If we have $a(x)$, $b(x)$ which are invariant distributions on a Markov chain $X_n$ with state space $S$, how can I prove that $|a(x)-b(x)|$ is also invariant?
I know that I must show that:
$\sum_{x \in S} |a(x)-b(x)|p(x,y) = c(y)$
where $p(x,y)$ is the transition function for $X_n$.
So what I did was I partitioned $S$ such that $A \subset S$ where if $a(x) \geq b(x)$ then $x \in A$. So that gets me:
$\sum_{x \in A}a(x)p(x,y) - \sum_{x\in A}b(x)p(x,y) - \sum_{x\in A^c}a(x)p(x,y)+\sum_{x\in A^c}b(x)p(x,y)$
But I do not know where to go from here... My first thought was to combine summations from complementary sets but I am not sure how to do that to make the whole thing either constant or constant as a function of $y$.
Thank you.
For a vector $h = (h_1,…,h_n)$, define $|h|$ as the entrywise absolute value: $$ |h| = (|h_1|, …, |h_n|) $$
Now, for example, if $z=(1/2,1/2,0,0,0)$ and $w=(0,0,1,0,0)$ then for any scalars $\alpha, \beta$ we have: $$ |\alpha z + \beta w| = |\alpha| z + |\beta| w $$ and if $z^T = z^T P$ and $w^T = w^T P$ then: $$ |\alpha z + \beta w|^T P = |\alpha|z^T P + |\beta| w^T P = |\alpha | z^T + |\beta| w^T = |\alpha z + \beta w|^T $$
This example can be generalized to see that the absolute value of linear combinations of invariant distributions with disjoint support indices are also invariant under multiplication by $P$.