Find $\mathbb{P}_{\mathcal{F}}:2^{\{x_1,x_2,\dots\}}\times\Omega\rightarrow[0,1]$ s.t. $\mathbb{P}_{\mathcal{F}}(\cdot,\omega)$ is a prop.-measure

Question

Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space, $\mathcal{F}\subseteq\Sigma$ a $\sigma$-Algbra and $X:\Omega\rightarrow\mathcal{X}$ a random variable with a countable set $\mathcal{X}=\{x_1,x_2,\dots\}$. Show that there exists a mapping $\mathbb{P}_{\mathcal{F}}:\mathcal{P}(\mathcal{X})\times\Omega\rightarrow[0,1]$, such that ...

1. $\mathbb{P}_{\mathcal{F}}(\cdot,\omega)$ is a propability measure on $\mathcal{P}(\mathcal{X})$ for all $\omega\in\Omega$ and...

2. $\mathbb{P}_{\mathcal{F}}(A,\cdot)$ is a conditinal expectation of $\mathbb{I}_{\{X\in A\}}$ for each $A\in\mathcal{P}(\mathcal{X})$.

Im racking my head on how to properly define such a mapping and solve the Problem.

I think we can say a.s. that $E[1_{\{X=x_i\}}|\mathcal{F}]$ is a positiv probability-sequence, which adds up to 1. So i tried to use this sequence to construct a probability measure $\mathbb{P}_{\mathcal{F}}(\cdot,\omega)$ for almost all $\omega\in\Omega$. But so far I haven't got anywhere.

I tried to read more literature to get closer to the problem but without success. Any assistance or thoughts would be much appreciated.

Answer 1

There exists a $\Omega'\in\mathcal{F}$ with $\mathbb{P}(\Omega')=1$, such that for all $\omega\in\Omega'$:

$$\sum_{x_i\in\mathcal{X}}E[\mathbb{I}_{\{X=x_i\}}|\mathcal{F}]=E[\sum_{x_i\in\mathcal{X}}\mathbb{I}_{\{X=x_i\}}|\mathcal{F}]=E[\mathbb{I}_{\{X\in\mathcal{X}\}}|\mathcal{F}]=1$$

(It should be noted that we used that $\mathcal{X}$ is countable to switch the sum and the integration of the expected value). Because $E[Y|\mathcal{F}]\geq0$ holds for $Y\geq0$ a.s. the sequence $E[\mathbb{I}_{\{X=x_i\}}|\mathcal{F}]$ (with $i\in\mathbb{N}$) is a propability sequence defined on $\Omega'$.

Now we define $\mathbb{P}_{\mathcal{F}}(A,\omega')=\sum_{x_i\in A}E[\mathbb{I}_{\{X=x_i\}}|\mathcal{F}](\omega')$ for all $\omega'\in\Omega'$ and $\mathbb{P}_{\mathcal{F}}(A,\omega)$ as a arbitrary propability measure on $2^{\mathcal{X}}$ for all $\omega\in\Omega\setminus\Omega'$ (for example $\mathbb{P}_{\mathcal{F}}(A,\omega)=\mathbb{I}_{\{x_1\}}(A)$). Thus $\mathbb{P}_{\mathcal{F}}(A,\omega)$ fulfills the property (1).

For property (2) we have to show that $E[\mathbb{I}_{F}\mathbb{P}_{\mathcal{F}}(A)]=E[\mathbb{I}_{F}\mathbb{I}_{\{X\in A\}}]$for all $F\in\mathcal{F}$. This is true because ...

$$E[\mathbb{I}_{F}\mathbb{P}_{\mathcal{F}}(A)]=E[\mathbb{I}_F\sum_{x_i\in A}E[\mathbb{I}_{\{X=x_i\}}|\mathcal{F}]]=E[\mathbb{I}_FE[\mathbb{I}_{\{X\in A\}}|\mathcal{F}]]=E[\mathbb{I}_F\mathbb{I}_{\{X\in A\}}]$$

... holds in $\Omega'$ and $\mathbb{P}(\Omega\setminus\Omega')=0$. Again the fact that $\mathcal{X}$ is countable is used.