This is an exercise in Durrett's probability book.

$p$ is the transition probability for a markov chain on a countable space. $f$ is said to be superharmonic if $f(x)\geq\sum_y p(x,y)f(y)$, or equivalently $f(X_n)$ is a supermartingale. Suppose $p$ is irreducible. If every nonnegative superharmonic function is constant, show that $p$ is recurrent.

It's not that easy to use the statement "every nonnegative superharmonic function is constant", so I tried 2 ways to reformulate the statement.

1.For all $f\geq 0$ nonconstant, there exists an $x$ s.t. $f(x)<\sum_y p(x,y)f(y)$. Show that $p$ is recurrent.

2.If $p$ is transient, show that there exists a nonnegative superharmonic function which is nonconstant.

But I haven't got a clue about how to prove this.


I think the easiest way is through strategy 2). Let $X$ be a transient irreducible Markov Chain. Consider an arbitrary $x_{0}$ and define:

$$\tau = \inf\{n \geq 0: X_{n} = x_{0}\}$$


$$f(x) = P(\tau < \infty|X_{0}=x)$$

By construction, $f(x) \in [0,1]$ and $f(x_{0}) = 1$. Since $X$ is transient, there exists $y$ such that $f(y) < 1$. Finally, by Markovianity, for any $x \neq x_{0}$,

$$f(x) = P(\tau < \infty|X_{0}=x) = \sum_{y}{p(x,y)P(\tau < \infty|X_{1}=y)} = \sum_{y}{p(x,y)f(y)}$$


$$f(x_{0}) = 1 \geq \sum_{y}{p(x,y)f(y)}$$

Hence, $f$ is super-harmonic and non-constant.

ps: the condition is actually necessary and sufficient. Can you prove the reverse?

  • 4
    $\begingroup$ Yes the condition is necessary and sufficient. I have proved the reverse and that's why I only asked the first part. In fact, if $p$ is recurrent and let $f$ be a nonnegative superharmonic function, then $f(X_n)$ is a nonnegative supermartingale thus $f(X_n)\rightarrow Y, a.s.$ As $p$ is recurrent, for all $x, P(f(X_n)=f(x), i.o.)=1$, thus $Y=f(x) a.s.$ As this is true for all $x$, $f$ is constant. $\endgroup$ – Iew Jul 3 '12 at 16:24
  • $\begingroup$ How do we take the last equality at the line after $x\neq x_0$? $\endgroup$ – perlman Nov 28 '17 at 0:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.