# Detailed balance distribution reflecting a random walk

This above example happens in an infinite state space $S = {i}_{i=1}$

By the definition of detailed balance condition,

Definition: $\pi\left ( x \right )p\left ( x,y \right ) = \pi\left ( y \right )p\left ( y,x \right )$

The author asserts that the initial distribution given in the example satisfies the detailed balance condition but this is not clear and when I try to attempt, I was unable to show this is true.

Secondly, there is an exponent $i$ in the explicit form for $\pi$; where does this exponent come from?

• Is there no one who knows? The stochastic/ markov tag seems to be very slow moving... – Physkid Mar 8 '18 at 10:36
• What initial distribution are you talking about? The author says that the chain is a birth-death process and thus will satisfy the detailed balance equations. Also, it's been only 3 hours since you asked the question. You must have more patience than that. – Greenparker Mar 8 '18 at 12:05
• @Greenparker That explains. May I know why the exponent i exists? – Physkid Mar 8 '18 at 14:07
• Welcome to CV @Physkid Sometimes good questions can literally take months or years to receive good answers on this site. Please have patience. :) – Alexis Mar 8 '18 at 16:53
• @Alexis Thank you for telling me that. I was under the impression that the question was "unanswerable" due certain constraints. – Physkid Mar 9 '18 at 6:37

Thus for all states $i, j$, $\pi(i) P(i,j) = \pi(j) P(j,i)$, where $\pi$ is the unique stationary distribution of this process defined for all $i \geq 0$.
For $j = i+1$, $i \geq 0$ we have that \begin{align*} \pi(i) P(i,i+1) &= \pi(i+1) P(i+1,i) \\ \Rightarrow p \pi(i) &= (1-p)\pi(i+1) \\ \Rightarrow \pi(i+1) & = \pi(i) \left(\dfrac{p}{1-p}\right)\,. \end{align*}
This is what they do in the example. But note that $i$ is chosen generically, so this holds for all $i \geq 0$. That is, \begin{align*} \pi(i) & = \pi(i-1) \left(\dfrac{p}{1-p}\right)\\ & = \pi(i-2) \left(\dfrac{p}{1-p}\right)^2 \\ & =\pi(i-3) \left(\dfrac{p}{1-p}\right)^3\\ & \vdots\\ & = \pi(0) \left(\dfrac{p}{1-p}\right)^i\\ & = c\left(\dfrac{p}{1-p}\right)^i\,. \end{align*} So the exponent $i$ comes from an iterative argument.