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can does anybody know if the following expectations are available in closed for...

Let $\{ X_t : t = 1, 2, 3 \dots \}$ be a random variable defined on a Markov chain with m -step transition matrix $P_m^{i,j}$. I'm trying hunting for a closed form expression for the following expectations which are telescoping:

$m = 1 : E_t \Big [ \frac{1}{X_{t+1}} \Big ]$

$m = 2 : E_t \Big [ \frac{1}{X_{t+1}} \frac{1}{X_{t+2}} \Big ]$

$m = 3 : E_t \Big [ \frac{1}{X_{t+1}} \frac{1}{X_{t+2}} \frac{1}{X_{t+3}}\Big ]$

A clue would be lovely - thanks.

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ok, or even more generally, how would one find : $E_t [\prod_{i=0}^k f(x_{t+i})] $ –  Luap Nalehw Oct 11 '12 at 16:57
    
The problem formulation is not clear. If $X$ is a Markov Chain? Does it take values over reals (otherwise, what is $1/X$)? What is $E_t$? –  S.D. Dec 4 '12 at 7:37
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1 Answer

By definition, $$ E\Big [ \frac{1}{X_{t+1}} \frac{1}{X_{t+2}} \Big ]=\sum_{x,y,z}P[X_t=x]\frac{P_1^{x,y}P_1^{y,z}}{yz}, $$ where $$ P[X_t=x]=\sum_{u}P[X_0=u]P_t^{u,x}. $$

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