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Consider the space $(\Omega,\mathcal{F},P)$. Show that $E(c)=c,\forall c\in \mathbb{R}$.

I thought about two distinct ways to show that.

  1. $E(c)=\int_\Omega c\ dP=c\int_\Omega dP=cP(\Omega)=c$;

  2. Let $f=c$ a.e.. Then $f=c.\chi_1 (\{f=c\})+b.\chi_2(\{f\neq c\})$ as a simple function. Then $E(f)=\int_\Omega f dP=c .P(\{f=c\})$. Clearly, $P(\{f=c\})=1$, otherwise the sum of this probability with the probability of its complement wouldn't be 1.

Are both acceptable?

Thanks!

Celine Harumi
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    Part of the answer at https://stats.stackexchange.com/questions/386649 directly implies this result. Part of (1) is erroneous: it is axiomatic that $\Pr(\Omega)=1,$ not $c.$ The first line in $2$ is not true, because it implies that the values of $f$ on a set of measure zero must be constantly $b,$ which is not necessarily the case. – whuber Sep 04 '19 at 14:55
  • @whuber What part in (1) contradicts $\Pr(\Omega)=1$? – Juho Kokkala Sep 07 '19 at 11:08
  • @Juho: (1) ends with the assertion "$\Pr(\Omega)=c.$" – whuber Sep 07 '19 at 13:25
  • @whuber I see (1) ending with the equation $c\dot P(\Omega)=c$ where I believe the $.$ denotes multiplication (possibly $\cdot$ was intended). If not, I don't know how to make sense of the $.$s appearing in (2), either. – Juho Kokkala Sep 07 '19 at 15:01
  • @Juho: Thank you: that's a better interpretation of the period. I wasn't reading the line creatively enough to see that. – whuber Sep 07 '19 at 17:22
  • That was a multiplication. – Celine Harumi Sep 08 '19 at 19:54
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    Use `\cdot` in the $\TeX$ markup, as in $c\cdot \Pr(f=c).$ – whuber Sep 09 '19 at 15:22

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