Question 11.5.19: Show that, for an ergodic Markov chain (see Theorem 11.16), ......
Show that, for an ergodic Markov chain (see Theorem 11.16),
\sum\limits_{j}{m_{ij}w_j} = \sum\limits_{j}{z_{jj}-1}= K.
The second expression above shows that the number K is independent of i. The number K is called Kemeny’s constant. A prize was offered to the first person to give an intuitively plausible reason for the above sum to be independent of i. (See also Exercise 24.)
Recall that
m_{ij}= \sum\limits_{j}{\frac{z_{jj}-z_{ij}}{w_j} }.
Multiplying through by wj summing on j and, using the fact that Z has row sums 1, we obtain
m_{ij}= \sum\limits_{i}{z_{jj}}-\sum\limits_{j}{z_{ij}}= \sum\limits_{j}{z_{jj}}-1 = K,
which is independent of i.
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