2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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 '18 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 '18 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 $S^1$. $\endgroup$ – Daneel Olivaw Nov 11 '18 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 '18 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.