# E[log(Z_t^2)], proof of convergence with Law of Large Numbers

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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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. –  George Lowther Jun 6 '10 at 17:13
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. –  George Lowther Jun 6 '10 at 17:22

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.
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.