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

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 $$0. Show that the market is arbitrage-free if and only if it is : $$a

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

• In words, this condition says that the risk-free outcome is somewhere in between the worst risky outcome and the best risky outcome... Makes sense no? Nov 14, 2018 at 19:49
• @noob2 It makes, but we need a rigorous mathematical proof. Nov 14, 2018 at 19:51
• It would really be much simpler to prove if you use the mathematical definition of arbitrage instead of the fundamental theorem. Assuming the condition you are trying to prove does not hold, try to construct a portfolio for which the value is null or negative at $t=0$ and whose value at $t=1$ is always positive. Nov 14, 2018 at 20:02
• @Daneel Olivaw How would that be derived? I would really appreciate an elaboration of a solution! I'm still a beginner so I get things too confused. Nov 14, 2018 at 20:05

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:

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

• Thanks, simply elaborated ! My way around using martingales was way too complicated mathematically ! Nov 15, 2018 at 17:50
• One question only, what does the phrase : "the risky asset is shorted" mean ? Nov 15, 2018 at 18:12
• It means you sell it. In financial markets you can equivalently "go long" (buy) or "go short" (sell) an asset even if you don't own it. How do you sell an asset that you don't have? In practice, suppose an investor $A$ has a position in a risky asset that he wants to hold for the long term. In the meantime and to generate an extra return, he can lend the asset to you for a period $[T_0,T_1]$ in exchange for some interest. Once you have borrowed the security at $T_0$, you can sell it but you need to buy it back before $T_1$ in order to return it to $A$. Nov 15, 2018 at 18:17
• Copied this answer and posted as his own under the cross-posted question on Math SE - s://math.stackexchange.com/questions/2998811. Nov 17, 2018 at 11:26