Is p(x)dx equal to dp(x)?
I'm confused with the definition of the expectation operator. Assume a random variable $X$ having a probability distribution $p(x)$. Then the expected value of $X$ can be computed as $\int xp(x)dx$.
It is noted in 1 that, given a probability space $(\Omega, \Sigma, P)$ as defined in 2, the general definition of the expected value is
$\int_{\Omega} X dP = \int_{\Omega} X(w)P(dw) $
where $P$ is the probability measure returning an events probability in $\Sigma$. Is this probability measure the same as the distribution $p(x)$. Additionaly. what does $X(w)$ and $P(dw)$ mean?
Also, the Eq. 1 in states that $E[f(x)] = \int f(x)p(x) = \int f(x)dp(x)$, is this correct or simply there is a notation error?
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$\begingroup$I think when you write $EX = \int_\omega x p(x) dx$, the $p(x)$ is the p.d.f. However, $EX = \int_\omega X dP(X)$. The $P$ is probability measure. So they are different.
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