EDIT: I have found the answer on my own, so you can consider this as an exercise. ;-)

Consider iid random variables $(X_n)_{n\ge 1}$ with $P(X_n=1)=p$, $P(X_n=-1)=1-p$ for $\frac{1}{2}<p<1$ and a predictable process $(\kappa_n)_{n\ge 1}$ with $\kappa_n\in[0,V_{n-1})$, where $V_n:=C+\sum_{i=1}^n \kappa_iX_i$ for $n\ge 1$ and $C>0$. Define $\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)$.

I want to show that

$$(\log(V_n)-nc)_{n\ge 1}$$

is a super martingale with respect to $(\mathcal{F}_n)_{n\ge1}$, where


So what I have done so far is showing that the process is integrable by using the fact that

$$0<\theta(n)\le V_n\le 2^n V_0,$$

where $\theta(n):=V_0-\sum_{i=1}^n\kappa_i\in(0,\infty)$ where as already mentioned $\kappa_i\in[0,V_{i-1})$ and taking

$$\max_{x\in[\theta(n),2^n V_0]}\{|\log(x)|\}.$$

It is left to show that $$E[\log(V_n)-nc|\mathcal{F}_{n-1}]\le \log(V_{n-1})-(n-1)c.$$

So I tried following:

\begin{align} E[\log(V_n)-nc|\mathcal{F}_{n-1}]&=E[\log(V_{n-1}+\kappa_n X_n)-nc|\mathcal{F}_{n-1}]\\ &=E[\log(V_{n-1}+\kappa_n X_n)]-nc, \text{ since $\kappa_n$ predictable, $X_n$ independent.}\\ &=p\cdot\log(V_{n-1}+\kappa_n)+(1-p)\cdot\log(V_{n-1}-\kappa_n)-nc\\ &=p\cdot\log\Big(\frac{V_{n-1}+\kappa_n}{V_{n-1}-\kappa_n}\Big)+\log(V_{n-1}-\kappa_n)-nc\\ &=\log\Big((\frac{V_{n-1}+\kappa_n}{V_{n-1}-\kappa_n})^p\Big)+\log(V_{n-1}-\kappa_n)-nc\\ &=\log((V_{n-1}+\kappa_n)^p(V_{n-1}-\kappa_n)^{(1-p)})-nc, \end{align} but I do not know where I have to go from here. I also wonder if all of the steps are proper reasoning. I also see some kind of a relation of




where you can factorize $V_{n-1}$ and then maybe using concavity of $\log$, but I do not succeed.

So any hint would be very nice. :)

Thanks in advance!

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