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Daniele Tampieri
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In this paper by Diaconis and Zabell from 1982, Theorem 2.1 and the remark after essentially stated that

Given two probability measures $P$ and $Q$ on the same probability space $\Omega$. If $Q\ll P$ and their RN-derivative $\frac{dQ}{dP} \in L^\infty$ , then $Q$ can be obtained from $P$ by conditioning.

TheTheir proof is very easy and the assumption $\frac{dQ}{dP} \in L^\infty$ is essential, for it. However I was wondering if it is possible to relax the assumption $\frac{dQ}{dP} \in L^\infty$ or if this is a necessary condition, is it? I was looking at concrete cases of this problem, for example between $\text{Exp}(a)$ and $\text{Exp}(b)$.

I know very little about the literature on this topic, any suggestion is also appreciated.

In this paper by Diaconis and Zabell from 1982, Theorem 2.1 and the remark after essentially stated that

Given two probability measures $P$ and $Q$ on the same probability space $\Omega$. If $Q\ll P$ and their RN-derivative $\frac{dQ}{dP} \in L^\infty$ , then $Q$ can be obtained from $P$ by conditioning.

The proof is very easy and the assumption $\frac{dQ}{dP} \in L^\infty$ is essential, I was wondering if it is possible to relax the assumption $\frac{dQ}{dP} \in L^\infty$ or if this is a necessary condition? I was looking at concrete cases of this problem, for example between $\text{Exp}(a)$ and $\text{Exp}(b)$.

I know very little about the literature on this topic, any suggestion is also appreciated.

In this paper by Diaconis and Zabell from 1982, Theorem 2.1 and the remark after essentially stated that

Given two probability measures $P$ and $Q$ on the same probability space $\Omega$. If $Q\ll P$ and their RN-derivative $\frac{dQ}{dP} \in L^\infty$ , then $Q$ can be obtained from $P$ by conditioning.

Their proof is very easy and the assumption $\frac{dQ}{dP} \in L^\infty$ is essential for it. However I was wondering if it is possible to relax the assumption $\frac{dQ}{dP} \in L^\infty$ or if this is a necessary condition, is it? I was looking at concrete cases of this problem, for example between $\text{Exp}(a)$ and $\text{Exp}(b)$.

I know very little about the literature on this topic, any suggestion is also appreciated.

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Xiao
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Radon-Nikodym derivative and conditional probability

In this paper by Diaconis and Zabell from 1982, Theorem 2.1 and the remark after essentially stated that

Given two probability measures $P$ and $Q$ on the same probability space $\Omega$. If $Q\ll P$ and their RN-derivative $\frac{dQ}{dP} \in L^\infty$ , then $Q$ can be obtained from $P$ by conditioning.

The proof is very easy and the assumption $\frac{dQ}{dP} \in L^\infty$ is essential, I was wondering if it is possible to relax the assumption $\frac{dQ}{dP} \in L^\infty$ or if this is a necessary condition? I was looking at concrete cases of this problem, for example between $\text{Exp}(a)$ and $\text{Exp}(b)$.

I know very little about the literature on this topic, any suggestion is also appreciated.