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Let $P\left ( X_{i}=1 \right )=p$ and $P(X_{i}=-1)=1-p$ where $p \in \left ( 0,1 \right )$ and $p\neq \frac{1}{2}$.

Let $S_{n}=S_{0}+X_{1}+\cdot \cdot \cdot X_{n}$. Define $M_{n}=\left ( \frac{1-p}{p} \right )^{S_{n}}$

Recall: if $h\left ( x \right )=\sum_{y}p\left ( x,y \right )h\left ( y \right )$ then $h\left ( X_{n} \right )$ is a martingale.

The author begins with

$\left ( p\left ( \frac{1-p}{p} \right )^{S_{n}+1} +\left ( 1-p \right )\left ( \frac{1-p}{p} \right )^{S_{n}-1}\right )$

which I can understand that the author takes into account the fact that the Markov chain could either move forward or backward and multiply that by its associated probability.

However, I do not understand why the summation vanishes. enter image description here

Any help is appreciated.

Mathematicing
  • 6,253
  • The summation vanishes because $p(x,y) = 0$ for any $y \notin {x-1,x+1}$. – saz Apr 16 '18 at 14:55
  • @saz The second equality looks problematic. I have tried the distributive law over subtraction while also breaking the exponent on x. – Mathematicing Apr 17 '18 at 13:10
  • Not sure what you mean by "problematic". You just have to use $(a/b)^q = a^q/b^q$... that's it. – saz Apr 17 '18 at 13:15

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