What is $E[X_{n+1}|X_0 = i_0, . . . , X_{n−1} = x_{n−1}, X_n = i]??$ Given that $\sum_j^n jp_{i,j}=i$.
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As written, the quantity whose value you're asking about is not clear. Is it $\ E\left[X_{n+1}|X_0 = i_0, \dots , X_{n−1} = x_{n−1}, X_n = i\right]\ $? – lonza leggiera Mar 09 '20 at 05:00
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@lonzaleggiera sorry edited for clarity. But yes that is what I am asking! – confused doughnut Mar 09 '20 at 05:02
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Hint: Well, since it's a Markov chain, $$\ E\left[X_{n+1}\,|\,X_0 = i_0, \dots , X_{n−1} = x_{n−1}, X_n = i\right]= E\left[X_{n+1}\,\left|\, X_n = i\right.\right]\ .$$ That is, you can ignore all the conditioning equations except the last. And $$ E\left[X_{n+1}\,|\, X_n = i\right]=\sum_{j=1}^njP\left[X_{n+1}=j\,|\,X_n=i\right] $$ Do you know how to express $\ P\left[X_{n+1}=j\,|\,X_n=i\right]\ $ in terms of the transition matrix (whose entries you've implicitly given as $\ p_{i,j}\ $)?
lonza leggiera
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thank you so much for the explanation. I see the answer is just i now. But to clarify you last statement about expressing isn't it just $p_i,j$ in the transition matrix? – confused doughnut Mar 09 '20 at 05:24
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Yes, by definition, $\ p_{i,j}=P\left[X_{n+1}=j,|,X_n=i\right]
$. – lonza leggiera Mar 09 '20 at 05:35