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Learning abstract conditional expectations has me a bit confused. If we take a process $Y$, let $\mathcal{F}$ equal its internal filtration, then why is $E[Y_t | \mathcal{F_s}] = E[Y_t | Y_s]$?

The first is the expectation of $Y_t$ given ALL the information obtained by observing the process until time $s$, yet the right hand side is the expectation of $Y_t$ given ONLY the information avaliable at time $s$. Clearly they are two different objects, no?

An example of this in use:

https://www.stat.cmu.edu/~cshalizi/754/notes/lecture-17.pdf

On page 3, in the proof of prop 178, in the first line, (17.6), it goes from conditioning on a $\sigma$-algebra to conditioning on a single variable.

ImeanH
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1 Answers1

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In general $ E[Y_t | \mathcal{F}_s] \neq E[Y_t|Y_s] $.

As, $ \mathcal{F}_s = \sigma(\{ Y_r, r \leq s \}) $ depends in the knowledge of all previous states.

Nevertheless when a process satisfies this relation it is said to satisfy the Markov property.

In the example you show, they are considering a Wiener process (also called standard Brownian Motion), which is a process known to satisfy the Markov property.

Edit: I was a little off in the def of the Markov Property, check the comments.

Joaquin San
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  • By the way, $ E[Y_t | \mathcal{F}_s] = E[Y_t|Y_s] $ is not at all what one calls the Markov property. – Did Feb 28 '17 at 22:08
  • Right!, you require $ \mathbb{P}( Y_t \in A | \mathcal{F}_s ) = \mathbb{P}( Y_t \in A | Y_s ) $ – Joaquin San Mar 01 '17 at 22:17