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For a positive random variable we define

$\mathbb{E}\{X\} = \sup\{\mathbb{E}\{Y\}: 0 \le Y \le X\}$, where $Y$ is a simple random variable (and we have already defined expectations for them)

Then it is written:

We can have $\mathbb{E}\{X\} = \infty$ even when $X$ is never equal to $\infty$.

Why is that so?

If $X$ is never equal to $\infty$ then $\exists M: X \le M$, and thus every simple random variable $Y$ such that $Y \le X$ is also $Y \le M \Rightarrow \mathbb{E}\{Y\} \le M$

So also $\mathbb{E}\{X\} = \sup\{\mathbb{E}\{Y\}\} \le M$

Why am I wrong?

Ant
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  • Look at for instance the Cauchy distribution, or at the example given here. – Clement C. May 15 '14 at 19:27
  • @ClementC. thanks but it's not clear to me why the above definition (which is a the sup of a set) can be calculated in the "usual" way with that series. And anyhow, where is the error in the proof ? thank you in advance. – Ant May 15 '14 at 19:35
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    The error is when introducing $M$: the fact that $X$ is never $\infty$ does not mean that is it bounded, it can take arbitrarily large (finite) values, and even if this happens with very small probability this is enough to make the expectation infinite. – Clement C. May 15 '14 at 19:37
  • @ClementC. The last part of your comment is a bit missleading. A random variable can take arbitrarily large values and have finite expectation. But if the probability of taking larger and larger values doesn't drop off "fast enough", then the expectation is infinite, even though all the values individually are finite. Saying "... and even if this happens with very small probability this is enough to make the expectation infinite" OTOH gives the impression that all unbounded random variables have infinite expectation. – fgp May 15 '14 at 19:43
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    Indeed, it can be misinterpreted; but I guess it's safe to assume that the fact there exists unbounded r.v.'s with finite expectation (eg, normal r.v., to mention the most common one) is known... The goal was to explain what was going on in the two examples I linked, and where the mistake in the question was. – Clement C. May 15 '14 at 19:47
  • @ClementC. I can see that. Thank you. But can you link to a paper or online resource that shows how the above definition of expectation translates to the one that we actually use (ie the classic series?) I would be very grateful. – Ant May 17 '14 at 16:59
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    In terms of measure theory, a series is "only" an integral with relation to a counting measure: i.e., a measure $\mu$ on $X$ such that there exists a set $S\subseteq X$ with $$\forall A\subseteq X,\quad \mu(A) = \lvert A\cap S\rvert \in\mathbb{N}\cup{\infty}$$ For instance, taking $X=\mathbb{R}$ and $S=\mathbb{N}$, you get that for any (non-negative in a first time, for the sake of being sure the sum exists in $[0,\infty]$) function $f\colon\mathbb{R}\to\mathbb{R}$ $$\int_{\mathbb{R}}f(x)\mu(dx) = \sum_{n=0}^\infty f(n)$$ – Clement C. May 17 '14 at 22:41
  • @ClementC. I can intuitively understand what you wrote; $\mu(A)$ is counting the number of integers that are in $A$, and so $\mu(dx) = 0$ except when $x = n \in \mathbb{N}$, hence it is a series. But (1) I would like to understand it a little better, in a more rigorous way. and (2) the notation $\int_\mathbb{R} f(x) \mu(dx)$ has only been introduced to me as a shorthand for the mean value of a random variable $f(x)$ with respect to a probability $\mu(x)$. That is all I know about it, to me it's not an "integral", it's just a symbol. Can you point to some paper where it is explained better? :-) – Ant May 20 '14 at 10:55
  • @Ant: You can have a shot at one of these. I particularly like Çinlar's teaching style, but these lecture notes look good, at first glance (the definition of integral is p.21). – Clement C. May 20 '14 at 12:34
  • @ClementC. I'm sorry for the late reply. Thank you very much for the patience and the links ;-) I'll try to work it out! :) – Ant May 26 '14 at 14:36

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