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At the bottom he writes, "the stationary distribution is actually independent of adding a constant diagonal (identity) matrix," but I'm not sure how that could
by frei 10y ago
At the bottom he writes, "the stationary distribution is actually independent of adding a constant diagonal (identity) matrix," but I'm not sure how that could be true (intuitively it doesn't make sense, but I don't know the math)
edit: A identity matrix wouldn't affect the stationary distribution, but if you had the actual "stay" probabilities they wouldn't all be the same, and thus not an identity matrix at all.
- The_suffocated 10y agoAdding a constant diagonal matrix indeed would not affect the equilibrium distribution. Mathematically, if the original transition matrix is A and you add a multiple of the identity matrix to it, then after normalisation by row sums, the new transition matrix becomes tA+(1-t)I for some 0<=t<=1. Since the Perron vector x of A corresponds to the eigenvalue 1, it must also be an eigenvector of tA+(1-t)I, because (tA+(1-t)I)x=tAx+(1-t)x=tx+(1-t)x=x. Hence the stationary distribution remains unchanged. But as you said, the stationary distribution does change if the missing diagonal is not constant. And there is no reason to believe it is constant in the first place. In the end, what the author measures is still different from what he thinks he measures.
- tgb 10y agoI agree with you, but your last line is wrong: the footnotes make it clear the author knows the omission.