# Meaning of symbol product (cross) in a circle: ⨂

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).)

• How familiar are you with measure theory? – Michael McGovern Feb 15 '17 at 16:21
• @MichaelMcGovern Novice. But I can read and learn :) `currently heading towards Wikipedia. Cheers for the pointer! – iago-lito Feb 15 '17 at 16:22
• Here, $\otimes$ stands for a [product measure][1]. [1] en.m.wikipedia.org/wiki/Product_measure – Michael McGovern Feb 15 '17 at 17:05
• @MichaelMcGovern heading to this now. Thank you :) – iago-lito Feb 15 '17 at 17:25
• @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 $\int_{A\times B}{\Phi(a, b)\, \mu(a)\, \nu(b)\ \mathrm{d}a\,\mathrm{d}b}$.. – iago-lito Feb 15 '17 at 18:01

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 $$\tilde{\Phi}=\int_{A\times B}\Phi(a,b)\mu(a)\nu(b)dadb\,.$$