I am stuck on the following problem.
Your winnings per unit stake on game $n$ are $\epsilon_n$, where the $\epsilon_n$ are IID RVs with $$ P(\epsilon_n=1)=p, \ P(\epsilon_n=-1)=q, \text{ where } .5<p=1-q < 1. $$ Your stake $C_n$ on game n must lie between $0$ and $Z_{n-1}$, where $Z_{n-1}$ is your fortune at time $n-1$. Let $\mathcal{F}_n=\sigma(\epsilon_1,\dots,\epsilon_n)$ be your history up to time $n$. Show that if $C$ is any predictable strategy, then $\log(Z_n)-n\alpha$ is a supermartingale, where $\alpha=p\log p + q\log q + log 2.$ For what strategy $C$ is $\log(Z_n)-n\alpha$ a martingale?
This is what I have so far.
$$ E(\log(Z_{n-1}+\epsilon_n C_n)-n\alpha|\mathcal{F}_{n-1})=\log(Z_{n-1})+E(\log(1+\epsilon_n \frac{C_n}{Z_{n-1}})|\mathcal{F}_{n-1})-n\alpha, $$ using Jensen's inequality I would get, $$ \leq \log(Z_{n-1})+\log(E(1+\epsilon_n \frac{C_n}{Z_{n-1}}|\mathcal{F}_{n-1}))-n\alpha=\log(Z_{n-1})+\log(1+\frac{C_n}{Z_{n-1}}E(\epsilon_n))-n\alpha $$ $$ =\log(Z_{n-1})+\log(1+ \frac{C_n}{Z_{n-1}}(p-q))-n\alpha. $$
Following this approach it remains to show that $\log(1+ \frac{C_n}{Z_{n-1}}(p-q))-n\alpha\leq -(n-1)\alpha$, or rather $\log(1+ \frac{C_n}{Z_{n-1}}(p-q))\leq \alpha$.
I'm not quite sure how to get this.
We have $\frac{C_n}{Z_{n-1}}\leq 1$, so we can bound above by $\log(2p)$, so we would need to show that when $p>1/2$ we have $2p\leq 2 p^p q^q$, which I'm not sure is possible.
Any help would be appreciated.