2
$\begingroup$

Background Information:

A market satisfies the Law of One Price if every two self-financing strategies that replicate the same claim have the same initial value.

An inconsistent pricing strategy is a self-financing strategy $\phi$ with $V_T(\phi)\equiv 0$ and $V_0(\phi) < 0$.

Question:

Prove the Law of One Price holds if and only if there does not exist an inconsistent pricing strategy.

Attempted proof - Suppose we have two self financing strategies $\phi$ and $\psi$ that replicates some claim $X$ such that $V_0(\phi) = V_0(\psi)$. Hence we cannot satisfy the condition of $V_0(\phi) < 0$ nor $V_0(\psi) < 0$ so there is no inconsistent pricing strategy.

I am not sure how to show the converse and whether this is rigorous enough. Any suggestions are greatly appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

Assume the law of one price. We show that there does not exist an inconsistent pricing strategy. Suppose that $\phi$ is an inconsistent self-financing trading strategy, that is, $V_T(\phi)\equiv 0$ and $V_0(\phi) < 0$. Consider another self-financing trading strategy $\psi$ that does not nothing, that is, without holding any of the underlying assets. Then $V_T(\psi)\equiv 0$ and $V_0(\psi) = 0$. This contradicts the law of one price, since both $\phi$ and $\psi$ replicate the same claim, but the initial prices are different.

On the other hand, assuming that there does not exist an inconsistent pricing strategy, we show that the law of one price holds. Consider any two self-financing trading strategies $\phi_1$ and $\phi_2$ such that $V_T(\phi_1) = V_T(\phi_2)$. Note that $\phi=\phi_1-\phi_2$ is also a self-financing trading strategy, and $V_T(\phi) = V_T(\phi_1) - V_T(\phi_2)\equiv 0$. Since there does not exist an inconsistent pricing strategy, $V_0(\phi) \equiv 0$. That is, $V_0(\phi_1) =V_0(\phi_2)$. Therefore, the law of one price holds.

$\endgroup$
1
  • $\begingroup$ Than you Gordon, looks really good! $\endgroup$
    – Wolfy
    Commented Nov 3, 2016 at 20:11

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.