# Formula for expected value of continuous random variable without using density

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?

• There is one symbol without any explanation/definition in your writing. – user158565 Jun 26 '19 at 18:01
• Yes, sorry, it is corrected now. – TEX Jun 26 '19 at 18:50
• If you search for the "law of the unconscious statistician" on this forum you will find many answers. – Xi'an Jun 26 '19 at 18:59
• 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)? – whuber Jun 26 '19 at 19:04
• Without using $\Omega$, is something like $\int_{\mathcal{V}}u(v) P(dv)$ admissible? I don't want to use $\Omega$ if possible. – TEX Jun 27 '19 at 9:33