1
$\begingroup$

Let's $\mu$ a probability measure on $(\Omega, \mathcal{F})$. Let's $\mathcal{J}_1\supset\dots\supset\mathcal{J}_n\supset\dots$ a sequence of sub-$\sigma$-fields of $\mathcal{F}$. If $A\in\mathcal{J}_n$ for all $n\in\mathbb{N}$ is well known that $ \mu(A|\mathcal{J}_n)=1_{A} $ Here $1_{A}$ is the caracteristic of $A$.

Question. How to prove that if $A\in\displaystyle\bigcap_{n\in\mathbb{N}}\mathcal{J}_n$ then
$ \mu(A|\displaystyle\bigcap_{n\in\mathbb{N}}\mathcal{J}_n)=1_A\;? $

  • 0
    Notice that $\mathcal{J} = \bigcap_{n \in \mathbb{N}} \mathcal{J}_n$ is a sub-$\sigma$-field. By your first statement, it follows that it is well known that $\mu(A|\mathcal{J}) = 1_A$.2012-04-12

1 Answers 1

1

We have to use and show the following:

If $\mathcal A\subset\mathcal F$ is a $\sigma$-algebra and $A\in\mathcal A$ then $\mu(A\mid\mathcal A):=E[\mathbb 1_A\mid\mathcal A]=\mathbb 1_A$. (we can write the condition expectation as $\mathbb 1_A$ is bounded, hence integrable)

As $\mathbb 1_A$ is $\mathcal A$-measurable, it's the best candidate to be a version of its conditional expectation.