# Turn two probability spaces with $\mathbb {P_1,P_2}$ into one probability space with one $\mathbb P$

Assume we have two arbitrary probability measures $\mathbb P_1, \mathbb P_2$ on the same arbitrary measurable space $(\Omega, \mathcal B)$. (Or more generally, $n$ such probability measures).

I want to generate a new probability space that captures the same information, but that captures the information in $\mathbb {P_1,P_2}$ as conditional distributions generated by information $I_1,I_2$ on the same prior distribution $\mathbb P$.

My intuition is that we can generate a new probability space $(\Omega',\mathcal B',\mathbb P)$, and for any variable $V:\Omega\to X$, we can define $V':\Omega'\to X$, and then choose $I_1, I_2$ to partition $\Omega'$ such that for any event in the original space $E\in \mathcal B$, there is a natural corresponding $E'\in\mathcal B'$, where we have $$\mathbb P_i(E)=\mathbb P(E'|I_i)$$

Edit: I want the information in $\mathbb P_1,\mathbb P_2$ to be represented in $I_1,I_2$, not in $\mathbb P$. That is $\mathbb P$ should induce a "uniform" distribution on $\Omega$: e.g. if $\Omega=[0,1]^{10}$, then $\mathbb P$ should induce a uniform distribution on $[0,1]^{10}$, and the $I_1,I_2$ should capture the "non-uniformity" in $P_1,P_2$.

This would mean that we can fully represent the two probability distributions as conditional distributions with the same prior.

• Is this indeed possible? I am not sure how to actually do this. It seems to me that $\Omega'$ needs to be much bigger than $\Omega$ in order to capture the two distributions.

• What is the cleanest way to do this?

If I've understood your question, yes, this is possible.

Let $\Omega' = \Omega \times \{i_1,i_2\}$. Let $\mathcal{B}' = \mathcal{B} \otimes 2^{\{i_1,i_2\}}$.

Let $A \in \mathcal{B}$, and define $Q(A \times \{i_1\})=1/2 P_1(A)$ and $Q(A \times \{i_2\}) = 1/2 P_2(A)$. You can verify that $Q$ so defined extends to all of $\mathcal{B}'$.

Associate with each event $A \in \mathcal{F}$, the event $A'= A \times \{i_1, i_2\} \in \mathcal{B}'$. Let $I_j = \Omega \times \{i_j\}$, $j=1,2$.

Finally, verify that $Q(A' \mid I_j) = P_j(A)$, $j=1,2$.

• Ah I see, I didn't think of this. Technically it answers my question as stated, but this is not what I intended, because $P_1$ and $P_2$ are being represented in $Q$. i.e. you're constructing $Q$ specifically to capture the information in $P_1,P_2$. Instead I intended $Q$ (what I call $P$) to be a vanilla, simple probability measure that is "uniform on $\Omega'$ ". e.g. if $\Omega= [0,1]^10$, then $P$ should be uniform on $[0,1]^10$. – user585926 Sep 6 '18 at 4:13
• @user585926 Can you please use standard notation to describe what is meant by "$[0,1]^10$"? – grndl Sep 6 '18 at 8:42
• Sorry, I meant $[0,1]^{10}$ – user585926 Sep 6 '18 at 15:11