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Your winnings per unit stake on game $n$ are given by independent random variables $\epsilon_n$ such that $P(\epsilon_n=1)=p$, $P(\epsilon_n=-1)=q$ with $1/2<p=1-q<1$. Let $C_n$ be your stake on game $n$ which must lie between $0$ and $Z_{n-1}$, where $Z_n$ is your fortune at time $n$ and $Z_0$ is a constant. Let $\mathcal{F}_n=\sigma(\epsilon_1,...,\epsilon_n)$.

Show that if $C_n$ is any previsible strategy then $\log Z_n-n\alpha$ is a supermartingale with $\alpha=p\log p+q\log q+\log2$ but for a certain strategy it is a martingale and find that strategy.

I' m not sure if I am even starting correctly. It seems to me that $Z_n$ is given by $Z_{n+1}=Z_n+\epsilon_{n+1}C_{n+1}$ because the fortune at the next time is given by the fortune at the present time plus (or minus) the stake on the next game. Now I have $$ E(\log Z_{n+1}-(n+1)\alpha | \mathcal{F}_n)=E(\log(Z_{n}+\epsilon_{n+1}C_{n+1})| \mathcal{F}_n)-\alpha-n\alpha $$ so I should prove $E(\log(Z_{n}+\epsilon_{n+1}C_{n+1}| \mathcal{F}_n)-logZ_n-\alpha\leq 0$.

In the conditional expectation I can factor $C_{n+1}$ and since it is measurable wrt $\mathcal{F}_n$ I get $$ E(\log(Z_{n}+\epsilon_{n+1}C_{n+1})| \mathcal{F}_n) =\log(C_{n+1})+E(\log(\frac{Z_{n}}{C_{n+1}}+\epsilon_{n+1})| \mathcal{F}_n) \leq \log Z_n +\log E(\frac{Z_{n}}{C_{n+1}}+\epsilon_{n+1}| \mathcal{F}_n)$$ by using Jensen. So it remains to show $$ \log E(\frac{Z_{n}}{C_{n+1}}+\epsilon_{n+1}| \mathcal{F}_n) \leq \alpha $$ or equivalently $$E(\frac{Z_{n}}{C_{n+1}}+\epsilon_{n+1}| \mathcal{F}_n)\leq 2p^pq^q.$$ What I can get is that by measurability and independence $E(\frac{Z_{n}}{C_{n+1}}+\epsilon_{n+1}| \mathcal{F}_n)=\frac{Z_{n}}{C_{n+1}}+E\epsilon_{n+1}=\frac{Z_{n}}{C_{n+1}}+p-q$ but I don't think this is less than $2p^pq^q$.

Edit: By factoring $Z_n$ and repeating the steps I get $$ E(\log(Z_{n}+\epsilon_{n+1}C_{n+1})| \mathcal{F}_n)\leq \log Z_n+\log(1+\frac{C_{n+1}}{Z_n}E\epsilon_{n+1})\leq \log Z_n+\log(2p)$$ because $E\epsilon_{n+1}=p-q$. So the supermartingale condition would be verified if $2p\leq 2p^pq^q$ and equivalently if $1\leq (\frac{1-p}{p})^{1-p}$, however this is true if $1>2p$ unless I didn't reverse some inqualities.

In this way it seems that if we choose $C_{n+1}=\beta Z_n$ for $\beta$ such that $1+\beta(p-q)=2p^pq^q$ we get equalities in the previous steps and therefore a martingale.

Do you think it is correct?

bibo
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  • In which textbook are this coming from? – Victor Mar 11 '15 at 21:55
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    Williams, Probability with martingales. Actually I think if instead of factoring $C_{n+1}$ I factor $Z_{n}$ I get to verify the condition$ 1+\frac{C_{n+1}}{Z_{n}}E(\epsilon_{n+1}) \leq 2p^pq^q$ which should be true for $1<2p$ – bibo Mar 11 '15 at 22:03
  • I don't think that works, I worked through this problem and following what you have after the Edit $1\leq\left( \frac{1-p}{p}\right)^{1-p}$ holds if $1>2p$, but the problem tells us that $p>1/2$ so the inequality can never hold using this approach. – User112358 Mar 05 '16 at 22:08

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