2
$\begingroup$

I have a problem understanding the following statement from Klenke, p. 234:

If we write $\Xi_n(\omega) := \xi_n \bigl(X(\omega)\bigr) = \frac{1}{n} \sum^n_{i=1} \delta_{X_i(\omega)}$ for the $n$-th empirical distribution, then, by the result of this exercise, $\mathcal{E}_n = \sigma(\Xi_n)$.

Random variables $X_i$ have values in a Polish space $E$.

$\mathcal{E}_n$ is defined here, it's like the exchangeable $\sigma$-algebra, but just for permutations in $S(n)$.

Now again, the random variable $\Xi_n$ is a map $\Xi_n\colon \Omega \rightarrow \mathcal{M}_{\mathrm{fin}}(E)$ where $\mathcal{M}_{\mathrm{fin}}(E)$ means the space of all finite measures (= "a random variable giving us random measures").

But to get $\sigma(\Xi_n)$, I have to know what the $\sigma$-algebra on $\mathcal{M}_{\mathrm{fin}}(E)$ should be.

What is it?


tl;dr

Is there some canonical choice for the $\sigma$-algebra on the space of finite measures $\mathcal{M}_{\mathrm{fin}}(E)$ of a Polish space $E$?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, there is. It is the $\sigma$-field generated by all $\textit{projection maps}$: $$\pi_B:\mu\rightarrow\mu(B),\:\:\:\: B\in \mathcal{E},$$ where $(E,\mathcal{E})$ is the Polish space under consideration.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .