This is a question continued from here... Proof about Steady-State distribution of a Markov chain

I have a stochastic matrix $P_\delta$ of dimensions $n\times n$. I look at the matrix as a transition matrix of some Markov chain $M$. Matrix $P_\delta$ is dependent of $\delta\in(0,1]$. For this $P_\delta$ I would like to compute the steady state distribution $\pi_\delta$.

I am interested in elements of $\pi_\delta$ with respect to $\delta$. From my experiments I concluded, that for each $i$, the $\pi_{\delta}(i)$ is monotonic function of $\delta$ (either increasing or decreasing).

So for some $i\in[1,n]$ I can write a formula for $\pi_\delta(i)$:

$$ \pi_\delta(i)=\sum_{j=1}^n \pi_\delta(j) \frac{k(j)}{1+\delta k(j)} (a(j,i)+\delta I(j,i)) $$

Here $k(j)\in [1,10]$, $a(j,i)\in[1,10]$ and $I(j,i)\in\{0,1\}$ are constants. Also note that the right-hand side of the equation includes $\pi_\delta(i)$.

How can I conclude that for each i, the $\pi_\delta(i)$ is monotonic function of $\delta$? I think it is not enough to derive the right side by $\delta$ and show that sign of the derivative does not change, because $\pi_\delta(j)$ are dependent of $\delta$.

  • $\begingroup$ You also have to keep in mind, and use that $\sum_i \pi_\delta(i) = 1$ $\endgroup$
    – SBF
    Jan 9, 2013 at 13:30
  • $\begingroup$ Right. I totally missed that. So I can get rid of the $\pi_\delta(j)$ on the right-hand side. But is it enough to show that the derivative does not change sign? $\endgroup$
    – Nejc
    Jan 9, 2013 at 13:33


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