I'm reading about Random Measures.

As a definition, the book I'm reading gives that:

Let $(E, \mathcal{E})$ be a measurable space. A random measure on $(E, \mathcal{E})$ is a transition kernel from $(\Omega, H)$ into $(E, \mathcal{E}).$

More explicitly, a mapping $M:\Omega\times\mathcal{E}\rightarrow\mathbb{R}^{+}$ is called a random measure if $\omega → M(\omega, A)$ is a random variable for each $A \in\mathcal{E}$ and if $A\rightarrow M(\omega,A)$ is a measure on $(E, \mathcal{E})$ for each $\omega\in\Omega.$

Then, comes the following definition:

If $M$ is a random measure on $(E,\mathcal{E}),$ or each $A\in\mathcal{E},$ we define the $\sigma-$algebra $$\mathcal{F}(A)=\sigma(\{M(B):B\in\mathcal{E}, B\subset A\}).$$

And finally the next observation: If $\{A_{i}\}_{i=1}^{n}$ are disjoint sets on $\mathcal{E},$ then $\sigma-$algebras $\mathcal{F}_{A_{1}},\ldots,\mathcal{F}_{A_{n}}$ are independent.

I don't get why the random variables $M(A_{1}),\ldots,M(A_{n})$ are independent.

I've tried to use the definition above but I don't get any useful.

Any kind of help is thanked in advanced.

  • 1
    $\begingroup$ This independence is a property of some random measures, most notably Poisson measures, but not of all. $\endgroup$
    – Did
    Mar 12 '18 at 7:00
  • $\begingroup$ Thanks @Did. I've seen this in definition of Poisson Random Measures. My doubt now comes from the counterexample given by Kavi Rama Murthy. Could you give me a hand with this? $\endgroup$
    – Squird37
    Mar 12 '18 at 7:14
  • $\begingroup$ Not a counterexample. The set of Poisson measures is a strict subset of the set of random measures, so one cannot expect properties of the former to be satisfied by every member of the latter. (Simply repeating my first comment, which you may not have read very carefully.) $\endgroup$
    – Did
    Mar 12 '18 at 7:16

This is obviously false: take $M(\omega,A)=\mu(A) X(\omega)$ where $X$ is a fixed positive random variable. A random measure which satisfies the independence condition stated above is called an independently scattered random measure.

  • $\begingroup$ Thanks to answer @KaviRamaMurthy. I see in your counterexample that $M$ is a random measure, but why this doesn't satisy the observation? Is there another definition of "random measure" which satisfies the observation? $\endgroup$
    – Squird37
    Mar 12 '18 at 6:30
  • $\begingroup$ Well, $M(A_j)$ are all multiples of the same random variable $X$ and so they are not independent. $\endgroup$ Mar 12 '18 at 8:28
  • $\begingroup$ Poisson random measure referred to by user Did is independently scattered.There is no contradiction between he is saying and what I am saying. $\endgroup$ Mar 12 '18 at 8:30
  • $\begingroup$ Thanks @KaviRamaMurthy. Now is clear. $\endgroup$
    – Squird37
    Mar 12 '18 at 13:48

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