2
$\begingroup$

The following quantity $\tilde\pi$ is defined in the textbook Markov chains and mixing times by David A. Levin. Here $\tau_z^+ = \min \left\{t\geq 1| X_t = z\right\}$.

Let $z \in \mathcal{X}$ be an arbitrary state of the Markov chain. We will closely examine the average time the chain spends at each state in between visits to $z$. To this end, we define $$ \begin{aligned} \tilde{\pi}(y) &:=\mathbf{E}_{z}(\text { number of visits to } y \text { before returning to } z) \\ &=\sum_{t=0}^{\infty} \mathbf{P}_{z}\left\{X_{t}=y, \tau_{z}^{+}>t\right\} \end{aligned} $$

I am having trouble understanding why the second equality above is true. I tried to use the fact that for non-negative integer valued random variables, $\mathbb{E}T = \sum_{t\geq0} \mathbb{P}(T>t).$ But I couldn't prove that. Any help is appreciated.

$\endgroup$

1 Answer 1

4
$\begingroup$

Let $N$ denote the number of visits to $y$ before returning to $z$. Let $I_t$ be the random variable that is 1 if $X_{t}=y, \tau_{z}^{+}>t$ and zero otherwise. Then by definition, $N = \sum_t I_t$.

This is an ubiquitous trick in probability. Express a random variable that counts something as a sum of 0/1 random variables (also called indicators).

$\endgroup$
1
  • $\begingroup$ Thank you. I get it now. $\endgroup$
    – Mathaddict
    May 23 at 11:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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