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If $p(i)$ is the probability of state $i$ with $p(i)>0$ and $p_{ij}=p(j.i^{-1})$ are transition probabilities where product and inverse refer to group operations, how can I show that the transition probability matrix is doubly stochastic?

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Note that to show that the matrix $P$, with coordinates $p_{ij}$, is doubly stochastic, it is enough to show that the values of each rows and each column sum to $1$.

For this, observe that $\sum\limits_i p(i) = 1$.

Now consider the $j'$th column, the sum of of elements in this column is given by $$\sum\limits_i p_{ij} = \sum\limits_i p(j \cdot i^{-1}) = \sum\limits_{j^{-1}\cdot i^{-1}|i \in G} p(i) = 1,$$ where the second equality is just a change of variables.

The same logic also holds for the sum of rows, which gives the desired result.

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  • $\begingroup$ I'm not sure I understand the second equality. Could you please break it down or explain for me? Should the range of the sum be $j^{-1}\cdot i|i \in G$? How does it sum to 1? I think I'm quite naive! $\endgroup$ – Blain Waan Oct 17 '18 at 21:21
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    $\begingroup$ The main idea is that to go over all elements in the group of the type $j \cdot i^{-1}$ (where $j$ is fixed and $i$ is every possible element in $G$), is exactly like simply going over all elements in the group. To see this, for $g \in G$, just take $i = g^{-1} \cdot j$, so that $j\cdot i^{-1} = g$. $\endgroup$ – Cain Oct 17 '18 at 21:29

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