Show that the market is arbitrage free 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 : $$a<S_0^1(1+r)<b$$

A value process is defined as :

$$V_t = V_t^\bar{\xi} = \bar{\xi}\cdot \bar{S}_t = \sum_{i=0}^d \xi_t^i\cdot \bar{S}_t^i, \quad t \in \{0,1\}$$

where $\xi = (\xi^0, \xi) \in \mathbb R^{d+1}$ is an investment strategy where the number $\xi^i$ is equal to the number of pieces from the security $S^i$ which are contained in the portfolio at the time period $[0,1], i \in \{0,1,\dots,d\}$.

If a market has arbitrage, then the following hold. In case they do not, the market is arbitrage free.

$$V_0 \leq 0, \quad \mathbb P(V1 \geq 0) = 1, \quad \mathbb P(V_1 > 0) > 0$$

Question : How would one proceed to proving this using the mathematical definition of arbitrage by constructing a certain strategy ?

Answer 1

In your answer you proved the contrapositive and, hence, the forward implication that no arbitrage implies $a < S_0^1(1+r) < b$.

To prove the converse we assume that $a < S_0^1(1+r) < b$ and show that there can be no possible arbitrage. A potential arbitrage opportunity arises either from a portfolio that is short the risk-free asset and long the risky asset (with weights appropriately scaled) where

$$\tag{1}V_0 = \xi S_0^0 + S_0^1 \leqslant 0, \\ \implies \xi \leqslant - \frac{S_0^1}{S_0^0},$$

or a portfolio that long the risk-free asset and short the risky asset where

$$\tag{2}V_0 = S_0^0 + \xi S_0^1 \leqslant 0, \\ \implies \xi \leqslant - \frac{S_0^0}{S_0^1}$$

It then follows that there can be no arbitrage if in either case there exists at least one future state $\omega$ with non-zero probability such that $V_1(\omega) < 0$. Under the given assumptions the risky asset attains nonnegative minimum and maximum values $a = S_1^1(\omega_{min})$ and $b= S_1^1(\omega_{max})$, respectively, where $\mathbb{P}(\{\omega_{min}\}) > 0 $ and $\mathbb{P}(\{\omega_{max}\}) > 0 $.

Thus, in case (1)

$$\min V_1(\omega) = \xi S_0^0(1+r) + S_1^1(\omega_{min}) = \xi S_0^0(1+r) + a \\\leqslant -\frac{S_0^1}{S_0^0} S_0^0(1+r)+a = -S_0^1(1+r) + a < 0,$$

and in case (2)

$$\min V_1(\omega) = S_0^0(1+r) + \xi S_1^1(\omega_{min}) = S_0^0(1+r) + \xi b \\\leqslant S_0^0(1+r) - \frac{S_0^0}{S_0^1}b = \frac{S_0^0}{S_0^1}\left(S_0^1(1+r) - b\right) < 0$$

Therefore, it is impossible to construct a portfolio such that $V_0 \leqslant 0$ and $\mathbb{P}(V_1 \geqslant 0) = 1$.

Answer 2

Note : The following answer was provided by Daneel Olivaw from a post regarding a similar question of mine for finance mathematics at Quantitive Finace Stack exchange : https://quant.stackexchange.com/questions/42637/equivalent-martingale-measure-exists-if-and-only-if-a-s-011r-b . The user Daneel Olivaw providing a hint and a semi-complete answer made me realise that my martingale approach was too complicated and something more practical regarding arbitrage was needed, thus he yielded the elaboration above, which proves the one "side" of the statement.

Answer of Daneel Olivaw using the arbitrage approach :

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:

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