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

Does this answer your question?

  • $\begingroup$ 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 $[0,1]^10$. $\endgroup$ – user585926 Sep 6 '18 at 4:13
  • $\begingroup$ @user585926 Can you please use standard notation to describe what is meant by "$[0,1]^10$"? $\endgroup$ – grndl Sep 6 '18 at 8:42
  • $\begingroup$ Sorry, I meant $[0,1]^{10}$ $\endgroup$ – user585926 Sep 6 '18 at 15:11
  • $\begingroup$ @user585926 I think your most recent edit made your question less clear. You should try to formulate something that is mathematically precise. $\endgroup$ – grndl Sep 6 '18 at 17:41
  • $\begingroup$ I have edited again, but not sure if this is what you meant? $\endgroup$ – user585926 Sep 7 '18 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.