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I'm struggling with the understanding of the properties of a Radon-Nykodym Derivative process. In my example we defined the probability measure $\mathbb{Q}[A]=\int_A Z(\omega) d\mathbb{P}(\omega)$. Then Z is the Radon-Nykodym derivative of $\mathbb{Q}$ w.r.t. $\mathbb{P}$ and $Z_t=E_\mathbb{P}[Z|\mathcal{F}_t]$ is a Radon-Nykodym derivative process, I saw that then the conditional expectation: $$E_\mathbb{Q}[Y|\mathcal{F}_t]=\frac{1}{Z_t} E_\mathbb{P} [YZ_s|\mathcal{F_t}]$$ asuming here that Y is $\mathcal{F}_s$ measurable where $s \geq t$.

I startet with a event $A \in \mathcal{F}_t$. Then: $$ E_\mathbb{Q}[Y|\mathcal{F}_t]=E_\mathbb{Q}[Y 1_A]=E_P[YZ_t1_A]=E_P[E_P[YZ_t1_A|\mathcal{F_t}]$$ But then I couldn't do it further, because then I would take out $Z_t$ but this would make no sense, when looking at the equation. Could somebody help me? Thank you!

Thank you!

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  • $\begingroup$ This question is missing a lot of important context. How are the measures $\mathbb{P}$ and $\mathbb{Q}$ related? What is $Z_t$? I can make guesses based on what you're trying to do but you should really include those details in the question. $\endgroup$ Commented Apr 13, 2018 at 10:58
  • $\begingroup$ Hi nobody thanks for your reply, I did some changes, I hope the question is now more clear. Thank you :) $\endgroup$ Commented Apr 13, 2018 at 12:41
  • $\begingroup$ Shouldn't $A \in \mathcal F_t$? $\endgroup$ Commented Apr 13, 2018 at 12:56
  • $\begingroup$ exactly, sorry. I‘ve corrected it $\endgroup$ Commented Apr 13, 2018 at 13:11
  • $\begingroup$ You should define all the notations. For example, let $(\Omega,\mathcal{F}, \mathbb{P})$ be a probability space and let $\{\mathcal{F}_t\mid t\geq 0\}$ be a filtration... Moreover, whether $\mathbb{Q}$ is given at the very beginning or it is defined via $Z$. What is $Z$? Is $Z:\mathcal{\Omega}\rightarrow \mathbb{R}$ and $\mathcal{F}/\mathcal{B}(\mathbb{R})$ measurable? or $\mathcal{G}/\mathcal{B}(\mathbb{R})$ for some sub $\sigma$-algebra $\mathcal{G}\subseteq \mathcal{F}$. $\endgroup$ Commented Apr 13, 2018 at 16:05

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First notice that $$\mathbb{E}_\mathbb{P}[YZ_s | \mathcal{F}_t] = \mathbb{E}_\mathbb{P}[ \mathbb{E}_\mathbb{P}[YZ | \mathcal{F}_s] | \mathcal{F}_t] = \mathbb{E}_\mathbb{P}[YZ | \mathcal{F}_t]$$ by the tower law, where the first equality follows from the definition of $Z_s$ and the fact that $Y$ is $\mathcal{F}_s$ measurable.

Let $C_t = \frac{1}{Z_t} \mathbb{E}_\mathbb{P}[YZ | \mathcal{F}_t]$. Then $C_t$ is $\mathcal{F}_t$ measurable so $$\mathbb{E}_\mathbb{P}[ Z C_t | \mathcal{F}_t] = C_t \mathbb{E}_\mathbb{P}[Z | \mathcal{F}_t] = C_t Z_t = \mathbb{E}_\mathbb{P}[YZ | \mathcal{F}_t].$$

We would like to show that $\mathbb{E}_\mathbb{Q}[Y | \mathcal{F}_t] = C_t$. That is, we need to check that for every $A \in \mathcal{F}_t$ we have $ \mathbb{E}_\mathbb{Q}[ {1}_A C_t] = \mathbb{E}_\mathbb{Q} [ {1}_A Y]$. Notice that we do not expect to have $\mathbb{E}_\mathbb{Q}[Y | \mathcal{F}_t] = \mathbb{E}_\mathbb{Q}[Y 1_A]$ as you have written since the left hand side is a random variable whilst the right hand side is just a number.

So we compute for $A \in \mathcal{F}_t$, $$\mathbb{E}_\mathbb{Q}[1_A C_t] = \mathbb{E}_\mathbb{P} [1_A Z C_t] = \mathbb{E}_\mathbb{P} [ 1_A \mathbb{E}_\mathbb{P} [ ZC_t | \mathcal{F}_t]] = \mathbb{E}_\mathbb{P} [ 1_A \mathbb{E}_\mathbb{P} [ZY | \mathcal{F}_t]] = \mathbb{E}_\mathbb{P} [ YZ 1_A ] = \mathbb{E}_\mathbb{Q} [ 1_A Y ]$$ as desired.

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  • $\begingroup$ @TheoreticalEconomist whoops, that was just a typo (now fixed). Thanks for pointing it out! $\endgroup$ Commented Apr 13, 2018 at 15:52
  • $\begingroup$ Looks good now! +1 $\endgroup$ Commented Apr 13, 2018 at 15:54
  • $\begingroup$ Thank you for the great response, I think I get it know $\endgroup$ Commented Apr 14, 2018 at 7:13
  • $\begingroup$ @raphim If you're happy with the answer, you should accept the answer by clicking on the check mark next to it. $\endgroup$ Commented Apr 22, 2018 at 15:37
  • $\begingroup$ did that now, sorry didn't know about this feauture :) $\endgroup$ Commented Apr 23, 2018 at 16:52

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