2
$\begingroup$

I came across this expression:

$\tilde{\Phi}(\mu, \nu)\ \dot{=}\ \int_{A\times B}{\Phi(a, b)\ \mathrm{d}\mu \otimes \mathrm{d}\nu}$

In a context where:

  • $A$ and $B$ are compact metric spaces
  • $\mu$ and $\nu$ are probability distribution over $A$ and $B$, resp.
  • $\Phi$ is a continuous function $A \times B \rightarrow \mathbb{R}$
  • $\tilde{\Phi}(\mu, \nu)$ is said to be the expected value of $\Phi$

I understand that you need to integrate over $A \times B$ to get this expected value, and to take $\mu$ and $\nu$ distributions into account while doing this. But..

How am I supposed to understand the $\otimes$ symbol here? What is this operation? How does $\mathrm{d}\mu$ relates to $a$ and $\mathrm{d}\nu$ relates to $b$ within this integrand?

(To get the full context, I've found this in these pretty neat notes introducing differential game theory (equation 2.8 page 13).)

  • 0
    How familiar are you with measure theory?2017-02-15
  • 0
    @MichaelMcGovern Novice. But I can read and learn :) `currently heading towards Wikipedia. Cheers for the pointer!2017-02-15
  • 1
    Here, $\otimes$ stands for a [product measure][1]. [1] https://en.m.wikipedia.org/wiki/Product_measure2017-02-15
  • 0
    @MichaelMcGovern heading to this now. Thank you :)2017-02-15
  • 0
    @MichaelMcGovern okay, so this is just a way to write that the integral is computed using the product measure $\mu\otimes\nu$ over $A\times B$ without loss of generality concerning the form and nature of $A,\ B,\ \mu$ and $\nu$, right? In trivial cases, it may just read as $\int_{A\times B}{\Phi(a, b)\, \mu(a)\, \nu(b)\ \mathrm{d}a\,\mathrm{d}b}$..2017-02-15
  • 0
    $A$ and $B$ are sets $\mu$ and $\nu$ are functions.2017-02-15

0 Answers 0