Why $EN < \infty $ implies $ N < \infty $? (It's part of Borel-Cantelli lemma's proof).
Why EN < \infty implies N < \infty ?
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measure-theory
probability-theory
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0What is $E$ and $N$? (In particular, could any of them _fail_ to be finite and if so how?) – 2011-09-30
1 Answers
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If $\mathbb{E}[N] < \infty$, then necessarily $\mathbb{E}[N^+] < \infty$ and $\mathbb{E}[N^-] < \infty$, where $N^+(\omega): = \max\{N(\omega),0\}$ and $N^-(\omega): = \max\{-N(\omega),0\}$, the positive part and negative part of $N$ respectively. This is because of how expectation is defined, as the sum of the expectations of the positive part and negative part. Now, If $N$ (and thus $N^+$) took the value positive infinity on a set of positive measure $A$, then $\mathbb{E}N^+ \geq \mathbb{E}[N^+ \chi_A] = \infty$.
So since the expectation of $N$ is finite, the set where $N$ is infinite must have zero measure, or in other words, $N < \infty$ a.s. The same argument applies to $N^-$, implying that the $N > -\infty$ a.s. as well.
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0Thank you Andre, you are right. I have edited my answer to include this. – 2011-10-17