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We know that by the law of one price: in a one-period model $(\overline{\pi},\overline{S})$ for an arbitrage-free market model it follows that for two strategies $\overline{\rho}$ and $\overline{\xi}\in \mathbb R^{d+1}$: $$ \overline{S}\cdot\overline{\xi}=\overline{S}\cdot\overline{\rho}\; \; \mathbb P-\text{a.s.}\implies \overline{\pi}\cdot\overline{\xi}=\overline{\pi}\cdot\overline{\rho}\; \; \mathbb P-\text{a.s.}$$

Question: Find a market model that is not arbitrage-free but still obeys the law of one price.

My idea:

Consider the space $\Omega:=\{\omega_{0},\omega_{1}\}$, with $\mathbb P(\{\omega_{i}\})>0,\; i =0,1$ and two assets (one risk-free, the other risky) such that $S^{1}(\omega_{0})=0$ and $S^{1}(\omega_{1})=1$. It can be proven that for $r>0$ as the interest rate on the risk-free asset, i.e. $(\pi^{0},S^{0})=(1,1+r)$ that the model is arbitrage free if and only if:

$$ S^{1}(\omega_{0})=0< \pi^{1}(1+r)<1=S^{1}(\omega_{1})$$

Now consider two strategies $\overline{\rho}=(\rho^{0},\rho^{1})$ and $\overline{\xi}=(\xi^{0},\xi^{1})$ such that

$$\overline{S}\cdot\overline{\xi}=\overline{S}\cdot\overline{\rho}\; \; \mathbb P-\text{a.s.}$$ i.e.

on $\{ \omega_{1}\}$: $(1+r)\xi^{0}+\xi^{1}=(1+r)\rho^{0}+\rho^{1}$

and on $\{ \omega_{0}\}$: $(1+r)\xi^{0}=(1+r)\rho^{0} \implies \xi^{0}=\rho^{0}$

and thus $\xi^{1}=\rho^{1}$. Thus fundamental idea here is that we have not had to place any restrictions on the risk-free rate, $r >0$. Thus for a fixed $\pi^{1}$ we choose $r> \max\{\frac{1}{\pi^{1}}-1,\varepsilon\}$ for $\varepsilon >0$ such that $\pi^{1}(1+r)> 1$.

Therefore $\overline{\xi}=\overline{\rho}$ and thus trivially $\overline{\pi}\cdot\overline{\xi}=\overline{\pi}\cdot\overline{\rho}$. But the market model admits arbitrage.

I feel like I may be "cheating" by setting $S^{1}(\omega_{0})=0$ since aren't are asset prices assumed positive (does this mean strictly positive?). Is this indeed cheating? Are there other examples where the prices are strictly positive?

Thanks for the help!

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