# 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?

• No, I've not mistranslated the article. Please notice, that I distinguish between invariant measures and distributions in my question. And I don't equivocate any terms. The only mistake I've made, is that I thought $\mu_1$ would be a probability measure. For some reason, I thought it would assign a constant value of $1$ to any measurable set, but that's obviously not the case. I've defined $\mu_1$ only for the unit sets (since that determines this measure uniquely) and I think, that this led to the confusion. – 0xbadf00d Oct 20 '15 at 10:21
• @Did I've no idea what you mean. As I've said before, I thought $\mu_1$ is a probability measure (which is obviously wrong). Since $\mu_1$ is invariant, it would be called an invariant distribution. – 0xbadf00d Oct 20 '15 at 11:26