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What does $d \mathbb{P}(\omega)$ mean in expectation of r.v. $f$?

$$\mathbb{E} f = \int_{\Omega} f(\omega) d \mathbb{P}(\omega)$$

Yes sure it's some "infinitesimal", but should this mean that $\mathbb{P}(\omega)$ is a variable of $f$? Since in elementary integrals:

$$\int_X f(x) dx$$ and $x$ is a variable of $f$.

$\omega \in \Omega$ by def. but why is it $d \mathbb{P}(\omega)$ then and now just $d \omega$?

mavavilj
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  • It's notation; it says the integration variable is $\omega$ and the measure with respect to which you integrate is $\mathbb{P}$. Some people prefer to write $\mathbb{P}(d \omega)$ to mean the same thing. – Ian May 03 '19 at 12:02
  • @Ian But is $\mathbb{P}(\omega)$ then part of $f$, how? $f$ is $f: \Omega \rightarrow \mathbb{R}$. – mavavilj May 03 '19 at 12:04
  • No. $\int_\Omega f(\omega) d\mathbb{P}(\omega)$ says "integrate $f(\omega)$ over $\Omega$ and against the measure $\mathbb{P}$". – Ian May 03 '19 at 12:05
  • @Ian What does integrating against measure mean? – mavavilj May 03 '19 at 12:05
  • Unless that's just a linguistic question (in other words you don't know the word in English), I can't teach you measure theory from scratch on MSE. – Ian May 03 '19 at 12:05
  • @Ian You can give me a pointer as to where to look for. – mavavilj May 03 '19 at 12:06
  • Probably any introductory book on measure theory (written in English) will use the term "integrate against ...", but they may not use this notation. Usually we don't actually write it exactly like this except when doing multiple integrals against measures other than the Lebesgue measure (especially in probability theory where the explicit dependence of $f$ on $\omega$ is rarely spelled out). – Ian May 03 '19 at 12:08

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Part of what might be confusing to you already shows up in discrete probability. Here is an example: we roll a fair die, and let $X$ be the resulting number. We (that is, math teachers) say things like, the expected value of $X$ is $EX=\sum_{i=1}^6 i (1/6)$ and the expected value of $X^2$ is $EX^2=\sum_{i=1}^6 i^2 (1/6)$. Suppose we also roll an unfair die, and call the result $Y$, for which $P(Y=i) = p_i$ where not all the $p_i$ values are $1/6$. We say the expectations of $Y$ and $Y^2$ are $EY=\sum_{i=1}^6 i p_i$ and $EY^2 = \sum_{i=1}^6 i^2p_i$. In all cases the expectation is a weighted sum of the value we are talking about, and the system of weights we assign to the fundamental outcomes. In more advanced textbooks you will often see expressions like $EX=\int_\Omega X(\omega) P(d\omega)$, where $\omega$ plays the role of $i$ in my examples in this paragraph, where $P(\omega)$ plays the role of $p_i$ and so on.

In this way, we think of expectations as arising from two ingredients: a random quantity and the corresponding probability distribution. In your case, the expression $f(\omega)$ is the random quantity whose expectation we seek, and $d \mathbb P(\omega)$ is the corresponding probability distribution.

You might be also confused by the use of the letter $f$, which in textbooks is often used more-or-less exclusively as part of the probability law specification. Your example, however, is not an instance.

kimchi lover
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