probability theory - I think I've found an invariant distribution for a transient discrete Markov chain - Where is my mistake?

Let

• $E$ be an at most countable set equipped with the discrete topology $\mathcal E$



• $X=(X_n)_{n\in\mathbb N_0}$ be a discrete Markov chain with values in $(E,\mathcal E)$, distributions $(\operatorname P_x)_{x\in E}$ and transition matrix $$p=\left(p(x,y)\right)_{x,y\in E}:=\left(\operatorname P_x\left[X_1=y\right]\right)\;.$$


• $\tau_x^1:=\inf\left\{n\in\mathbb N:X_n=x\right\}$ and $$\varrho(x,y):=\operatorname P_x\left[\tau_y^1<\infty\right]$$

A measure $\mu$ on $(E,\mathcal E)$ is called invariant $:\Leftrightarrow$ $$\mu p=\mu\;,\tag 1$$ where $$\mu p\left(\left\{x\right\}\right):=\sum_{y\in E}\mu\left(\left\{y\right\}\right)p(y,x)\;\;\;\text{for }y\in E\;.$$ If $\mu$ is a probability measure with $(1)$, it's called an invariant distribution. Moreover, $x\in E$ is called transient $:\Leftrightarrow$ $$\varrho(x,x)<1\;.$$

At the German Wikipedia page, they state, that if $X$ is transient, i.e. all states are transient, then there exists no invariant distribution.

However, I think, that I've found a counterexample:

• Consider the random walk on $\mathbb Z$ with transition probabilities $$p(x,x+1)=r\;\text{and}\;p(x,x-1)=1-r\;\;\;\text{for }x\in\mathbb Z$$ for some $r\in (0,1)$


• Let $$\mu_1\left(\left\{x\right\}\right):=1\;\text{and}\;\mu_2\left(\left\{x\right\}\right):=\left(\frac r{1-r}\right)^x\;\;\;\text{for }x\in\mathbb Z$$ and $\mu_1(\emptyset):=\mu_2(\emptyset):=0$

It's easy to see, that $\mu_1$ and $\mu_2$ both satisfy $(1)$. Moreover, if $r\ne 1/2$, the random walk is transient and $\mu_1\ne \mu_2$. In addition, each non-negative linear combination of $\mu_1$ and $\mu_2$ is an invariant measure, too.

While these combinations are no invariant distribution in general, at least $\mu_1$ should be one. So, I think I've found an invariant distribution of a transient discrete Markov chain. Is this the world's end or did I made a mistake?

Answer

You made more than one mistake. You mistranslated the German Wikipedia article – it doesn't talk about measures (Maß) but about distributions (Verteilung). You also seem to be equivocating between measures, probability measures and distributions in your arguments in English. The counting measure is indeed invariant under these transitions, but it's neither a probability measure nor a distribution.