2
$\begingroup$

I am working on the following problem:

I want to show that for a Markov chain with state space $S$, $\sum_0^\infty P^n(x,y) \le \sum_0^\infty P^n(y,y)$ holds $\forall x,y \in S$. I am able to prove this for the case where $y$ is recurrent, by using the result $\sum_1^\infty P^n(x,y) = \frac{\rho_{xy}}{1-\rho_{yy}}$. However, I am having difficulties proving it for the case where $y$ is transient. It was suggested that we use the equation $P^n(x,y) = \sum_{m=1}^n P_x(T_y=m)P^{n-m}(y,y)$ (for both the recurrent and transient cases), but I am having trouble figuring out how this is useful. Thanks!!!

$\endgroup$
  • 1
    $\begingroup$ You should have used a self-study tag to avoid getting the answer straight away. $\endgroup$ – Xi'an Feb 2 '17 at 21:37
2
$\begingroup$

\begin{align*} \sum_{n=0}^\infty P^n(x,y)&=\sum_{n=0}^\infty \sum_{m=1}^nP_x(T_y=m)P^{n-m}(y,y)\\ &=\sum_{m=1}^\infty \sum_{n=m}^\infty P_x(T_y=m)P^{n-m}(y,y)\\ &= \sum_{m=1}^\infty P_x(T_y=m) \sum_{n=m}^\infty P^{n-m}(y,y)\\ &= \sum_{m=1}^\infty P_x(T_y=m) \sum_{n=0}^\infty P^{n}(y,y)\\ &=[1-P_x(T_y=\infty)]\sum_{n=0}^\infty P^{n}(y,y). \end{align*}

$\endgroup$

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.