Turning an Inconsistent Pricing Strategy into Strong Arbitrage
Summary
The document shows how a self-financing strategy with nonzero initial value and zero terminal value can be combined with a risk-free asset position to construct a strong arbitrage. If the original strategy starts with a negative value, add enough of the money-market account to offset that initial cost. The combined strategy then begins at zero and ends with positive value, since the added risk-free holding grows while the original strategy finishes at zero.
If the original strategy instead has positive initial value, reverse its direction first, then apply the same construction. This is an algebraic implication of the stated definitions, assuming the risk-free asset is available and its terminal value is positive. The document presents no market data or empirical evidence; the result is a theoretical argument under those assumptions.
Key ideas
- A pricing inconsistency is defined as a self-financing strategy with nonzero initial value and zero terminal value.
- Reversing the strategy handles the case where its initial value is positive.
- A risk-free asset position can offset the initial value of the original strategy.
- The combined strategy starts at zero and has positive terminal value, meeting the definition of strong arbitrage.
Tags
Full text
# If there is an inconsistent pricing strategy then by defintion we have strong arbitrage
# If there is an inconsistent pricing strategy then by defintion we have strong arbitrage
Background Information:
An Inconsistent pricing strategy is a self financing strategy $\phi$ with $V_T(\phi)= 0$ and $V_0(\phi) \neq 0$
A strong arbitrage is a self-financing strategy $\phi$ with $V_0(\phi) = 0$ and $V_T(\phi) > 0$
Question:
> Suppose there exists an Inconsistent pricing strategy. Prove from the definition that there must exist a strong arbitrage.
Attempted proof - Let $\phi$ be a self-financing strategy such that $V_0(\phi)\neq 0$ and $V_T(\phi) = 0$.
I am confused how this is possible to prove seems like we have a direct contradiction. Any suggestions are greatly appreciated.
## Answer by Gordon (score 2, accepted)
https://quant.stackexchange.com/a/31217
We assume that $V_0(\phi)<0$; otherwise, we can consider the strategy $-\phi$. Then, we buy extra $-V_0(\phi)/S_0^0$ share of the risk-free asset $S^0$, from the $k+1$ assets $S^0, S^1,\ldots, S^k$, which is the deposit or money-market account, and hold until maturity $T$, that is, we consider the trading strategy $\psi$, where \begin{align*} \psi_i = \begin{cases} -V_0(\phi)/S_0^0, & \text{ if } i=0,\\ 0, & \text{ otherwise}, \end{cases} \end{align*} without any intermediate adjustment. It is then clear that $V_0(\psi+\phi)=0$, and $V_T(\psi+\phi)>0$. In other words, there exists a strong arbitrage strategy (e.g., $\psi+\phi$).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.