# Pricing when arbitrage is possible through Negative Probabilities or something else

Assume that we have a general one-period market model consisting of $d+1$ assets and $N$ states.

Using a replicating portfolio $\phi$, determine $\Pi(0;X)$, the price of a European call option, with payoff $X$, on the asset $S_1^2$ with strike price $K = 1$ given that

$$S_0 =\begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, S_1 = \begin{bmatrix} S_1^0\\ S_1^1\\ S_1^2 \end{bmatrix}, D = \begin{bmatrix} 1 & 2 & 3\\ 2 & 2 & 4\\ 0.8 & 1.2 & 1.6 \end{bmatrix}$$

where the columns of $D$ represent the states for each asset and the rows of D represent the assets for each state

What I tried:

We compute that:

$$X = \begin{bmatrix} 0\\ 0.2\\ 0.6 \end{bmatrix}$$

If we solve $D'\phi = X$, we get:

$$\phi = \begin{bmatrix} 0.6\\ 0.1\\ -1 \end{bmatrix}$$

It would seem that the price of the European call option $\Pi(0;X)$ is given by the value of the replicating portfolio

$$S_0'\phi = 0.5$$

On one hand, if we were to try to see if there is arbitrage in this market by seeing if a state price vector $\psi$ exists by solving $S_0 = D \psi$, we get

$$\psi = \begin{bmatrix} 0\\ -0.5\\ 1 \end{bmatrix}$$

Hence there is no strictly positive state price vector $\psi$ s.t. $S_0 = D \psi$. By 'the fundamental theorem of asset pricing' (or 'the fundamental theorem of finance' or '1.3.1' here), there exists arbitrage in this market.

On the other hand the price of $0.5$ seems to be confirmed by:

$$\Pi(0;X) = \beta E^{\mathbb Q}[X]$$

where $\beta = \sum_{i=1}^{3} \psi_i = 0.5$ (sum of elements of $\psi$) and $\mathbb Q$ is supposed to be the equivalent martingale measure given by $q_i = \frac{\psi_i}{\beta}$.

Thus we have

$$E^{\mathbb Q}[X] = q_1X(\omega_1) + q_2X(\omega_2) + q_3X(\omega_3)$$

$$= 0 + \color{red}{-1} \times 0.2 + 2 \times 0.6 = 1$$

$$\to \Pi(0;X) = 0.5$$

I guess $\therefore$ that we cannot determine the price of the European call using $\Pi(0;X) = \beta E^{Q}[X]$ because there is no equivalent martingale measure $\mathbb Q$

So what's the verdict? Can we say the price is 0.5? How can we price even if there is arbitrage? What's the interpretation of 0.5?

## 2 Answers

I believe there is not a unique price if you can't short. Say, instead of buying the option you spent 0.5 on a half a unit of the asset $S^2_1$ This asset pays out $[0.4, 0.6, 0.8]$ which first order stochastically dominates the option. So, no matter your probability beliefs about the states, in that setting you'd never pay $0.5$ for the option which pays less in every state. This suggests the right price is less than $0.5$. Similarly, buying $0.25$ units of the $S^0_1$ asset or $0.167$ units of the $S^1_1$ asset would likewise stochastically dominate the option. In fact, because for $0.375$ units of asset $S^1_2$, which costs on $0.375$, you can still have an asset that pays out $[0.3, 0.45, 0.6]$, it seems unlikely that the price could even be as high as $0.375$. Asset 0 implies a price below $0.4$ and asset 1 below $0.45$

Some python code to solve:

import numpy as np
S0 = np.array([,,])
D = np.array([[1,2,3], [2,2,4], [0.8, 1.2, 1.6]])
X = np.array([[0.0],[0.2],[0.6]])
phi = np.dot(np.linalg.inv(D.transpose()), X)
print('The weights of the portfolio that replicates payoff X are: \n', phi)
P_X = np.dot(S0.transpose(), phi)
print('With a price: ', P_X)
print('Normalizing to pay a fixed price P_X for each of the three assets, what payoffs can you get?')
D_norm = D/(2*S0)
print(D_norm)
print('Notice that all three first order stochastically dominate the option for a price of: ', P_X)
print(D_norm - X.transpose())
print('Using each of the base assets, what\'s the minimum quantity that dominates?')
D_relative = X.transpose() / D
print(D_relative)
Min_dominating_fraction = np.max(D_relative,axis=1)
print('Minimum fraction of each of the assets that dominates X\n', Min_dominating_fraction)
P_Min_dominating_fraction = S0.transpose() * Min_dominating_fraction
print('At prices of: ', P_Min_dominating_fraction)
print('Therefore the option price should be less than: ', np.min(P_Min_dominating_fraction))


This code doesn't spell out the price of the option, it just show my calculations for the paragraph above. I believe the real price of this option would actually be zero if shorting is permitted. If you buy three units of asset $S^2_0$ and short one unit of $S^1_0$ you get an asset with payouts $[ 0.4, 1.6, 0.8]$. This position costs nothing to take, has positive payouts for all states, and first order stochastically dominates the option itself. Since it is possible to make a better than replicating portfolio at zero cost the price should be zero. Oh the insanity at work when arbitrages are present!

• BKay, in your opinion what's the interpretation of 0.5? – BCLC Oct 4 '17 at 10:26
1. You cannot use negative probabilities in this context. When there is no unique probability measure, there can be no unique price. You only know that it is in [0, 0.6] range, if you want to tighten this interval you need to make further assumptions/tweak inputs
2. I agree with your conclusion that there no suitable probability measure. But I am not sure about your computation. In 3 state world, you only have 2 variables (probabilities of 2 states of your choice), the probability of the third state can be determined from the fact that the sum of probabilities is $1$ - you do not seem to be using this fact! In your case, you have 3 assets, thus no arbitrage 3 equations. Solving 3 equations with 2 variables almost always fail! It would make sense to introduce the third variable, the interest rate. Let the probability of state $1$ be $p$, the probability of state $3$ be q and the interest rate be $r$. Thus for assets $1$ and $3$ we have

$p + 2(1 - p - q) + 3q = 2(1+ r)$ and $0.8 p + 1.2(1-p-q) + 1.6 q = 1+r$

which is equavalent to $q-p=2r$ and $0.8(q-p) +0.4 = 2r$; this implies that $q-p=2$, which is not solvable if $p,q \in [0;1]$!

• Thanks Yulia V. 1 How do you know negative probabilities cannot be used? Of course there does not exist a equivalent martingale probability measure but how do we know there might not be a equivalent martingale quasiprobability measure? Where did you get $[0,0.6]$? 2 What do you mean? Solving $S_0 = D \psi$ indeed gives such vector as a solution – BCLC Dec 16 '15 at 19:00
• $[0, 0.6]$ comes from the fact that, in the given 3 states, $0$ is the smallest price of the option (state $0$) and $0.6$ is the highest possible price of the option (state $2$). Trading outside these bounds would be unreasonable. – Yulia V Dec 16 '15 at 19:10
• Equation $S_0 = D \psi$ seems to be wrong because 1. It does not take into account the fact that the sum of probabilities to end up in state 0, 1 and 2 must be $1$; 2. It does not take into account the interest rate (actually, the upper bound of the interval should be a discounted value of $0.6$, but, since we cannot work out the interest rate, I have left it as $0.6$ – Yulia V Dec 16 '15 at 19:14
• I would not consider something as esoteric as negative probabilities for this very simple problem. In the real world, I would look into data error/noise or bid/ask. – Yulia V Dec 16 '15 at 19:15
• no, it is $1/4$ :) Not sure why you divide by 2. In any case, if yo do not impose "=1"constraint when solving the equation, your solution will not, in all likelihood, satisfy it. – Yulia V Dec 16 '15 at 19:46