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Are the following two the same:




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


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What are $t_k$, $t_{k+1}$ and how are they linked? –  Davide Giraudo Jun 6 '12 at 11:50
$X$ 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 $\ldots,t_k,t_{k+1},\ldots$ –  Shash Jun 6 '12 at 11:57
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 $V(S) = S$) –  mike Jun 6 '12 at 11:58
I changed the notation now. –  Shash Jun 6 '12 at 12:03
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. –  John Engbers Jun 6 '12 at 14:41
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1 Answer

The first random variable carries more information that the second.

To see this, note that your setting is equivalent to the following: three random variables $U$, $W$ and $Z$, defined on the same probability space, $U$ integrable, are given and one is interested in $$ R=\mathrm E(U\mid W,Z),\qquad T=\mathrm E(\mathrm E(U\mid W)\mid Z). $$ Obviously, $R$ determines $T$ since $T=\mathrm E(\mathrm E(R\mid W)\mid Z)$, but there is no way to reconstruct $R$ from $T$.

Consider for example the case where $U=W+Z$ with $W$ and $Z$ independent and integrable. Then, $R=U$ and $T=\mathrm E(U)$.

Note: In your setting, $U=V(X_{t_{k+1}})$, $W=g(X_{t_{k+1}})$ and $Z=X_{t_k}$.

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