<p>Right, <span class="math-container">$\xi$</span> is just a portfolio, so some amount for each of the securities. If the portfolio was all <span class="math-container">$S^0$</span>, 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).</p>
<p>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 <span class="math-container">$\omega$</span>. 1 and 3 both have S1 going up, great, but 2 has it going down by one.</p>
<p>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 <span class="math-container">$\xi=(1.5,1,-0.5)$</span>.</p>
<p>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.</p>
<p>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.</p>
<p>The initial portfolio value is <span class="math-container">$\xi S_0$</span>, and the valuations for the scenarios are given by <span class="math-container">$\xi S_1$</span>, where this time S_1 is a matrix made up of the <span class="math-container">$\omega$</span> as columns.</p>
<p>If we replace this with a normalised version <span class="math-container">$U_n=S_n/S_0$</span>, then we can view the changes as a matrix <span class="math-container">$V$</span> where the columns are <span class="math-container">$V_n=U_1(\omega_n)-U_0$</span>. In this matrix the top line are just 0 (because the cash security doesn't change value).</p>
<p>V, then:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">w1</th>
<th style="text-align: center;">w2</th>
<th style="text-align: center;">w3</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">0</td>
<td style="text-align: center;">0</td>
<td style="text-align: center;">0</td>
</tr>
<tr>
<td style="text-align: center;">1/2</td>
<td style="text-align: center;">-1/2</td>
<td style="text-align: center;">3/2</td>
</tr>
<tr>
<td style="text-align: center;">2/7</td>
<td style="text-align: center;">-2/7</td>
<td style="text-align: center;">3/7</td>
</tr>
</tbody>
</table>
</div>
<p>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.</p>

CC BY-SA 4.0
2023-08-16T20:02:09.590Z
2023-08-16T20:51:46.107Z
2023-08-16T20:51:46.107Z
Phil H
<p>This is an exercise which I came upon while studying an introduction to financial mathematics.</p>

<p><strong>Exercise :</strong></p>

<blockquote>
  <p>Consider the finite sample space <span class="math-container">$\Omega = \{\omega_1,\omega_2,\omega_3\}$</span> and let <span class="math-container">$\mathbb P$</span> be a probability measure such that <span class="math-container">$\mathbb P[\{\omega_1\}] &gt; 0$</span> for all <span class="math-container">$i=1,2,3$</span>. We define a financial market of one period which is consisted by the probability space <span class="math-container">$(\Omega,\mathcal{F},\mathbb P)$</span> with <span class="math-container">$\mathcal{F} := 2^\Omega$</span> and the securities <span class="math-container">$\bar{S} = (S^0,S^1,S^2)$</span> which are consisted of the zero-risk security <span class="math-container">$S^0$</span> and two securities <span class="math-container">$S^1,S^2$</span> which have risk. Their values at the time <span class="math-container">$t=0$</span> are given by the vector 
  <span class="math-container">$$\bar{S}_0 = \begin{pmatrix} 1\\2\\7 \end{pmatrix}$$</span>
  while their values at time <span class="math-container">$t=1$</span>, depending whether the scenario <span class="math-container">$\omega_1,\omega_2$</span> or <span class="math-container">$\omega_3$</span> happens, are given by the vectors 
  <span class="math-container">$$\bar{S}_1(\omega_1) = \begin{pmatrix} 1\\3\\9\end{pmatrix}, \quad \bar{S}_1(\omega_2) = \begin{pmatrix} 1\\1\\5\end{pmatrix}, \quad \bar{S}_1(\omega_3) = \begin{pmatrix} 1\\5\\10 \end{pmatrix}$$</span>
  (a) Show that this financial market has arbitrage.</p>
  
  <p>(b) Let <span class="math-container">$S_1^2(\omega_3) = 13$</span> 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.</p>
</blockquote>

<p><strong>Attempt :</strong></p>

<p><strong>(a)</strong> We have that a value process is defined as :</p>

<p><span class="math-container">$$V_t = V_t^\bar{\xi} = \bar{\xi}\cdot \bar{S}_t = \sum_{i=0}^d \xi_t^i\cdot \bar{S}_t^i, \quad t \in \{0,1\}$$</span></p>

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

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

<p><span class="math-container">$$V_0 \leq 0, \quad \mathbb P(V1 \geq 0) = 1, \quad \mathbb P(V_1 &gt; 0) &gt; 0$$</span></p>

<p>I understand that the different <span class="math-container">$S$</span> vectors will be plugged in to calculate <span class="math-container">$V_t$</span> but I really can't comprehend <span class="math-container">$\xi$</span>. What would the <span class="math-container">$\xi$</span> vector be ? </p>

<p>Any help for me to understand what <span class="math-container">$\xi$</span> really is based on the problem and how to complete my attempt will be much appreciated.</p>

<p>For <strong>(b)</strong>, showing that it does not have arbitrage is similar to <strong>(a)</strong> 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.</p>

<mathematical-economics><finance><financial-economics><financial-markets><martingales>
Showing that a market model has arbitrage and describing martingales
Rebellos
# Showing that a market model has arbitrage and describing martingales

(Asked by: username_0 on 2018-11-11)

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.

**Tags:** mathematical-economics, finance, financial-economics, financial-markets, martingales

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## 1 answers

### Answer by username_1 on 2023-08-16. Score: 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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