2

Where $P$ is a probability measure on the space $\Omega$ and $X$ is a real-valued random variable? Specifically what does $P(d\omega)$ mean and how is it different from $dP$ together with the integral for $\Bbb{E}(X)$ involving supremum of approximative simple functions?

I'm going through the book A Basic Course in Probability Theory and some things they don't explain. That's fine though, other than that it's a great book, and I already know some measure theory.

So I'm at the part about simple functions (which I've read about / worked with before) and the change of variables formula, where the above notation is used:

$\Bbb{E}(h(X)) \equiv \int_{\Omega} h(X(\omega))P(d\omega) = \int_{S} h(x) Q(dx)$,

provided one of the two integrals exists, and $h: \Bbb{S} \to \Bbb{R}$ is Borel measurable.

Daniel Donnelly
  • 21,342
  • @Jack, so $P(d\omega) = dP$ in this context? – Daniel Donnelly Jul 25 '17 at 00:23
  • 1
    Yes. See also: https://en.wikipedia.org/wiki/Expected_value#General_definition –  Jul 25 '17 at 00:23
  • 1
    very roughly: $P(\operatorname d \omega) = \dfrac{\operatorname d P(\omega)}{\operatorname d \omega}\operatorname d \omega$ – Graham Kemp Jul 25 '17 at 00:24
  • 2
    @GrahamKemp: What does you comment mean? What is $\frac{dP(\omega)}{d\omega}$? –  Jul 25 '17 at 00:26
  • 1
    I just meant that the equivication of $\int_\Omega g(\omega) \operatorname d P\color{silver}{(\omega)} = \int_\Omega g(\omega) P(\operatorname d \omega)$ is essentially the chain rule. – Graham Kemp Jul 25 '17 at 00:48
  • 1
    See also: https://math.stackexchange.com/questions/5230/notation-question-integrating-against-a-measure –  Jul 25 '17 at 00:59
  • @GrahamKemp Are you trying to confuse things or what? Mentioning the chain rule to justify what is simply a notation, honestly... – Did Jul 25 '17 at 08:02
  • 1
    We can write $\int X dP=\int X(\omega) dP(\omega)=\int X(\omega) P(d\omega)$ interchangeably. The first is the shortest and cleanest. When you need to refer to a specific variable, the second and third are used. Personally I prefer the third because it is clearer in certain contexts, for example if you have a function $\mu$ on $\Omega\times\mathcal F$ such that $\mu(x,\cdot)$ is a probability measure for all $x\in\Omega$. If you want to integrate $X$ with respect to $\mu(x,\cdot)$, $\int X(\omega)P(x,d\omega)$ seems far clearer to me than $\int X(\omega)dP(x,\omega)$. – Jason Jul 25 '17 at 14:22

0 Answers0