1

Assume there is a multi period discrete market i.e. a finite Probability space, finite time periods $\lbrace 0,1\ldots, n\rbrace$, a stochastic process $(S(t))_{t\in\lbrace 0,1,\ldots,n\rbrace}\subseteq \mathbb{R}^d$ with $S_0(t)>0$ for all $t\in\lbrace 0,\ldots,n\rbrace$, that is adapted to the Filtration $\mathscr{F}_t:=\sigma(S(k), k\leq t)$ ($t\neq 0, n$), $\mathscr{F}_n:=2^{\Omega}$, $\mathscr{F}_0:=\lbrace 0,\Omega\rbrace$.

How can I prove:
If there exists some $t\in \lbrace 1,\ldots, n \rbrace$ such that $P(\phi(t)^T S(t)<0)>0$ for all predictable, selffinancing trading strategies $(\phi(t))_{t\in\lbrace 1,\ldots, n\rbrace}\subseteq\mathbb{R}^d$, i.e. $\phi(t)$ is $\mathscr{F}_{t-1}$ measurable and satisfies $\phi(t)^T S(t)= \phi(t+1)^TS(t)$ for all $t\in \lbrace 1, \ldots, n-1\rbrace$, then there can't be an arbitrage possibility in this market.

Where an arbitrage opportunity means:
There is a self financing trading strategy $\psi(t)$ such that $\psi(1)^T S(0)=0$ and $P(\psi(n)^T S(n)\geq 0)=1$ and $P(\psi(n)^T S(n)>0)>0$

MackeyTopology
  • 786
  • There is something missing in the no-arbitrage condition that you mention. In particular, notice that $\phi(t)=0$ for all $t$ is a predictable, selffinancing trading strategy, but $P(\phi(t)^T S(t)<0)=0$ trivially without implying anything about arbitrage. – Pavel Kocourek Dec 30 '22 at 05:38
  • Thank you very much for youre comment. I forgot to mention that we're also assuming: $P(\phi(n)^TS(n)>0)>0$ and $\phi(1)^TS(0)=0$. – MackeyTopology Dec 30 '22 at 07:08
  • Maybe also the following might be usefull: a trading strategy $\phi$ is selffinancing if and only if $\phi(t)^TS(t)=\phi(1)^TS(0)+\sum_{k=1}^t \phi(k)^T (S(k)-S(k-1))$ for all $t\in\lbrace 0, \ldots, n\rbrace$ i.e. the portfolio value is always a sum of the initial value and the increment process. – MackeyTopology Dec 30 '22 at 07:24
  • Please update your question for the additional assumption. However, I think that standardly this problem would be formulate differently, because here you only need to take into account that $P(\psi(n)^T S(n)\geq 0)=1$ is equivalent to $P(\psi(n)^T S(n)<0)=0$. – Pavel Kocourek Dec 30 '22 at 14:32

0 Answers0