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Background Information:

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

Question:

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.

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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 $V_T(\psi+\phi)> 0$.

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