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Are the following two the same:

$E[V(X_{t_{k+1}})|g(X_{t_{k+1}}),X_{t_k}]$

and

$E[E[V(X_{t_{k+1}})|g(X_{t_{k+1}})]|X_{t_k}]$

Where $X$ is Markov chain

$X_{t_k} \in \mathcal{R}^n$

$V: \mathcal{R}^n \rightarrow \mathcal{R}$

$g: \mathcal{R}^n \rightarrow \mathcal{R}$

V is some value function.

If not, what is the difference ? Any help would be really appreciated, even in reframing the question

Thanks!

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    What are $t_k$, $t_{k+1}$ and how are they linked?2012-06-06
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    $X$ is a stochastic path, which takes value $X_{t_k}$ at time $t_k$ and $X_{t_{k+1}}$ at $t_{k+1}.$ The events happen only at discrete time $\ldots,t_k,t_{k+1},\ldots$2012-06-06
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    No, the first is $V(X_{t_{k+1}})$. The second is a function of $X_{t_k}$. If your process was random walk, with $t_k = k$ the first would be $S_{k+1}$ and the second $S_k$.( and $V(S) = S$)2012-06-06
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    I changed the notation now.2012-06-06
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    Your edit, replacing $X_{t_{k+1}}$ by $g(X_{t_{k+1}})$ doesn't affect the counterexample, because you can take $g(x)=x$.2012-06-06
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    Thanks Nate, but here $X$ could be in $\mathcal{R}^n$ and both $g$ and $V$ are $\mathcal{R}^n\rightarrow\mathcal{R}$2012-06-06
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    The counterexample should still work, by allowing $g(x_1, \ldots, x_n) = x_1$, and $V(X_{t_{k+1}})$ to be a random walk in the first coordinate (or something similar). @mike's comment should then apply.2012-06-06

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