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In general if I have that \begin{equation} \mathbb{E}(X)<\infty \end{equation} does this imply that $|X|<\infty$ a.s? An attempted proof:

Let $(\Omega,\mathbb{F},\mathbb{P})$ be a probability space and $K>0$ be a constant, then \begin{equation} \mathbb{E}(X)=\int_{\Omega}XdP\,=\int_{\{|X|\leq K \}}XdP\,+\int_{\{|X|>K\}}XdP\, <\infty \end{equation} Taking the limit as $K\rightarrow \infty$ \begin{equation} \int_{\{|X|>K\}}XdP\, \rightarrow 0 \end{equation} Struggling how to finish from here or posssibly this isn't true? Thanks!

Michael
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  • But what if $X$ takes on the value $\infty$ on a set of measure zero? – mathse May 11 '14 at 09:52
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    @mathse measure zero doesn't count ("$|X|<\infty$ a.s.") – Hagen von Eitzen May 11 '14 at 09:53
  • Yes, I overlooked the "almost surely". – mathse May 11 '14 at 10:26
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    How about using Markov's Inequality? $P(X=\infty)=P(\cap_{n=1}^{\infty} X\geq n)=\lim_{k} P(\cap_{n=1}^{k}X\geq n)=\lim_{k} P(X\geq k) \leq \lim_{k} \frac{E[x]}{k}=0$ – Mael Nov 10 '14 at 17:38
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    Note ... $|X|<\infty$ a.s. should not be called bounded. Instead, call it finite perhaps. I would say bounded means $\exists m<+\infty$ so that $|X| \le m$ a.s. – GEdgar Dec 29 '24 at 09:32

2 Answers2

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If not $|X|<\infty$ a.s., then there is a set of positive measure where $|X|=\infty$. The integral $$ \int_\Omega X\,\mathrm dP= \int_\Omega \max\{0,X\}\,\mathrm dP-\int_\Omega \max\{0,-X\}\,\mathrm dP$$ is defined only when at most one of the summands is infinite. If $X=+\infty$ on a set of positive measure, then $X=-\infty$ only on a zero-set, and then $\mathbb E(X)=+\infty$. Similarly, if $X=-\infty$ on a set of positive measure, then $\mathbb E(X)=-\infty$. Thus we have the following possibilities:

  • $|\mathbb E(X)|<\infty$ and $|X|<\infty$ a.s.
  • $\mathbb E(X)=+\infty$ and $X>-\infty$ a.s.
  • $\mathbb E(X)=-\infty$ and $X<\infty$ a.s.
  • $X$ is not integrable ($\mathbb E(X)$ does not exixts)

(So specifically, as you wrote $\mathbb E(X)<\infty$ without absolute value, it is possible to have $|X|=\infty$ with positive probability)

Hagen von Eitzen
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    I think you've answered my question but I have two side queries. Firstly why does $X=+\infty$ on a set of positive measure, imply $X=-\infty$ only on a zero-set. Secondly, at the risk of sounding stupid, I don't fully understand your final comment. Surely when $\mathbb{E}(X)<\infty$ it is just a real number and therefore it's absolute value is also finite? Assuming $X$ is integrable, $\mathbb{E}(X)<\infty$ corresponds to bullet points 1 and 3, in which cases $X<\infty$ a.s.? – Michael May 11 '14 at 10:30
  • Wait never mind the second point, I see that $\mathbb{E}(X)=-\infty$ then with positive probability $X=-\infty$ which gives with positive probability $|X|=\infty$. – Michael May 11 '14 at 10:42
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$E|X| \geq E|X| 1_{|X| = \infty} = \infty P(|X| = \infty)$. As the left hand side is finite, so $P(|X| = \infty) = 0$, implying the almost sure finiteness.

For the almost sure bounded part, just use the continuity of probability measure and the Markov's inequality.$P(\cap_{k=1}^{\infty}|X|\geq k) = P(|X| = \infty) =\lim_{n\to \infty}P( \cap_{k=1}^{n} X\geq k) = \lim_{n \to \infty}P(X\geq n) \leq \limsup_{n}\frac{EX}{n} = 0$

Kai
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