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Let $(X_n)_{n\in\mathbb{N}}$ be a Markov chain with State space $E=\{1,2,3\}$ and transission matrix

$$P=\begin{bmatrix} 0 & 1/3 & 2/3 \\ 1/4 & 3/4 & 0 \\ 2/5 & 0 & 3/5 \end{bmatrix}$$

How can I compute $E[X_{n+1}^3|X_n=j]$, for $j=1,2,3$.

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  • $\begingroup$ Looks like there's an error in your matrix as the rows should sum to $1$ $\endgroup$ – Simon S Nov 24 '14 at 18:51
  • $\begingroup$ Sorry, I corrected it! $\endgroup$ – Dimitri C Nov 24 '14 at 18:52
  • $\begingroup$ Just expand Expected value definition and as you know the condintional probability distribution function, you'll finish it : $$E[X_{n+1}^3|X_n=1]=\sum_{X_{n+1}=1}^3 X_{n+1}^3.P(X_{n+1}|X_n=1)$$ $\endgroup$ – Fardad Pouran Nov 24 '14 at 18:56
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Can't you just calculate it? For example,

$E[X_{n+1}^3 | X_n = 1] $

$= 1^3.Pr(X_{n+1} = 1 | X_n = 1) + 2^3.Pr(X_{n+1} = 2 | X_n = 1) + 3^3.Pr(X_{n+1} = 3 | X_n = 1)$

$ = 1 \cdot 0 + 8 \cdot (1/3) + 27 \cdot (2/3) $

$ = ...$

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  • $\begingroup$ Thanks, I was just not sure how to handle the power of 3. $\endgroup$ – Dimitri C Nov 24 '14 at 19:01

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