Proof with probability inequalities and infinite sequences

Question

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 \begin{equation} \mathrm{E}\log^{+}|\xi_0|<\infty, \tag{1} \end{equation} 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 $\zeta>1$, \begin{equation} \sum_{k=1}^{\infty}P\{|\xi_{k}|>\zeta^{k}\}<\infty. \tag{2} \end{equation}

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:

• I. Berkes, L. Horváth and P. Kokoszka. GARCH processes: structure and estimation. Bernoulli 9 (2003), no. 2, 201--227. doi:10.3150/bj/1068128975.

Answer 1

I think I asked a dumb question. The proof is right. The missing step in the proof is as follows.

\begin{align} &~~~~\sum_{k=1}^{\infty} P\{\frac{\log^{+}|\xi_0|}{\log\zeta}>k\}\\ &\leq \sum_{k=1}^{\infty} \int_{k-1}^k P\{\frac{\log^{+}|\xi_0|}{\log\zeta}>s \}ds\\ &=\mathrm{E} \log^{+} |\xi_0|/\log \zeta, \end{align}

Answer 2

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.