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I am currently working on a problem associated with the random variable $Z_n$ such that $Z_n = Z_{n-1}X_n$, where $X_i's$ are i.i.d. random variable with the calculated expectation to be greater than 1.

So after calculation I found that $E[Z_n] \to \infty$ as $n \to \infty$, how ever $Z_n$ converges to 0 almost surely after I taking log and apply the SLLN. / I am confused on why the results are different but still compatible? Thanks!

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$L_1$ convergence (which means $E(|X_n-X|)\to 0) $and almost sure convergence are incomparable conditions where neither implies the other. Yours must be an example where almost sure convergence doesn't imply $L_1$ convergence. There are simpler examples, like

$$ X_n = \left\{\begin{array}{ll}0&\mbox{with probability $1-1/n^2$}\\n^3&\mbox{with probability $\frac{1}{n^2}$}\end{array}\right.$$

Essentially what can happen is the variable can be usually close to some value (the value it converges to a.s.), but very rarely very far away. Even if the rareness is such that it is guaranteed only far away finitely often (so that convergence is almost sure) it can be so large in the rare event when it is far away that the expected value is pushed away from the value to which it converges almost surely.

There can also be cases where the sequence strays infinitely often so as not to converge but nonetheless doesn't stray very far so doesn't impact the mean. For instance $$ X_n = \left\{\begin{array}{ll}0&\mbox{with probability $1-1/n$}\\1&\mbox{with probability $\frac{1}{n}$}\end{array}\right.$$ almost surely does not converge, but $E(|X_n|)\to 0.$

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  • $\begingroup$ Thank you so much for the clarification. It really helps! $\endgroup$ – Chloe Sep 21 '17 at 2:10

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