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I am struggling to find an arbitrage in the following configuration. I know how to prove that there is an arbitrage (using the fundamental theorem of asset pricing). So I ve proven there is an arbitrage. But how to find it?

I have two assets and a bond in a market where, the risk free asset has as interest $r$, and the two assets are defined as, where the lower script is the time of the asset, super script represents which asset (first or second) :

$$S_0^1 = 10 ; \qquad S_1^1 = \begin{bmatrix}12 \\ 8 \\ 6 \end{bmatrix}$$

$$S_0^2 = 5 ; \qquad S_1^2 = \begin{bmatrix}10 \\ 4 \\ 5 \end{bmatrix}$$

In that configuration, an obvious choice (given by the starting prices) is long asset 2 and short twice asset 1.

I know that if $S_0^2 = 6$, there is still an arbitrage. However, I can't find anymore how many shares of each asset I should take. In fact, actually I can if $r = 0$. In that case scenario, there is an arbitrage consisting in zero positions in the bond, a short position in stock 1 and two long positions in stock 2. This is not true anymore when $r > 0 $.

So I have two questions, can someone find an arbitrage here when I changed the price of the second asset, as well as what should be the method in general?

Also, is it possible that for creating an arbitrage, one has to invest in the bond ? I think it shouldn't change the arbitrage opportunities because the bond scales every output identically. Maybe this is the reason why I can't find a solution to my problem.

Cheers.


EDIT

My proof that there exists an arbitrage for $S_0^2 = 6$.

Using the fundamental theorem of asset pricing stating that there is an equivalent martingale measure iff the market is free of arbitrage, I am building up an EMM.

In order to do that, I search the solution to the following equation, where $p$ is the price of the second asset:

$$ \begin{bmatrix}10 \\ p \\ 1 \end{bmatrix} = \left ( \begin{matrix}12, 8 , 6 \\ 10, 4 , 5 \\ 1,1,1 \end{matrix} \right ) \begin{bmatrix}q_1 \\ q_2 \\ q_3 \end{bmatrix} $$

here $r$ is taken as equal to $0$. However the matrix is still invertible if for instance $r = 0.05$. I did the computations, it should be right, however it is a pain to write down in latex as those numbers are no longer integers...

the last line of the matrix comes from the fact that the sum of the probabilities has to be equal to $1$. Finally, using the implicit condition that all probabilities are positive, one get the three following conditions for the existence of the probabilities (iff existence of an EMM iff no arbitrage):

$$ 3 \leq p $$ $$ p \leq \frac{25} {3} $$ $$ 7 \leq p $$

thus, my conclusion is that when $ p \notin [7, 8 + \frac 1 3] \implies $ there exists an arbitrage. Am I making a mistake ?

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Let $S_t^k$ be the price of the $k^{th}$ risky asset at time $t$.

Let $x$ be your position in $S_t^0$ (risk-free bank account), $y$ your position in $S_t^1$ and $z$ your position in $S_t^2$.

You need to check two things to find an arbitrage strategy.

  1. Normally arbitrage strategies have zero initial cost, that is $$xS_0^0+yS_0^1+zS_0^2 =x+10y+6z\overset{!}=0.$$ Thus, $x=-10y-6z$.
  2. The payoff needs to be non-negative in every state and strictly positive in at least one state: $$xS_1^0+yS_1^1+zS_1^2 = \begin{bmatrix}x+12y+10z \\ x+8y+4z \\ x+6y+5z\end{bmatrix}= \begin{bmatrix}2y+4z \\ -2y-2z \\ -4y-z\end{bmatrix}.$$ As you see, if you chose $y<0$ and $z\in\left(-\frac{1}{2}y,-y\right)$, you'll get an arbitrage!

Example

Let $x=22$, $y=-4$ and $z=3$.

  • Your initial cost is $22-4\cdot10+3\cdot6=0$.
  • You payoff in state 1 is $22-4\cdot12+3\cdot10=4$.
  • Your payoff in state 2 is $22-4\cdot8+3\cdot4=2$.
  • Your payoff in state 3 is $22-4\cdot6+3\cdot5=13$.

So, you have zero cost but a positive payoff in every state $\implies$ arbitrage!

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    $\begingroup$ oh hi Alex ! Thanks for the answer. Coincidently I saw earlier your bounty question about the integral approximation. That's a good one ! For anyone wondering : math.stackexchange.com/questions/3638337/… I didn't realy understand why people were angry that you added a precision to the question. I guess it is because there is a reward at stake. $\endgroup$ – Marine Galantin Apr 25 at 23:04
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    $\begingroup$ Also, I am very sorry but the question wasn't about starting with initial prices 5 and 10, but with 6 and 10. This is where I have trouble finding a solution ! However your answer is well written, maybe a small change would solve the problem ? Do you see what I mean? I refer to the line " In that configuration, an obvious choice (given by the starting prices) is long asset 2 and short twice asset 1. " which corresponds to the solution you found I think. $\endgroup$ – Marine Galantin Apr 25 at 23:07
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    $\begingroup$ @MarineGalantin Thank you very much for your encouragement! :) It was really about a minor point, I didn't state the theorem precise enough. My fault, but yeah. I didn't expect it to be a big thing :o I would've thought that the main part of my question was quite clear. I hope you have more luck with your questions! Let me know if there's anything in my answer which needs further elaboration :) $\endgroup$ – Alex Apr 25 at 23:10
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    $\begingroup$ @MarineGalantin Oh sorry! I misread your question. I'll edit the answer :) Thanks for pointing that out $\endgroup$ – Alex Apr 25 at 23:12
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    $\begingroup$ I am. Please refer to my edit for the proof of the existence of an arbitrage. What do you think about it? Am I making somewhere a mistake ? $\endgroup$ – Marine Galantin Apr 25 at 23:34

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