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 elaboration.


1 Answer 1


Right, $\xi$ is just a portfolio, so some amount for each of the securities. If the portfolio was all $S^0$, then it would look like (1,0,0). The question is whether we can construct a set of weights which has a negative or zero value at t=0, but either zero or positive at t=1 under all circumstances (the given omegas).

Consider, then, if we went all in on security 1 and funded it by borrowing security 0 which is acting like cash. So we have 1 unit of S1, which had value 2 at t=0, so we need -2 of S0 (sorry entering latex on mobile is driving me a bit spare). Look at each $\omega$. 1 and 3 both have S1 going up, great, but 2 has it going down by one.

We could do the same with S3, but same problem, scenario 2 goes down. Note, however, that the ratios change. In scenario 2 the ratio goes up to 5, whereas the others the ratio goes down from 3.5 to 3 or 2. So if we entered a short position on S2 as well, we would curb our loss on scenario 2. In it, S2 drops by 2 when S1 only drops by 1, so if we are half as exposed to S2, then it would bring us back to zero valuation. So modify our portfolio to be -0.5 of S2, costing -3.5 in cash. So now $\xi=(1.5,1,-0.5)$.

Value zero initially (combining two net zero positions), in scenario 1 we gain 1 on S1, but lose it in S0 going up. In S2, same but in reverse, net 0. In scenario 3, though S1 goes up 5/2 and S2 only goes up 10/7. So our S1 position goes from 1 to 2.5, and our S2 position goes from -3.5 to -5. We gain 2.5-1.5=1, with Nonzero probability.

I've walked through this slowly (as I'm thinking about it), but since the portfolio is a weight of vector, you're asking for a solution to a matrix-vector equation.

The initial portfolio value is $\xi S_0$, and the valuations for the scenarios are given by $\xi S_1$, where this time S_1 is a matrix made up of the $\omega$ as columns.

If we replace this with a normalised version $U_n=S_n/S_0$, then we can view the changes as a matrix $V$ where the columns are $V_n=U_1(\omega_n)-U_0$. In this matrix the top line are just 0 (because the cash security doesn't change value).

V, then:

w1 w2 w3
0 0 0
1/2 -1/2 3/2
2/7 -2/7 3/7

Note that columns w1 and w2 are just a factor away from each other. If we were to change w3 to match, we would change it to be 3*w1, by making the bottom right entry 6/7. Converting back from V to S we would get (6/7+1)*7=13, and voila the answer to part b.


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