Background Information:

This question is from Lectures on Financial Mathematics: Discrete Asset Pricing.


Prove that for discrete models, the existence of a strong arbitrage is also equivalent to the existence of a self-financing strategy such that $V_0(\phi) < 0$ and $V_T(\phi) \geq 0$.

Attempted proof - Suppose $\psi$ is a self-financing strategy such that $$V_0(\psi) = 0 \ \ \text{and} \ \ V_T(\psi) > 0$$ Now suppose we also have a self-financing strategy $\phi$ such that $$V_0(\phi) > 0 \ \ \text{and} \ \ V_T(\psi) \geq V_T(\phi)$$ Now let $\Omega = \psi - \phi$, which is also a self-financing strategy since the difference of two self-financing strategies is a self-financing strategy. Then $$V_0(\Omega) = V_0(\psi - \phi) = V_0(\psi) - V_0(\phi) = 0-V_0(\phi) < 0$$ and $$V_T(\Omega) = V_T(\psi - \phi) = V_T(\psi) - V_T(\phi) \geq 0 $$ Thus we have a self-financing strategy such that $$V_0(\Omega) < 0 \ \ \text{and} \ \ V_T(\Omega) \geq 0$$ A similar arguments holds for the converse.

I am not sure if this is completely right, any suggestions are greatly appreciated.


1 Answer 1


We first assume the strong arbitrage, that is, there is a self-financing strategy $\phi$ such that $V_0(\phi)=0$ and $V_T(\phi)>0$. Then, on the finite sample space, there exists $\alpha >0$ such that \begin{align*} S_0^0\frac{V_T(\phi)}{S_T^0}\ge \alpha, \end{align*} where $S^0$ is the risk-free asset among the $k+1$ assets $S^0, S^1, \ldots, S^k$, which is the deposit or money-market account. We buy $-\alpha/S_0^0$ share of the risk-free asset $S_0$, and hold until maturity $T$, that is, we consider the trading strategy $\psi$, where \begin{align*} \psi_i = \begin{cases} -\alpha/S_0^0, & \text{ if } i=0,\\ 0, & \text{ otherwise}, \end{cases} \end{align*} without any intermediate adjustment. It is then clear that $V_0(\psi+\phi)=-\alpha <0$, and \begin{align*} V_T(\psi+\phi) &= V_T(\psi) + V_T(\phi)\\ &=-\alpha\frac{S_T^0}{S_0^0} + V_T(\phi)\\ &=\frac{S_T^0}{S_0^0}\left(S_0^0\frac{V_T(\phi)}{S_T^0} - \alpha \right) \ge 0 \end{align*} $$$$ For the other way around, where $V_0(\phi)<0$ and $V_T(\phi)\ge 0$, we can simply set $\alpha = -V_0(\phi)$ in the above strategy $\psi$. Then $V_0(\psi+\phi)=0$, and $V_T(\psi+\phi)> 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.