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I'm looking for some correct notation. Consider the random variable $V$ with support $\mathcal{V}$ and probability distribution $P_V$. Consider a function $u:\mathcal{V}\rightarrow \mathbb{R}$.

Let $\mathcal{V}$ be finite and suppose I want to compute the expected value of $u(V)$. This is

$$ \sum_{v\in \mathcal{V}} u(v)\times P_{V}(v) $$

Suppose now instead that $P_V$ is continuous. What is the correct notation to indicate the formula of the expected value of $u(V)$ without passing through the definition of density? I though about using $$ \int_{\mathcal{V}} u(v)\text{ } dP_{V}(v) $$

Is this correct?

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  • $\begingroup$ There is one symbol without any explanation/definition in your writing. $\endgroup$ – user158565 Jun 26 '19 at 18:01
  • $\begingroup$ Yes, sorry, it is corrected now. $\endgroup$ – TEX Jun 26 '19 at 18:50
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    $\begingroup$ If you search for the "law of the unconscious statistician" on this forum you will find many answers. $\endgroup$ – Xi'an Jun 26 '19 at 18:59
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    $\begingroup$ As suggested by @Xi'an, are you looking for an abstract expression as in stats.stackexchange.com/questions/411201 $$E[V] = \int_\Omega V(\omega)\,\mathrm{d}\mathbb{P}(\omega)$$ ($\mathrm{d}\mathbb{P}$ is a probability measure, not a density), or do you seek an alternative way to compute the expectation that is not based on the density (there are many)? $\endgroup$ – whuber Jun 26 '19 at 19:04
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    $\begingroup$ Without using $\Omega$, is something like $\int_{\mathcal{V}}u(v) P(dv)$ admissible? I don't want to use $\Omega$ if possible. $\endgroup$ – TEX Jun 27 '19 at 9:33

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