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Hi all, question: Let $Z_t$ be an iid sequence with $$\mathbb{E}\log(Z_t^2)<0 $$ Show that $$\sum_{j=0}^\infty Z_t^2 Z_{t-1}^2 ... Z_{t-j}^2 < \infty$$ almost surely

I am supposed to use LLN to solve this... but i can't make ends meet (this is exam preparation sheet question)

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  • $\begingroup$ This is just a straightforward exercise. Don't think it is really the kind of question this site is intended for. To answer, take the log of the product, divide by n and use LLN to deduce that the product is less than one (almost surely) for all large n. $\endgroup$
    – George Lowther
    Jun 6, 2010 at 17:13
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    $\begingroup$ Actually I misread it. The formula doesn't seem to make sense. Shouldn't the product be $Z_1\cdots Z_j$, in which case the LLN shows that the terms are almost surely bounded by a geometric series with ratio less than 1, so absoluty convergent. $\endgroup$
    – George Lowther
    Jun 6, 2010 at 17:22

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For every $t$, let $Y_t=\log(Z_t^2)$. Fix some $t$. The sequence $(Y_{t-k})_{k\geqslant0}$ is i.i.d. with $E[Y_t]\lt0$ hence the usual law of large numbers yields $\frac1j\sum\limits_{k=0}^{j-1}Y_{t-k}\to E[Y_1]$. Fix some negative $m\gt E[Y_1]$.

Then $\frac1j\sum\limits_{k=0}^{j-1}Y_{t-k}\leqslant m$ for every $j$ large enough, that is, for every $j\geqslant J$ where $J$ is random and almost surely finite. In particular, for every $j\geqslant0$, $\sum\limits_{k=0}^{j}Y_{t-k}\leqslant mj+X$, for some almost surely finite random $X$. This implies the pointwise convergence of the series since $$ \sum_{j\geqslant0}\exp\left(\sum\limits_{k=0}^{j}Y_{t-k}\right)\leqslant\sum_{j\geqslant0}\mathrm e^X\mathrm e^{mj}=\mathrm e^X(1-\mathrm e^m)^{-1}. $$ Note that the RHS above is almost surely finite since $m\lt0$ but is not (a priori) uniformly bounded since $X$ may be unbounded.

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If the term $Z_i^2$ of the independent and identically distributed random sequence is less than 1, then you can find a $q$ with $Z_i^2 \leq q < 1$ for almost all $i$ (acording to the law of large numbers). Then the $k$-fold product is less than $q^k$ such that the geometric series limits your sum by $\frac{1}{1-q}$.

I presume that you meant $Z_{t+j}^2$ instead of $Z_{t-j}^2$. Otherwise you'd get negative indices.

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  • $\begingroup$ Re-reading this three years later, I am surprised by the post (and by the votes): the only situation where $Z_i^2\leqslant q<1$ for every $i$ large enough (assuming this is what the strange mention "for almost all $i$" actually means) is when $P(Z_t^2\leqslant q)=1$. This condition is obviously not always implied by the hypothesis that $E(\log(Z_t^2))<0$, actually it is much stronger. Re-re-reading it again, I am inclined to suspect that "ac(c)ording to the law of large numbers" signals the OP is not sure of what they are talking about (the LLN says nothing about the single values $Z_t^2$). $\endgroup$
    – Did
    Oct 2, 2016 at 9:28

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