Suppose we have a probability space $(\Omega, \mathfrak{F}, P)$, and independent events $(E_n)_n$.
Consider $$M_n = \sum_{k=1}^n I_{E_k}$$ Is it correct to say that by the Kolmogorov $01$ law $M_n$ converges a.e or diverges a.e ?
Asked
Active
Viewed 106 times
2 Answers
2
Yes, it's a consequence of Kolmogorive 0-1 law. We can also derive it in the following way.
When $E(\sum_{k=1}^\infty I_{E_k}) < +\infty$, $\sum_{k=1}^\infty I_{E_k}$ is finite almost surely, so $M_n$ converges almost surely.
When $E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have
\begin{align} E e^{-\sum_{k=1}^\infty I_{E_k}} &= E\prod_{k=1}^\infty e^{-I_{E_k}} \\ &= \prod_{k=1}^\infty Ee^{-I_{E_k}} \\ &= \prod_{k=1}^\infty (1-P(E_k) + e^{-1}P(E_k))\\ &= \prod_{k=1}^\infty(1 - (1-e^{-1})P(E_k)) \end{align}
Since $\sum_{k=1}^\infty P(E_k) = E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have $\prod_{k=1}^\infty(1 - (1-e^{-1})P(E_k)) = 0$, (see this), i.e. $$E e^{-\sum_{k=1}^\infty I_{E_k}} = 0$$
So $e^{-\sum_{k=1}^\infty I_{E_k}} = 0$ almost surely, i.e. $M_n$ diverges almost surely.
Petite Etincelle
- 14,767
-
Petite Etincelle, how to do your last step without Markov's inequality and without Kolmogorov 0-1 Law? – BCLC May 03 '18 at 14:30
-
1@BCLC I think it's obvious that $E X=0$ for non negative would lead to $X=0$ almost surely. If you really want all the details, think of the set ${X \geq \frac{1}{n}}$, it has to be of zero probability, and X strictly positive is the union of all these sets, thus also of zero probability – Petite Etincelle May 04 '18 at 21:17
-
thanks! $ \ \ \ $ – BCLC May 04 '18 at 21:38
-1
A third proof: (First is by Kolmogorov 0-1 Law. Second is Petite Etincelle's answer)
When $E(\sum_{k=1}^\infty I_{E_k}) < +\infty$, same as in Petite Etincelle's answer
When $E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have for $A=1$,
$$\sum_{k=1}^\infty P(|I_{E_k}|\ge A) = +\infty$$
By Kolmogorov 3-Series Theorem, $\sum_{k=1}^\infty I_{E_k} = +\infty$
BCLC
- 13,459