# Direct proof that the stationary distribution of an irreducible chain is unique

Exercise 1.13 in the Levin-Peres-Wilmer book, http://pages.uoregon.edu/dlevin/MARKOV/markovmixing.pdf asks for a direct proof that the stationary distribution of an irreducible chain is unique. Note that Corollary 1.17 ibid. gives a very slick proof of this fact, but it's not "direct". They give a hint as follows: "Given stationary distributions $\pi_1$ and $\pi_2$, consider the state $x$ that minimizes $\pi_1(x)/\pi_2(x)$ and show that all $y$ with $P(x, y) > 0$ have $\pi_1(y)/\pi_2(y)=\pi_1(x)/\pi_2(x)$."

I couldn't get it to work and would appreciate additional hints.

• I think you need to use the stationary equation $\pi(x) = \sum_z \pi(z) P(z,x)$ and also the fact that $a = b\frac{a}{b}$ for $b>0$. – Michael Jun 20 '17 at 21:04
• One problem I have with this method is: (i) It assumes there exists such a minimizing state $x$ (this is not obvious for infinite state spaces), (ii) I find it easier to show this is true for all $y$ with $P(y,x)>0$, perhaps your hint had a typo or there is a parallel way to do it that uses the $P(x,y)$. – Michael Jun 20 '17 at 21:17
• The chain is assumed to be on a finite state space. – Aryeh Jun 20 '17 at 21:33
• @Michael I'm afraid I still don't follow. Where do you use the minimality of $\pi_1(x)/\pi_2(x)$? – Aryeh Jun 21 '17 at 8:01
• Let $r = \min_z[a(z)/b(z)]$ so $a(z) \geq rb(z)$ for all $z$. – Michael Jun 21 '17 at 8:16

Let us give the full proof, which is the same as Aryeh's, but extends it so that it does not require that $P(z,x) > 0$ for all states $z$.

Hence, as in Aryeh's proof (repeated below for the sake of completeness), let us consider $a$ and $b$, two stationary distributions. Also let $x$ the minimizer of $z \mapsto a(z)/b(z)$ and call $r=a(x)/b(x)$. Clearly it has to be the case that $r < \infty$ since by stationarity and irreducibility it holds that $b(z) > 0$ for all $z$. We get:

\begin{aligned} a(x) &= \sum_z a(z)P(z,x) \\ &= \sum_{z} \frac{a(z)}{b(z)}b(z) P(z,x) \\ &\geq \sum_{z} r b(z) P(z,x)\\ &= r \sum_z b(z) P(z,x) \\ &= r b(x)\\ &= a(x) \end{aligned}

But for equality to hold throughout the whole expression, we need equality to hold for all summands in the inequality ($\geq$) of line 3.

This implies that $\frac{a(z)}{b(z)}P(z,x) = rP(z,x)$ for all $z$. In particular, for all $z$ such that $P(z,x)>0$ we get that $\frac{a(z)}{b(z)}=r$.

However, to conclude, we need this to hold for all $z$, not just the ones with $P(z,x) > 0$.

Let us pick another $z$ such that $P(z,x)=0$. By irreducibility of the chain, there exists however some $n$ such that $P^n(z,x) > 0$, where $P^n$ denotes the $n$-step transition probabilities.

Repeating the same argument as above with the $n$-step Chapman-Kolmogorov equations (i.e. $a(x) = \sum_z a(z)P^n(z,x)$), we see that also $a(z)/b(z)=r$.

Hence $a(z)/b(z)=a(x)/b(x)$ holds for all states $z$, and since $\sum a(z)=\sum b(z)=1$ we conclude that $a=b$.

Following Michael's suggestions in the comments, I think no explicit assumptions of the type $P(x,y)>0$ are needed. Denoting $a:=\pi_1$ and $b:=\pi_2$, we have $$a(x) = \sum_z a(z)P(z,x)\ge\sum_z rb(z)P(z,x)=rb(x).$$ But in fact $a(x)=rb(x)$, since $x$ was chosen to minimize the ratio $a(x)/b(x)$. This means that none of the inequalities $a(z)\ge rb(z)$ can be strong and in fact must all be equalities, whence $a=b$.

Edit: ack, the $P(z,x)>0$ assumption is needed to make that final conclusion.