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Consider a random variable $X \colon\Omega \rightarrow \mathbb{R}$ for a probability space $(\Omega, \mathcal{F}, P)$.

We had the following definition for the expectation:

$$\mathbb{E}[X]= \int_{\Omega} X(\omega) d P(\omega). \quad (1)$$

For the discrete case we would probably write

$$\mathbb{E}[X]= \sum_{ \omega \in \Omega} X(\omega) P(\omega). \quad (2)$$

Now as non-mathematician (i.e. not having much knowledge about integrals and measure theory),

I am wondering why converting the sum in (2) to an integral does not result in

$$\mathbb{E}[X]=\int_{\Omega} X(\omega) P(\omega) d \omega\quad (3)$$

instead of (1).

(3) is what I would expect, because instead of the sum we just integrate, hence informally "replace sum in (2) by integral sign and add a $d \omega$ at the end".

What is the difference between (1) and (3)? Is it valid to write (3)? Is it just "notation" for measures $P$ or is there a reason for this?

Thank you very much and please have patient with a non-mathematician that never had a lecture in mass theory or integrals.

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  • $\begingroup$ Recall that $\Omega$ in $(\Omega,\mathcal F,P)$ can be any set. What would be $d\omega$ if $\Omega$ is, say, $\mathbb R^\mathbb R$? $\endgroup$
    – Did
    Feb 23, 2015 at 8:02
  • $\begingroup$ The discrete case can also be written as an integral. In this case $P = \sum_n p_n \delta_n$. Then $ E X = \int X dP = \sum_n p_n \int X d\delta_n = \sum_n p_n X(n)$. $\endgroup$
    – copper.hat
    Feb 23, 2015 at 8:08
  • $\begingroup$ So since I am already confused about this $d \omega$ in the most simple case of integrating over $\mathbb{R}$ (where I suspect it to mean an inifinitely small interval on the $\mathbb{R}$-axis), I am not really sure what this $d \omega$ could mean for other spaces, probably it means an infinitely small "volume" element. But what you write suggests tha this $d \omega$ is not properly defined for $\mathbb{R}^{\mathbb{R}}$. But then, what does it change to integrate over $d P(\omega)$? (Sorry, but I really have almost no idea about integrals, only from an applied (engineer) point of view) $\endgroup$
    – user136457
    Feb 23, 2015 at 8:23

1 Answer 1

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One writes $${\Bbb E}[X]=\int_\Omega X(\omega)\>{\rm d}P(\omega)$$ if only the probability measure $P:\>{\cal F}\to[0,1]$ is at stake, which assigns values to subsets $A\subset\Omega$, not to individual points $\omega\in\Omega$.

But often the probability space $\Omega$ has a "natural geometric measure", say length or area. This is particularly the case if $\Omega$ is a reasonable subset of some ${\mathbb R}^n$. Integration with respect to this geometric measure is then signalled by writing ${\rm d}\omega$. This has as yet nothing to do with probability. The latter is an additional feature which makes its appearance as a weight factor $\omega\mapsto f(\omega)$, called the probability density on $\Omega$. Such an $f$, if it exists, is completely determined by the given probablity measure $P$, and is related to $P$ via the formula $$P[A]=\int_A f(\omega)\>{\rm d}\omega\ .$$ There is no standard notation for this density; lets just call it $f$ for the moment. The standard normal $f(x):={1\over\sqrt{2\pi}}e^{-x^2/2}$ on ${\mathbb R}$ is an example. Given such a density the expectation of a random variable $X:\>\Omega\to{\mathbb R}$ is then given by $${\Bbb E}[X]=\int_\Omega X(\omega)\>f(\omega)\>{\rm d}\omega\ .$$

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  • $\begingroup$ thank you very much! This was extremely helpful! $\endgroup$
    – user136457
    Feb 23, 2015 at 16:47

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