This is an exercise which I came upon while studying an introduction to financial mathematics.

Exercise :

Consider the finite sample space $\Omega = \{\omega_1,\omega_2,\omega_3\}$ and let $\mathbb P$ be a probability measure such that $\mathbb P[\{\omega_1\}] > 0$ for all $i=1,2,3$. We define a financial market of one period which is consisted by the probability space $(\Omega,\mathcal{F},\mathbb P)$ with $\mathcal{F} := 2^\Omega$ and the securities $\bar{S} = (S^0,S^1,S^2)$ which are consisted of the zero-risk security $S^0$ and two securities $S^1,S^2$ which have risk. Their values at the time $t=0$ are given by the vector $$\bar{S}_0 = \begin{pmatrix} 1\\2\\7 \end{pmatrix}$$ while their values at time $t=1$, depending whether the scenario $\omega_1,\omega_2$ or $\omega_3$ happens, are given by the vectors $$\bar{S}_1(\omega_1) = \begin{pmatrix} 1\\3\\9\end{pmatrix}, \quad \bar{S}_1(\omega_2) = \begin{pmatrix} 1\\1\\5\end{pmatrix}, \quad \bar{S}_1(\omega_3) = \begin{pmatrix} 1\\5\\10 \end{pmatrix}$$ (a) Show that this financial market has arbitrage.

(b) Let $S_1^2(\omega_3) = 13$ while the other values remain the same as before. Show that this financial market does not have arbitrage and describe all the equivalent martingale measures.

Attempt :

(a) We have that 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 $[0,1], i \in \{0,1,\dots,d\}$.

Now, I also know that to show that a market has arbitrage, I need to show the following :

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

I understand that the different $S$ vectors will be plugged in to calculate $V_t$ but I really can't comprehend $\xi$. What would the $\xi$ vector be ?

Any help for me to understand what $\xi$ really is based on the problem and how to complete my attempt will be much appreciated.

For (b), showing that it does not have arbitrage is similar to (a) as I will just show that one of these conditions will not hold. What about the martingale stuff though ? It's a mathematical substance we really haven't been into so, if possible, I would really appreciate an elaborations.


1 Answer 1


The parameter $\xi$ represents your strategy, namely the quantity you hold in your portfolio of each security $S^0$, $S^1$ and $S^2$. Consider the following strategy: $${\xi}=(\xi^1,\xi^2,\xi^3)=(1.5,1,-0.5)$$ Then: $$\begin{align} & t=0: && \xi\bar{S}_0=\xi^0S_0^0+\xi^1S_0^1+\xi^2S_0^2 = 1.5+2-3.5=0 \\ & t=1: && \xi\bar{S}_1(\omega_1)=1.5+3-4.5=0 \\ &&& \xi\bar{S}_1(\omega_2)=1.5+1-2.5=0 \\ &&& \xi\bar{S}_1(\omega_3)=1.5+5-5=1.5>0 \end{align}$$ Thus: $$\xi\bar{S}_0=0, \quad \mathbb{P}(\xi\bar{S}_1\geq0)=1, \quad \mathbb{P}(\xi\bar{S}_1>0)>0$$

Hence the market has arbitrage.

For question b), you need to generalize to prove that there is no portfolio $\xi$ that allows arbitrage (instead of just finding a counterexample as in a).

  • $\begingroup$ Hello, why is it legit to just take a random $\xi$ and show the conditions ? Wouldn't the $\xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ? $\endgroup$
    – Rebellos
    Nov 11, 2018 at 20:51
  • $\begingroup$ It is not a random $\xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $\xi$, and there is a logic to it. $\endgroup$
    – Alex C
    Nov 11, 2018 at 22:48
  • $\begingroup$ @Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(\omega_1)/S_1^2(\omega_1), S_1^1(\omega_3)/S_1^2(\omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(\omega_2)/S_1^1(\omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$. $\endgroup$ Nov 11, 2018 at 23:02
  • $\begingroup$ Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do? $\endgroup$
    – Rebellos
    Nov 11, 2018 at 23:06

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