Markov chain which at each transition either goes $+1$ with probability $p$ or $-1$ with probability $q$. $(q/p)^{S_n}, n \geq 1$ is a martingale.

Question

Consider the Markov chain which at each transition either goes up $1$ with probability $p$ or down $1$ with probability $q = 1 - p$. Argue that $(q/p)^{S_n}, n \geq 1$ is a martingale.

I tried to show $E[Z_{n+1}|Z_1,...,Z_n] = Z_n$ as follows \begin{align} E[Z_{n+1}|Z_1,...,Z_n]& = E[Z_{n+1}|Z_n] \\ &= E[(q/p)^{S_{n+1}}|Z_n]\\ &= E[(q/p)^{S_{n+1}}|Z_n]\\ &= (q/p)^{E[S_{n+1}|Z_n]} \\ &= (q/p)^{p(\log_{q/p}(Z_n)+1) + q(\log_{q/p}(Z_n)-1)}\\ &= Z_n (q/p)^{p-q} \end{align}

something seems to have gone wrong but I can't figure out what.

Answer 1

Your fourth equality is wrong. You cannot write $Er^{X}=r^{EX}$.

$$E[Z_{n+1}|Z_1,Z_2,..,Z_n]$$$$=E[(\frac q p)^{Z_{n+1}}|Z_n]$$$$=E[(\frac q p)^{Z_{n}+1}] P(Z_{n+1}=1|Z_n)$$$$+E[(\frac q p)^{Z_{n}-1}] P(Z_{n+1}=-1|Z_n).$$ This gives $(\frac q p)^{Z_n} \frac q p p+(\frac q p)^{Z_n} (\frac q p) ^{-1} q=(\frac q p)^{Z_n}$.