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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – 0xbadf00d Oct 20 '15 at 10:21
  • $\begingroup$ @0xbadf00d Why argue? The paragraph "Existenz und Eindeutigkeit" starts with: "Im Allgemeinen müssen keine stationäre Verteilungen existieren. Beispiel hierfür sind transiente Markow-Ketten. Diese besitzen nie stationäre Verteilungen." $\endgroup$ – Did Oct 20 '15 at 10:46
  • $\begingroup$ @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. $\endgroup$ – 0xbadf00d Oct 20 '15 at 11:26
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    $\begingroup$ @0xbadf00d Are you for real? You: "At the German Wikipedia page, they state, that if X is transient, i.e. all states are transient, then there exists no invariant measure." joriki: "You mistranslated the German Wikipedia article – it doesn't talk about measures (Maß) but about distributions (Verteilung)." You: "I've not mistranslated the article." WP: "Im Allgemeinen müssen keine stationäre Verteilungen existieren." O well... $\endgroup$ – Did Oct 20 '15 at 22:35
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    $\begingroup$ @0xbadf00d: A typo is when you type one or two letters wrong. Calling a distribution a measure is not a typo, and especially not when these are the central concepts that you're writing about and you're quoting a text in translation. $\endgroup$ – joriki Oct 21 '15 at 17:37

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