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In Independent events and Kolmogorov, it seems Petite Etincelle is trying to answer without Kolmogorov 0-1 Law.

Without using Kolmogorov 0-1 Law how do you prove the last step? I tried:

$$0 = E[e^{-M_{\infty}}] = E[e^{-M_{\infty}}1_{M_{\infty}=\infty}+e^{-M_{\infty}}1_{M_{\infty}<\infty}]$$

$$=E[0 \times 1_{M_{\infty}=\infty}+e^{-M_{\infty}}1_{M_{\infty}<\infty}]$$

Then what?

With Kolmogorov 0-1 Law, how do you prove the last step?

I tried to suppose on the contrary that $P(e^{-\sum_{k=1}^\infty I_{E_k}} = 0)<1$. Then by Kolmogorov 0-1 Law, $$P(e^{-\sum_{k=1}^\infty I_{E_k}} = 0)=0$$

$$\to P(\sum_{k=1}^\infty I_{E_k} < \infty) = 1$$

$$\to E[\sum_{k=1}^\infty I_{E_k}] < \infty$$

$$\to \sum_{k=1}^\infty E[I_{E_k}] < \infty$$

$$\to \sum_{k=1}^\infty P(E_k) < \infty$$

BCLC
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    To be honest, I am not sure what is your question. If $X$ is a non-negative random variable such that $\mathbf{E}[X]=0$, then $X=0$ holds $\mathbf{P}$-a.s. This is easily seen in view of the Markov's inequality: $$\forall\epsilon>0\ :\quad\mathbf{P}[X\geq\epsilon]\leq\frac{1}{\epsilon}\mathbf{E}[X] = 0.$$ – Sangchul Lee May 03 '18 at 14:11
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    @SangchulLee Lol thanks for the feedback and the answer. Post as answer? Happy 5th week of Easter! ^-^ – BCLC May 03 '18 at 14:19

2 Answers2

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(Expanded from comments)

Recall the fact that if $X$ is a non-negative random variable satisfying $\mathbf{E}[X] = 0$, then $X = 0$ $\mathbf{P}$-a.s. Indeed, by the Markov inequality

$$ \mathbf{P}[X \geq \epsilon] \leq \frac{1}{\epsilon}\mathbf{E}[X] = 0 $$

and hence $\mathbf{P}[X > 0] = \mathbf{P}[\cup_{n\geq 1}\{X \geq 1/n\}] \leq \sum_{n\geq1} \mathbf{P}[X\geq1/n] = 0 $.

Now you can apply this claim to $X = e^{-M_{\infty}}$.

Sangchul Lee
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  • Sangchul Lee, why not apply the same idea here? $\int_{\Omega} |f| d\lambda = 0 \Rightarrow {x \in \Omega: f(x) \neq 0}$ is a null set – BCLC May 03 '18 at 14:28
  • @BCLC, Again I am not sure what your asking. The proof in the link works, and the idea in my answer works. They are just different realizations of the idea that the indicator function $\mathbf{1}_{{X \neq 0}}$ is almost majorized by a constant multiple $X$ with some error that we can easily control. – Sangchul Lee May 03 '18 at 14:48
  • I mean Markov's inequality directly proves the proposition there right? – BCLC May 03 '18 at 14:54
  • @BCLE, Oh, now I understand your point. Of course we can do that. On the other hand, I see that the argument there is equally elementary and concise. :) Also, editing someone's answer to ask your own question seems not an appropriate way of posing a question. – Sangchul Lee May 03 '18 at 14:57
  • And of course, we have $\mathbf{P}[X \leq \epsilon] = 1$ and taking $\epsilon\downarrow0$ (or more precisely, choosing a sequence $\epsilon_n \downarrow 0$ and letting limit along this subsequence) gives $\mathbf{P}[X = 0] = 1$. – Sangchul Lee May 03 '18 at 14:59
  • I would ask you to restate without prepositions, but I think that would be rude or entitled. I guess I'll have a reread and try to deduce which thing you refer to which preposition. I suppose it (haha) is my fault for having too many things at once leading to the need for prepositions in the first place. – BCLC May 03 '18 at 14:59
  • Why all this need for choosing sequence $\epsilon_n$ or discretisation $\bigcup_{n \ge 1}$? I mean, $|a-b|\le \varepsilon \iff a=b$ sooo.... – BCLC May 03 '18 at 15:00
  • Basically the continuity of measure follows from its $\sigma$-additivity, which means that only 'countably-infinite limit operations' are valid for measures in principle. Of course, in the presence of monotonicity (such as in our case), this easily improves to limit along continuum. – Sangchul Lee May 03 '18 at 15:02
  • So are either of these answers wrong please? Answer to this question. Answer to another question. – BCLC May 03 '18 at 15:04
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    @BCLC, I would say they are incomplete. – Sangchul Lee May 03 '18 at 15:06
  • K thanks $ \ \ $ – BCLC May 03 '18 at 15:06
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Use Markov's inequality with $X = e^{-M_{\infty}}$:

$$ \mathbf{P}[X \geq \epsilon] \leq \frac{1}{\epsilon}\mathbf{E}[X] = 0 $$

$$ \to \mathbf{P}[X \leq \epsilon] = 1 $$

$$ \to \mathbf{P}[X = 0] = 1 $$

BCLC
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