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

$$c=p\log(p)+(1-p)\log(1-p)+\log(2).$$

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

$$\log\Big(p^p(1-p)^{(1-p)}\Big)=c-\log(2)$$

with

$$\log\Big((V_{n-1}+\kappa_n)^p(V_{n-1}-\kappa_n)^{(1-p)}\Big)$$

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!

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.