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Oct 26, 2015 at 18:28 comment added user940 See my answer here: math.stackexchange.com/questions/876399/…
Oct 26, 2015 at 18:25 history edited Ruth90 CC BY-SA 3.0
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Oct 26, 2015 at 18:24 answer added Robert Israel timeline score: 4
Oct 26, 2015 at 18:23 comment added Ant Isn't it $M_n = X_1 \cdots X_n$? Why do you write that way?
Oct 26, 2015 at 18:22 comment added gt6989b Just thinking out loud: if $X_i$ are iid, so are $\ln(X_i)$, and so $$\frac{1}{n} \sum_{i=0}^n \ln (X_i) \to \mathbb{E}[X_i] = 0$$ a.s. by KSSLN.
Oct 26, 2015 at 18:22 comment added shalin This is false. Just let $X_n=1$ for all $n$. You need the added assumption that $X$ is not a.s. 1, or similar.
Oct 26, 2015 at 18:19 history edited gt6989b CC BY-SA 3.0
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Oct 26, 2015 at 18:15 history asked Ruth90 CC BY-SA 3.0