Edit: I'm a dumbass. The thing below is supposed to be just the motivation of asking. I want to ask for below and in general, hehe.

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 $\mathbb Q$

I noticed that one of the probabilities, in what was attempted to be the equivalent martingale measure, is negative. I remember reading about negative probabilities in Wiki and here

Half of a Coin: Negative Probabilities

However the following links

Negative Probabilities in Financial Modeling

Why so Negative to Negative Probabilities?

mentioned by Wiki seem to assume absence of arbitrage so I think they are not applicable. Or are they?

Is it perhaps that this market can be considered to be arbitrage-free under some quasiprobability measure that allows negative probabilities?

  • 1
    $\begingroup$ Your issue is that two mathematically distinct notions often get conflated and called no arbitrage. The distinct assumptions are (1) that asset pricing function $f$ is linear and (2) that any security with a positive payoff has a positive price. (1) implies the existence of state prices (that are possibly negative). Adding assumption (2) implies that there exists a positive vector of state prices. Cochrane refers to the first assumption as the law of one price and the second as the absence of arbitrage. $\endgroup$ Commented Dec 30, 2017 at 6:45

2 Answers 2


Without looking at probabilities or quasiprobabilities, I think your market will always allow for arbitrage opportunities.

Suppose you have a security that pays [1;0;0], this arrow security can be replicated by a portfolio of inv(D')*[1;0;0] = [-2; 0.5; 2.5]

the price of this portfolio is 2*-2 + 0.5*3 +1*2.5 = 0; So regardless of the probabilities, you have a security which has initial price of 0 and offers a payout of 1 in one specific state of nature.

  • $\begingroup$ I said 'there exists arbitrage in this market', so when you say 'I think your market will always allow for arbitrage opportunities' .................?? $\endgroup$
    – BCLC
    Commented Dec 30, 2017 at 15:17
  • $\begingroup$ Because you were looking for a way to make it arbitrage free by working with 'quasi probabilities: "Is it perhaps that this market can be considered to be arbitrage-free under some quasiprobability measure that allows negative probabilities?" My comment is, that regardless of probability measure you will always have arbitrage. $\endgroup$
    – mbison
    Commented Dec 30, 2017 at 16:04
  • 1
    $\begingroup$ Maybe to understand your problem better, how do you define arbitrage free under a 'quasi probability' measure? Do you want the price of the security to equal its discounted expected value under your 'quasi prob' measure? $\endgroup$
    – mbison
    Commented Dec 30, 2017 at 16:08

I read the paper and I do not believe so. No, negative probabilities are not useful for pricing options. While I have a very long list of reasons, there are a couple of simple ones that suffice.

First, such contracts would not be coherent in the de Finetti sense. As such, they intrinsically represent an arbitrage opportunity for a clever actor. The second is pragmatic. What is really being attempted is to permit a Brownian motion with drift to have meaning in a region where it cannot exist. In fact, during the crisis liquidity collapsed and the cost of liquidity went to infinity. Prices were intrinsically out of equilibrium in the sense that the present value of physical capital could no longer match the prices of capital contracts. The system halted. If it is a Brownian motion with drift, it halted. A wall was hit. The use of negative probabilities is an attempt to allow the differential equation to continue to exist as if the model were real. Models are not real.

Finally, it should be avoided because such contracts would not be admissible in the Wald sense. Again, it would be a long discussion, but you could construct a Bayesian model of the unobservables using standard definitions of probability. Because Bayesian methods, under very mild requirements, are intrinsically admissible, and the methods described in the paper would have to map to the Bayesian solution either at the limit or at each sample to be admissible you are going to have a problem. The issue is that Bayesian methods handle unobservables and so things such as catastrophe sets or other sources of incompatible observations may not be an intrinsic issue implying a very different math. Since admissibility requires matching the Bayesian choice and it could be modeled in a simpler manner using Bayesian tools that break with the standard differential equation, it is reasonable to believe that a solution driven from negative probabilities is outside the complete class of solutions, in the Wald sense.

The fact that one can construct a measure, while interesting, doesn't mean the measure would have a use in all circumstances. Because option prices are gambles inside non-conserving systems it is dangerous to go to far afield from standard methods when standard methods will work.


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