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).)
Best Answer
I assume $\mu$ and $\nu$ are distributions with $d\mu$ and $d\nu$ the corresponding measures. Then $d\mu\otimes d\nu$ denotes the product measure. Distributions $\mu$ can be seen as generalised functions, where $\mu(x)$ doesn't have to be defined. However, I think in your case $\mu$ and $\nu$ are just functions, so you may write $\mu(x)$ and $\nu(y)$. In this case $d\mu(x)=\mu(x)dx$, and the product measure becomes simply $$(d\mu\otimes d\nu)(a,b)=\mu(a)\nu(b)dadb\,.$$ Now, to be (overly) precise, the author actually meant to write $$\tilde{\Phi}:=\int_{A\times B}\Phi(a,b)(d\mu\otimes d\nu)(a,b)\,,$$ but he assumed that it was clear that $a$ and $b$ are integrated over. So when $\mu$ and $\nu$ are functions, you have $$\tilde{\Phi}=\int_{A\times B}\Phi(a,b)\mu(a)\nu(b)dadb\,.$$