Equivalent martingale measure exists if and only if $a < S_0^1(1+r)< b$

Question

Exercise :

We consider a market of one period $(\Omega, \mathcal{F}, \mathbb P, S^0, S^1)$, where the sample space $\Omega$ has a finite number of elements and the $\sigma-$algebra $\mathcal{F} = 2^\Omega$. Furthermore, with $S^0$ we symbolize the zero risk asset with initial value $S_0^0=1$ at the time $t=0$ and interest rate $r>-1$ (which means $S_1^0 = 1+r$). With $S^1$ we symbolize an asset with risk with initial value $S_0^1 >0$ at the time $t=0$ and with value $S_1^1$ at the time $t=1$ which is a random variable.

Let $\mathbb{P}[\{\omega\}]>0$ for all $\omega \in \Omega$. We define : $$a:=\min S_1^1(\omega) \quad \text{and} \quad b:=\max S_1^1(\omega)$$ and we assume that $0<a<b$. Show that the market is arbitrage-free if and only if it is : $$a<S_0^1(1+r)<b$$

Attempt :

Since we have to find a iff condition for the market to be arbitrage-free, it is the same as showing that there exists an equivalent martingale measure. This comes from the following theorem :

The Fundamental Theorem of Asset Pricing : A financial market is arbitrage-free if and only iff there exists an equivalent martingale measure.

So, let $\Omega = \{\omega_1, \dots , \omega_n\}$. Consider $\mathbb{Q}$ to be a probability measure. For $\mathbb{Q}$ to be a martingalem it must be :

$$S_1 \in L^1(\mathbb Q) \quad \text{and} \quad S_0 = \mathbb{E}_\mathbb Q\bigg[\frac{S_1}{1+r}\bigg]$$

These conditions, mean that :

$$\|S_1\|_1 < + \infty \Rightarrow |S_1^1(\omega_1) + \cdots + S_1^1(\omega_n)| < + \infty$$

Also, we have :

$$S_0^1 = \frac{S_1^1(\omega_1)}{1+r}\mathbb{Q}(\omega_1) + \cdots + \frac{S_1^1(\omega_n)}{1+r}\mathbb{Q}(\omega_n)$$ $$\Rightarrow$$ $$S_0^1(1+r) = S_1^1(\omega_1)\mathbb{Q}(\omega_1) + \cdots + S_1^1(\omega_n)\mathbb{Q}(\omega_n)$$

Now, for $\mathbb{Q}$ to be an equivalent martingale measure, it must be $\mathbb{Q} \sim \mathbb{P}$, thus since $\mathbb{P}[\{\omega\}] >0$ it must also be $\mathbb{Q}(\omega) >0$.

Finally, for $\mathbb{Q}$ to be a legit probability measure, its components must sum up to $1$.

Thus, we yield the following system of conditions :

$$\begin{cases} S_1^1(\omega_1)\mathbb{Q}(\omega_1) + \cdots + S_1^1(\omega_n)\mathbb{Q}(\omega_n) &=S_0^1(1+r) \\ |S_1^1(\omega_1) + \cdots + S_1^1(\omega_n)| &< + \infty \\ \mathbb{Q}(\omega_1) + \cdots + \mathbb{Q}(\omega_n) &= 1 \\ \mathbb{Q}(\omega_1) &> 0 \\ \quad \vdots \\ \mathbb{Q}(\omega_n) &>0 \end{cases}$$

Question : How would one proceed now to showing that if $a = \min S_1^1(\omega)$ and $b = \max S_1^1(\omega)$ then for an equivalent martingale measure to exist, it should be :

$$a<S_0^1(1+r)<b$$

Answer 1

Assume that:

$$ S_0^1(1+r)\leq a,b $$

Arbitrage for a portfolio $V_t$ is defined as:

$$V_0\leq0, \quad P(V_1\geq0)=1, \quad P(V_1>0)>0$$

Consider borrowing at rate $r$ to buy the risky asset such that $V_0=0$. Then, assuming $a\not= b$:

$$\begin{align} \min_{\omega}V_1(\omega)=a-S_0^1(1+r)\geq 0 \\ \max_{\omega}V_1(\omega)=b-S_0^1(1+r)> 0 \end{align}$$

Thus there is arbitrage. The same argument can be made if $S_0^1(1+r)\geq a,b$ but in this case the risky asset is shorted and the money is lent at a rate $r$. Hence to prevent arbitrage the market has to enforce the following constraint:

$$ a< S_0^1(1+r)< b$$

The inequality does not necessarily need to be strict, we can equivalently have:

$$ a\leq S_0^1(1+r)\leq b$$