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If a r.v. $X < a$, does it imply $EX < a$?

  1. If not, why is it different from what I know: If a r.v. $X \leq a$, it implies $EX \leq a$, proved by replacing $X$ with $a$ as the integrand.
  2. Note that $a \in \mathbb R$. If it allows that $a= \infty$, the answer is no: if $X$ has a Pareto distribution with $α=1$, then $X < \infty$, but $EX = \infty$, from http://en.wikipedia.org/wiki/Pareto_distribution and https://stats.stackexchange.com/a/91515/1005
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  • $\begingroup$ @whuber: I saw some text about how $\infty$ is involved in Lebesgue integrals, and if i am correct, the meanings of $\infty$ as values taken by a random variable $X$ and taken by its expectation $EX$ are the same. So I am still confused. $\endgroup$
    – Tim
    Commented Nov 3, 2014 at 21:58
  • $\begingroup$ @whuber, thanks for reminding me about Rudin's book (I forgot where I saw about $\infty$ in Lebesgue integral). According to that book, the definition of Lebesgue integral of a measurable function $f$ is $\int f d \mu := \int f_+ d \mu - \int f_- d \mu$ where $f_+$ and $f_-$ are the positive and negative parts of $f$. $\int f d \mu$ is well-defined iff $\int f_+ d \mu$ and $\int f_- d \mu$ are not both $\infty$, because $\infty - \infty$ is undefined. So I don't see why you said $\infty$ has different meanings for $X$ and $EX$. I see that it has the same meaning in both cases. $\endgroup$
    – Tim
    Commented Nov 3, 2014 at 22:51
  • $\begingroup$ I have looked over that section again (1.22 in the second edition) and believe you are correct. I apologize for misleading you: Rudin (rather elegantly) does use the symbol "$\infty$" in a consistent way. $\endgroup$
    – whuber
    Commented Nov 3, 2014 at 23:02

1 Answer 1

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\begin{align} EX &= \int x\cdot \mathbf{1}_{x< a} + x\cdot\mathbf{1}_{x\geq a} \ dP \\ &= \int x\cdot \mathbf{1}_{x< a}\ dP\\ &< \int a\cdot \mathbf{1}_{x< a}\ dP \\ &= a, \end{align}

so yes.

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  • $\begingroup$ I guess I should add, strict inequality follows because $P\{x<a\}=1$, and this set can be rewritten: $\{x<a\}=\cup_{n> 0}\{x<a-1/n\}$. So some $\varepsilon > 0$ must have $P\{x<a-\varepsilon\}>0$, hence the one integral is strictly less. $\endgroup$ Commented Jun 18, 2017 at 18:20

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