figured it out by using $\mathrm{E}X=\int_{0}^{\infty} P(X>s)ds$ for any nonnegative random variable $X$. Commented Aug 5, 2016 at 15:23
• Could you edit the title to highlight the actual problem? I don't know the terminology of these inequalities, but probably you do, so you could include a relevant name. Having GARCH in the title is not useful, IMHO. Also, the GARCH tag seems irrelevant as the inequality is probably not intrinsically specific to GARCH models, is it? Commented Aug 5, 2016 at 17:00
• My original title is 'Is this proof wrong?' And the only tag I used is 'probability'. Some people help me make modifications. If you like you also can do it. Commented Aug 5, 2016 at 17:33

Your argument does not invalidate their argument. Basically, they say $A<B$, where $B<\infty$, while you are saying, $A<B\times C$, where $C=\infty$. Both arguments are compatible and your argument does not invalidate theirs.
• @JRBNB I don't know whether they used it or not. In fact, you should provide a link to the paper (specially because you are copy-pasting bits of it). My point is about the material you just posted. Your inequallity corresponds to $A< B\times C$, where $A < B$ is their inequality. Commented Aug 4, 2016 at 23:31