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I have the following stupid question in my mind while i am studying for exams. Does $X<\infty \ a.s$, implies that $\mathbb E(X)<\infty$?

Further on this, is the converse of the above statement true? Do give me a bit summary on this. Thanks very much.

I thought this was true until I realize the following example: Let's consider a simple symmetric random walk, we know that each state is null-recurrent. Let $\tau_L$ be a stopping time when the walk first hits $L$ started from $0$. then $$\mathbb P(\tau_L<\infty)=1$$ so $\tau_L<\infty \ a.s.$ but we also know that $$\mathbb E(\tau_L)=\infty$$ Is this a counter-example? Thanks, i am a bit weak on measure theory.

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    $\begingroup$ Perhaps an easier counterexample. We know that $\sum\limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ and so $\sum\limits_{n=1}^\infty \frac{6}{\pi^2n^2} = 1$. Consider then a discrete random variable who takes value $n\in\Bbb Z^+$ with probability $\frac{6}{\pi^2 n^2}$. The expected value of this however would have been a multiple of the harmonic series which we know to diverge. $\endgroup$ – JMoravitz Nov 15 '19 at 14:58
  • $\begingroup$ so only the converse is true? $\endgroup$ – Kenneth Nye Nov 15 '19 at 15:02
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    $\begingroup$ What does $\mathbb E[X]<\infty $ a.s. means ? Notice that $\mathbb E[X]$ is deterministic (it's a constant if you prefer). So, there is no randomness here. $\endgroup$ – Surb Nov 15 '19 at 15:10
  • $\begingroup$ @surb you are right, i need to drop a.s. $\endgroup$ – Kenneth Nye Nov 15 '19 at 15:13
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Take $(\Omega ,\mathcal F,\mathbb P)=([0,1], \mathcal B([0,1]), m)$ where $m$ denote the Lebesgue measure on $[0,1]$. Consider $X(\omega )=\frac{1}{\omega }$.

Then, $X(\omega )<\infty $ a.s. but $\mathbb E[X]=\infty $.

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    $\begingroup$ Very good clear and easy (+1) $\endgroup$ – John Nov 15 '19 at 15:26
  • $\begingroup$ so the $P(X=\infty)=m(\{0\})=0$, while the the area is infinity? it is indeed a simple example. $\endgroup$ – Kenneth Nye Nov 15 '19 at 15:31
  • $\begingroup$ Yes exactly. But in probability, it's not a very good idea to interpret the expectation as an area...@KennethNye $\endgroup$ – Surb Nov 15 '19 at 15:51
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An even simpler counterexample is to take $P(X=n)=C/n^2$ where n is a positive integer and $C$ is a normalizing constant.

However, the converse is true. Usually, the expectation is only defined when $X$ is absolutely integrable. We have $P(|X|>M)M\leq E(|X|)<\infty$ for any $M>0$. If $X$ is infinite with positive probability, then $P(|X|>M)\ge P(|X|=\infty)>0$ for any $M$, so the left hand side of the inequality diverges as $M\to\infty$, a contradiction.

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The simplest counter example I can think of a random variable $X$ which can take values $\{1, 2, 4, ...\}$ where $P(X=n)=\frac{1}{2^n}$ if $n$ is a power of $2$ and $0$ otherwise.. Then

$E(X)=\sum\frac{2^{n-1}}{2^{n}}=\infty $

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