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I am having some trouble conceptualizing and calculating a conditional RN derivative. When using this definition:

RN Theorem

I can see that if $\mathbb{Q} \ll \mathbb{P}$: $$\mathbb{E}_\mathbb{Q}(g) = \int g(\omega) \mathbb{Q}(\mathsf d\omega) = \int g(\omega) f(\omega) \mathbb{P}(\mathsf d\omega) = \mathbb{E}_\mathbb{P}(gf)$$

where $f$ is the RN derivative defined by: $$f = \frac{\mathsf d\mathbb{Q}}{\mathsf d\mathbb{P}}$$

Im having trouble with the following relation: $$\mathbb{E}_\mathbb{Q}(g\mid\mathcal{Y}) \stackrel{?}{=} \mathbb{E}_\mathbb{P}(g \tilde{f}\mid \mathcal{Y})$$

where $\tilde{f}$ is something like a conditional RN derivative: $$\tilde{f} \stackrel{?}{=} \frac{d\mathbb{Q}_{(\cdot \mid \mathcal{Y})}}{d\mathbb{P}_{(\cdot \mid \mathcal{Y})}} \stackrel{?}{=} \frac{\frac{d\mathbb{Q}}{d\mathbb{P}}}{?}$$

I tried searching for "conditional RN derivative" and the like but couldn't find an appropriate source. If someone could help me conceptually understand as well as mechanically calculate this I'd very much appreciate it!

EDIT: To clarify, If I start from $\mathbb{P}$ and $\mathbb{Q}$ and form the conditional measures (from Bayes rule) $\mathbb{P}(\cdot | \mathcal{Y})$ and $\mathbb{Q}(\cdot | \mathcal{Y})$ after gathering some evidence/observations, I know I can write:

$$\mathbb{E}_{\mathbb{Q}(\cdot|\mathcal{Y})}(g|\mathcal{Y}) = \int g(\omega) \mathbb{Q}( \cdot | \mathcal{Y})$$

What is the $\tilde{f}$ that makes the following true?

$$\mathbb{E}_{\mathbb{Q}(\cdot|\mathcal{Y})}(g|\mathcal{Y}) = \mathbb{E}_{\mathbb{P}(\cdot | \mathcal{Y})}(g \tilde{f} | \mathcal{Y})$$

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  • $\begingroup$ $\Bbb{Q}(\cdot \mid {\mathcal{Y}})$ is not a measure. It is a random variable. So one cannot treat it as a measure. Maybe I missed your point? $\endgroup$ – Conrado Costa Jul 23 '15 at 17:56
  • $\begingroup$ To compute$\mathbb{E}_{\mathbb{Q}}(g \mid \mathcal{Y})$, you do not need $\mathbb{Q}(\cdot \mid \mathcal{Y})$. Here, $\mathbb{Q}(\cdot \mid \mathcal{Y})$ is usually called the regular conditional probability, which does not necessarily exist. $\endgroup$ – Gordon Jul 23 '15 at 18:29
  • $\begingroup$ Which is why we can use the RN derivative described in your answer to compute it under $\mathbb{P}(\cdot|\mathcal{Y})$ correct? $\endgroup$ – Diego Jul 26 '15 at 21:04
  • $\begingroup$ That is correct. $\endgroup$ – Gordon Jul 26 '15 at 22:05
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Here, \begin{align*} \tilde{f} = \frac{f}{\mathbb{E}_{\mathbb{P}}(f \mid \mathcal{Y})}. \end{align*} That is, \begin{align*} \mathbb{E}_{\mathbb{Q}}(g \mid \mathcal{Y}) = \frac{\mathbb{E}_{\mathbb{P}}(f g\mid \mathcal{Y})}{\mathbb{E}_{\mathbb{P}}(f \mid \mathcal{Y})}. \end{align*} This is the "abstract Bayes formula".

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  • $\begingroup$ Thank you very much! I found this under some measure theoretic lecture notes. If you dont' mind, can you provide some intuition as to how that form relates to the "traditional Bayes" ideas of a "prior", "likelihood" and "posterior"? Thanks! $\endgroup$ – Diego Jul 26 '15 at 21:06
  • $\begingroup$ This result is proved in Robert J. Elliott, Lakhdar Aggoun, John B. Moore: Hidden Markov Models Estimation and Control as Theorem 3.2 (Conditional Bayes Theorem) on p23: users.cecs.anu.edu.au/~john/papers/BOOK/B05.PDF $\endgroup$ – Tom Ellis May 31 '17 at 20:30

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