1
$\begingroup$

Let $\left\{X_{n}\right\}_{n\geq0}$ be a homogeneous markov chain with state space E and transition matrix P. Let $\tau$ be the first time n for which $X_{n}$ $\neq$ $X_{0}$, where $\tau=+\infty$ if $X_{n}=X_{0}$ for all $n\geq0$. We need to compute $E[\tau|X_{0}=i]$ in terms of $p_{ii}$

My solution

For any $l>0$ let $l$ be the first for which $X_{n}$ $\neq$ $X_{0}$ which means that $\forall$ $m$ such that $1\leq m <l$ the value of $X_{m}=X_{0}$ but $X_{i} \neq X_{0}$. The probability of this happening is the product of probability of $P_{ii}^{m}$ for all $m$ multiplied by $(1-P_{ii}^{l})$. The resulting product is then again multiplied by the value of $l$ and then summed over for all possible values of $l$ i.e from 1 to $\infty$ to find expectation

The expected value therefore is $(1-P_{ii}^{1})+2(1-P^{2}_{ii})P_{ii}^{1}+3(1-P_{ii}^{3})P_{ii}^{2}P_{ii}^{1}+.....=\sum_{j=1}^{\infty}(j(1-P^{j}_{ii})\prod_{k=0}^{j-1}P^{k}_{ii})$ where $P^{0}_{ii}=1$

Is this correct ?

$\endgroup$
1
  • $\begingroup$ Your notation seems inconsistent. You begin by saying the chain is "homogeneous". However, examining your expression it seems you are using $P^k_{ii}$ to denote the transition probability at time step $k$ rather than the probability to transition $i\to i$ in $k$ steps. (I have assumed a time homogeneous chain for my answer below.) $\endgroup$ Commented Oct 22, 2017 at 9:37

1 Answer 1

1
$\begingroup$

You are thinking along the right lines but it seems your notation is obscuring things (see bottom of this answer).

The event $\tau = k$ means we remain in state $X_0$ for $k-1$ steps (which has probability $p_{ii}^{k-1}$) and then jump to another state on step $k$ (which has probability $1-p_{ii}$). Consequently, $$ \mathbb{P}(\tau = k) = p_{ii}^{k-1}(1-p_{ii}). $$

This "waiting for an event to happen (in a time homogenous setting)" time distribution is called the geometric distribution.

The expectation can be evaluated with a "differentiate under the integral" trick:

$$ \mathbb{E}[\tau] = \sum_k k p_{ii}^{k-1}(1-p_{ii}) \\ = (1-p_{ii})\sum_k \frac{d}{dp} \left. p^k \right|_{p=p_{ii}} \\ = (1-p_{ii})\frac{d}{dp}\left. \sum_k p^k \right|_{p=p_{ii}} \\ = \frac{1}{1-p_{ii}}, $$ the differentiation in the infinite series being ok because all terms are non-negative. Also we used the formula for a geometric series.


  • Usually "homogeneous" means time homogeneous for Markov chains.
  • With the usual notation for transition matrices $P = (p_{ij})_{i,j \in \mathcal{S}}$, $P^k$ is the $k$-step transition probability (just the matrix power). You appear to use $P^k$ for the transition probability at time step $k$ (which is different to the usual convention).

With this understanding, taking $P^j_{ii}=p_{ii}$ (in your notation) makes your attempt look right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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