Put–call parity and generalized neo-additive pricing rules

Abstract

We study price formulas suited for empirical research in financial markets in which put–call parity is satisfied. We find a connection between risk and the bid–ask spread. We further study the compatibility of the model with market frictions, and determine market subsets where the Fundamental Theorem of Asset Pricing applies. Finally, we characterize the price formula.

Appendix

Proof of Lemma 4.1

We first assume that the bid–ask spread is proportional to a constant. It follows immediately that the bid–ask spread of bets which yield 1 if some event occurs, and 0 if the complementary event occurs, is constant.

Now, we assume that the bid–ask spread of bets of the form $1_E0$, where E is an event of $\Omega$, is equal to a constant $\lambda \in \mathbb {R}$. We are going to show that the capacity values of complementary events sum to a constant. For all $E\notin \{\emptyset ,\Omega \}$, we have

$$\begin{aligned} B(1_E 0)=\nu (E)+\nu (E^c)-1. \end{aligned}$$

Thus,

$$\begin{aligned} \nu (E)+\nu (E^c)=\lambda +1 \quad \text { for all }E \notin \{\emptyset ,\Omega \}. \end{aligned}$$

Finally, we assume that the capacity values of complementary events sum to a constant $k\in \mathbb {R}$. We are going to show that the bid–ask spread is proportional to the range of revenues. We let $X\in \mathbb {R}^\Omega$. We denote $x_1,\dots ,x_n$ the n coordinates of X such that $x_1\ge x_2\ge \dots \ge x_n$. Up to reindexing the states of nature, we assume that the payoff X yields $x_i$ in $\omega _i$ for all $i\in \left\{ 1, \dots , n \right\}$. By definition, the bid–ask spread of X equals

$$\begin{aligned} \sum _{i=1}^{m} x_i[\nu (\left\{ \omega _j\in \Omega \mid j\le i\right\} )-\nu \left( \left\{ \omega _j\in \Omega \mid j< i\right\} \right) - \nu \left( \left\{ \omega _j\in \Omega \mid j\ge i\right\} \right) +\nu \left( \left\{ \omega _j\in \Omega \mid j> i\right\} \right) ] \end{aligned}$$

which simplifies to

$$\begin{aligned} x_1[\nu (\{\omega _1\})+\nu (\{\omega _2,\dots ,\omega _m\})-1]-x_m[\nu (\{\omega _1,\dots ,\omega _{m-1}\})+\nu (\{\omega _m\})-1]. \end{aligned}$$

By applying the above assumption and by substituting $\lambda =k-1$, we obtain the desired result

$$\begin{aligned} B(X)=\lambda (x_1-x_m). \end{aligned}$$

Proof of Proposition 4.1

We first assume that there is no arbitrage in the bid–ask spread. We are going to show that the capacity values of complementary events sum to a real greater than 1. By assumption, we have

$$\begin{aligned} \tilde{\pi }(X)\ge -\tilde{\pi }(-X), \text { for all } X\in \mathbb {R}^\Omega . \end{aligned}$$

In particular, we have

$$\begin{aligned} \tilde{\pi }(1_A0)\ge -\tilde{\pi }(-1_A0), \text { for all } A\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega ) \end{aligned}$$

which implies

$$\begin{aligned} \nu (A)+\nu (A^c)\ge 1, \text { for all } A\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega ). \end{aligned}$$

Now, we assume that the capacity values of complementary events sum to a real greater than 1. We are going to show that there is no arbitrage in the bid–ask spread. We let $X\in \mathbb {R}^\Omega$ be a payoff. We denote $x_1,\dots ,x_n$ the n coordinates of X such that $x_1\ge x_2\ge \dots \ge x_n$. Up to reindexing the states of nature, we assume that the payoff X yields $x_i$ in $\omega _i$ for all $i\in \left\{ 1, \dots , n \right\}$. Then by definition of a Choquet expectation, we have

$$\begin{aligned} \tilde{\pi }(X) = \sum _{i=1}^{m} x_i[\nu \left( \left\{ \omega _j\in \Omega \mid j\le i\right\} \right) -\nu \left( \left\{ \omega _j\in \Omega \mid j< i\right\} \right) ] \end{aligned}$$

which, by assumption, is greater than

$$\begin{aligned} \sum _{i=1}^m x_i [1-\nu \left( \left\{ \omega _{j}\in \Omega \mid j>i\right\} \right) -(1-\nu \left( \left\{ \omega _{j}\in \Omega \mid j\ge i\right\} \right) ]. \end{aligned}$$

This sum simplifies to

$$\begin{aligned} \sum _{i=1}^m x_i [\nu \left( \left\{ \omega _{j}\in \Omega \mid j\ge i\right\} \right) -\nu \left( \left\{ \omega _{j}\in \Omega \mid j>i\right\} \right) ] \end{aligned}$$

which is equal to $-\tilde{\pi }(-X)$. We hence obtain the desired result, for all $X\in \mathbb {R}^\Omega$,

$$\begin{aligned} \tilde{\pi }(X)\ge -\tilde{\pi }(-X). \end{aligned}$$

Proof of Proposition 6.1

We are going to show that an event E is frictionless if, and only if, the capacity is additive with respect to this event. We first assume that E is a frictionless event. We are going to show that the capacity is additive with respect to E. By assumption, we have

$$\begin{aligned} \tilde{\pi }(1_E0)+\tilde{\pi }(-1_E0) = \tilde{\pi }(1_E0+(-1_E0)) \end{aligned}$$

which implies

$$\begin{aligned} \nu (E)+\nu (E^c)=1. \end{aligned}$$

Now, we let $A\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega )$ such that $A\cap E\ne \emptyset$ and $A\cap E^c\ne \emptyset$. Then by assumptions, we have

$$\begin{aligned} \tilde{\pi }(1_E0+1_A0)=\tilde{\pi }(1_E0)+\tilde{\pi }(1_A0) \end{aligned}$$

and

$$\begin{aligned} \tilde{\pi }(1_{E^c}0+1_A0)=\tilde{\pi }(1_{E^c}0)+\tilde{\pi }(1_A0). \end{aligned}$$

It implies

$$\begin{aligned} \nu (A\cap E)+ \nu (E\cup A\cap E^c) =\nu (E)+\nu (A) \end{aligned}$$

(1)

and

$$\begin{aligned} \nu (A\cap E^c)+ \nu (E^c\cup A\cap E) =\nu (E^c)+\nu (A). \end{aligned}$$

(2)

We replace $\nu (E^c)$ by $1-\nu (E)$, and we combine Eqs. 1 and 2 to get

$$\begin{aligned} \nu (A\cap E^c)+ \nu (A\cap E)+ \nu (E^c\cup A\cap E) + \nu (E\cup A\cap E^c)=1+2\nu (A). \end{aligned}$$

We now substitute $\nu (E^c\cup A\cap E)$ with $\tilde{\pi }(1_{E^c\cup A\cap E}0)$ and $\nu (E\cup A\cap E^c)$ with $\tilde{\pi }(1_{E\cup A\cap E^c}0)$. By assumption, we get

$$\begin{aligned} \nu (A\cap E^c)+ \nu (A\cap E)+\tilde{\pi }(1_{E^c}0)+\tilde{\pi }(1_{A\cap E}0)+\tilde{\pi }(1_E0)+\tilde{\pi }(1_{A\cap E^c}0)=1+2\nu (A). \end{aligned}$$

Then, again by assumption, we get the desired result

$$\begin{aligned} \nu (A)=\nu (A\cap E)+ \nu (A\cap E^c). \end{aligned}$$

Now we assume that the capacity is additive with respect to an event E. We are going to show that E is frictionless: we are going to show that for all $a\in \mathbb {R}$ and all $X\in \mathbb {R}^\Omega$,

$$\begin{aligned} \tilde{\pi }(X+a_E0)=\tilde{\pi }(X)+a\tilde{\pi }(1_E0). \end{aligned}$$

(3)

We fix $X\in \mathbb {R}^\Omega$ and we denote $x_1,\dots ,x_n$ its n coordinates such that $x_1\ge x_2\ge \dots \ge x_n$. We denote $A_{2i-1}\cup A_{2i}$ the event in which the payoff yields $x_i$ with $(A_{2i-1}\cup A_{2i})\cap E = A_{2i-1}$, as in the following table

	$x_1$	$x_2$	...	$x_n$
E	$A_1$	$A_3$	...	$A_{2n-1}$
$E^c$	$A_2$	$A_4$	...	$A_{2n}$

so that all events in E have an odd subscript and all events in $E^c$ have an even subscript. Events $A_i$ can be empty. We denote $\mathcal {E}$ the set of even integers in $\{1,\dots ,n\}$ and $\mathcal {O}$ the set of odd integers in $\{1,\dots ,n\}$ and we fix $i\in \mathcal {O}$. We first show that Eq. 3 is satisfied for $a>0$. We denote $\rho$ the ranking associated with X, and $\mu$ the corresponding probability in the Weber set. We consider another payoff, $Y=X+a_E0$, denoting $\rho ^\star$ the ranking associated with this payoff, and $\mu ^\star$ the corresponding probability in the Weber set. We can now show that $\mu (A_{i}\cup A_{i+1})=\mu ^\star (A_{i}\cup A_{i+1})$. By assumption, we can decompose $\nu (\{A_j \mid Y(A_j)\le Y(A_i)\})$ with respect to E, that is with respect to its odd and even events. In other words, we have $\nu (\{A_j \mid Y(A_j)\le Y(A_i)\})$ equal to

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)\ge Y(A_i), \,\, j \in \mathcal {O}\}) +\nu (\{A_j \mid Y(A_j)\ge Y(A_i), \,\, j\in \mathcal {E}\}). \end{aligned}$$

(4)

Similarly, we can decompose $\nu (\{A_j \mid Y(A_j)> Y(A_i)\})$ with respect to E. It is equal to

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)>Y(A_i), \,\, j \in \mathcal {O}\})+\nu (\{A_j \mid Y(A_j)> Y(A_i), \,\, j \in \mathcal {E}\}). \end{aligned}$$

(5)

Since i is odd, we have

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)\ge Y(A_i), \,\, j \in \mathcal {E}\}) = \nu (\{A_j \mid Y(A_j)> Y(A_i), \,\, j \in \mathcal {E}\}). \end{aligned}$$

(6)

By definition, the probability $\mu ^\star (A_i\cup A_{i+1})$ is equal to

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)\ge Y(A_i)\}) -\nu (\{A_j \mid Y(A_j)> Y(A_i)\}) \end{aligned}$$

which, by Eq. 4, 5 and 6, is equal to

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)\ge Y(A_i), \,\, j\in \mathcal {O}\})\\ -\nu (\{A_j \mid Y(A_j)> Y(A_i), \,\, j\in \mathcal {O}\}). \end{aligned}$$

By construction, the equalities

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)\ge Y(A_i), \,\, j \in \mathcal {O}\})=\nu (\{A_j \mid X(A_j)\ge X(A_i), \,\, j \in \mathcal {O}\}) \end{aligned}$$

and

$$\begin{aligned} \nu (\{A_j \mid Y(A_j)> Y(A_i), \,\, j \in \mathcal {O}\})=\nu (\{A_j \mid X(A_j)> X(A_i), \,\, j \in \mathcal {O}\}) \end{aligned}$$

are satisfied. Thus, the probability $\mu ^\star (A_{i}\cup A_{i+1})$ is equal to

$$\begin{aligned} \nu (\{A_j \mid X(A_j)\ge X(A_i), \,\, j \in \mathcal {O}\}) -\nu (\{A_j \mid X(A_j)> X(A_i), \,\, j \in \mathcal {O}\}) \end{aligned}$$

which, in turn, by assumption, is equal to $\mu (A_{i}\cup A_{i+1})$, yielding

$$\begin{aligned} \tilde{\pi }(Y)=\tilde{\pi }(X)+a\tilde{\pi }(1_E0). \end{aligned}$$

We also have

$$\begin{aligned} \tilde{\pi }(X+a_{E^c}0)=\tilde{\pi }(X)+a\tilde{\pi }(1_{E^c}0). \end{aligned}$$

We replace $a_{E^c}0$ by $a(\mathbb {1}_\Omega -1_E0)$ and we use the assumption to replace $\tilde{\pi }(1_{E^c}0)$ by $1-\tilde{\pi }(1_E0)$ to get

$$\begin{aligned} \tilde{\pi }(X+a(\mathbb {1}_\Omega -1_E0))=\tilde{\pi }(X)+a(1-\tilde{\pi }(1_E0)). \end{aligned}$$

Hence,

$$\begin{aligned} \tilde{\pi }(X-a_E0)=\tilde{\pi }(X)-a\tilde{\pi }(1_E0). \end{aligned}$$

It follows that, for all $a\in \mathbb {R}$ and all $X\in \mathbb {R}^\Omega$,

$$\begin{aligned} \tilde{\pi }(Y)=\tilde{\pi }(X)+a\tilde{\pi }(1_E0), \end{aligned}$$

that is, E is a frictionless event.

Now, we can show that the capacity is additive with respect to an event E if, and only if, all probability values in the Weber set coincide with the value of the capacity for this event. We first assume that the capacity is additive with respect to an event E. We are going to show that all the probabilities in the Weber set coincide with the value taken by the capacity on E. We fix a probability $\mu$ in the Weber set. We consider a vector X associated with this probability, that is, there exists a ranking $\rho$ such that $\rho$ is associated with X and $\mu$ is associated with X. We denote $x_1,x_2, \dots , x_n$ the coordinates of X such that $x_1\ge x_2\ge \dots \ge x_n$. As shown in the following table, we denote $A_{2i-1}\cup A_{2i}$ the event in which the payoff yields $x_i$ such that $(A_{2i-1}\cup A_{2i})\cap E = A_{2i-1}$, so that all events in E have an odd subscript.

	$x_1$	$x_2$	...	$x_n$
E	$A_1$	$A_3$	...	$A_{2n-1}$
$E^c$	$A_2$	$A_4$	...	$A_{2n}$

The relationship

$$\begin{aligned} \nu (\{A_j\mid X(A_j)\ge X(A_i)\}) - \nu (\{A_j\mid X(A_j)> X(A_i)\}) \end{aligned}$$

simplifies to

$$\begin{aligned} \nu (\{A_j\mid X(A_j)\ge X(A_i), j \in \mathcal {O}\}) - \nu (\{A_j\mid X(A_j)> X(A_i), j \in \mathcal {O}\}) \end{aligned}$$

when i is odd and $\mu (E)$ is equal to

$$\begin{aligned} \sum _{\begin{array}{c} i=1\\ i \in \mathcal {O} \end{array}}^{2n} [ \nu (\{A_j\mid X(A_j)\ge X(A_i), j \in \mathcal {O}\}) - \nu (\{A_j\mid X(A_j)> X(A_i), j \in \mathcal {O}\})] \end{aligned}$$

and simplifies to $\nu (E)$.

Now, we assume that all probabilities in the Weber set coincide with the capacity value for an event E, and we will show that the capacity is additive with respect to E. We let $E_1, E_2$ be two distinct subsets of $\Omega$ such that $E= E_1\cup E_2$ and we consider two events A and B such that $A=B \cup E_1$ and $B\cap E = \emptyset$. We let $\rho$ be a ranking such that $\rho (E_1)>\rho (B)>\rho (E_2)>\rho (\Omega {\setminus } ( E_1 \cup B \cup E_2))$ with the convention that $\rho (A)>\rho (B)$ if $\rho (\omega _i)>\rho (\omega _j)$ for all $\omega _i\in A$ and all $\omega _j \in B$. We let $\mu$ be the probability associated with $\rho$. We have $\mu (E)$ equal to

$$\begin{aligned} \nu (E_1) + \nu (E_1\cup B \cup E_2) - \nu (E_1 \cup B) \end{aligned}$$

which is, in turn, equal to

$$\begin{aligned} \nu (A\cap E) + \nu (A \cup E) - \nu (A). \end{aligned}$$

We let $\rho ^\star$ be a ranking such that

$$\rho ^\star (B)>\rho ^\star (E_1)>\rho ^\star (E_2)>\rho ^\star (\Omega {\setminus } ( E_1\cup B\cup E_2)$$

and we let $\mu ^\star$ be the associated probability. We have $\mu ^\star (E)$ equal to

$$\begin{aligned} \nu (B\cup E_1 \cup E_2) - \nu (B) \text { which is equal to } \nu (A\cup E) - \nu (A \cap E^c) \end{aligned}$$

and we get the desired result:

$$\begin{aligned} \nu (A) = \nu (A\cap E) + \nu (A \cap E^c)\text {, for all } A\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega ). \end{aligned}$$

Proof of Proposition 6.2

We assume that X is a frictionless payoff. We are, therefore, going to show that we can decompose it as a sum of frictionless events, in part, by contradiction. We write X with the following form:

$$\begin{aligned} X=\sum _{i=1}^n {x_i}_{E_i}0. \end{aligned}$$

We are going to prove that the $E_i$ are frictionless. We assume that there exist some $A_i$ $i\in \left\{ 1,\dots , n \right\}$ that are not frictionless. Up to reindexing, we decompose X into two sums. The left sum groups all $x_i$’s on frictionless events and the right one groups $x_i$’s on events with frictions:

$$\begin{aligned} X=\sum _{i=1}^k {x_i}_{E_i}0 +\sum _{i={k+1}}^n {x_i}_{E_i}0. \end{aligned}$$

We have $\tilde{\pi }(X+Y)$ equal to

$$\begin{aligned} \tilde{\pi }\left( X-\sum _{i=1}^k {x_i}_{E_i}0+\sum _{i=1}^k {x_i}_{E_i}0 +Y\right) . \end{aligned}$$

By assumption, this is not equal to

$$\begin{aligned} \tilde{\pi }\left( \sum _{i={k+1}}^n {x_i}_{E_i}0\right) +\tilde{\pi }\left( \sum _{i=1}^k {x_i}_{E_i}0 +Y\right) . \end{aligned}$$

By additivity, the preceding equation is equal to

$$\begin{aligned} \tilde{\pi }\left( \sum _{i={k+1}}^n {x_i}_{E_i}0+\sum _{i=1}^k {x_i}_{E_i}0\right) +\tilde{\pi }(Y). \end{aligned}$$

We can now recognize $\tilde{\pi }(X)+\tilde{\pi }(Y)$, a contradiction.

Now, we assume that X can be decomposed as a sum of frictionless events. We are going to show that X is frictionless. We have

$$\begin{aligned} X=\sum _{i=1}^n {x_i}_{E_i}0, \end{aligned}$$

where for all $i\in \left\{ 1,\dots ,n\right\}$ $x_i\in \mathbb {R}$, the events $E_i$ are frictionless and

$$\begin{aligned} \sum _{i=1}^n1_{E_i}0=\mathbb {1}_\Omega . \end{aligned}$$

If we let $Y\in \mathbb {R}^\Omega$, we have $\tilde{\pi }(X+Y)$ equal to

$$\begin{aligned} \tilde{\pi }\left( \sum _{i=1}^n {x_i}_{E_i}0+Y\right) . \end{aligned}$$

This is, by assumption, equal to

$$\begin{aligned} \sum _{i=1}^n \tilde{\pi }({x_i}_{E_i}0)+\tilde{\pi }(Y) \end{aligned}$$

We get the desired result: $\tilde{\pi }(X)+\tilde{\pi }(Y)$ for all $Y\in \mathbb {R}^\Omega$.

Proof of Lemma 6.1

We assume that there exists an event $A\notin \{\emptyset ,\Omega \}$ such that $\nu (A)+\nu (A^c)=1$. We are going to show that the bid–ask spread is nil. From Lemma 4.1, we have $\lambda = \nu (A)+\nu (A^c)-1$. Thus, $\lambda =0$ which entails $B(X)=0$ for all $X\in \mathbb {R}^\Omega$.

Now, we assume that the bid–ask spread is null. By definition, the bid–ask spread, $B(1_A0)=0$ implies $\lambda =0$ with $\lambda =\nu (A)+\nu (A^c)-1$.

Proof of Proposition 6.3

We assume that E is a frictionless event; we can show that $\tilde{\pi }$ is frictionless. We consider an event $A\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega )$, we have

$$\begin{aligned} \nu (A)=\nu (A\cap E)+\nu (A\cap E^c). \end{aligned}$$

This implies

$$\begin{aligned} a+bp(A)=2a+bp(A). \end{aligned}$$

Hence, $a=0$. Moreover,

$$\begin{aligned} \nu (E)+\nu (E^c)=b=1. \end{aligned}$$

Thus $a=0$ and $b=1$. Therefore, for all $A\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega )$,

$$\begin{aligned} \nu (A)=p(A). \end{aligned}$$

Now, if we assume that $\tilde{\pi }$ is frictionless, then $\nu$ is additive.

Proof of Proposition 7.1

First, we assume that the capacity is pairwise additive for payoffs with matching extreme revenues. Then it is, in particular, additive for bets with matching extreme revenues. We will now show that the capacity is a GNAC. To do so, we consider the following property, which we call Property A.

Definition 8.1

(Property A, Eichberger et al. 2012) $\nu (E\cup F)-\nu (F)=\nu (E\cup G)-\nu (G)$ is satisfied for all events $E,F,G\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega )$ such that $E\cup F\ne \Omega$, $E\cup G\ne \Omega$, $E\cap F= \emptyset = E\cap G$, $F\ne \emptyset$, $G\ne \emptyset$.

Eichberger et al. (2012) showed in Lemma 3 that Property A is satisfied if, and only if, the capacity is a GNAC. We will show that Property A is satisfied. We let $A, B\in {{\,\mathrm{\mathcal {P}}\,}}(\Omega )$, such that $A\cap B \ne \emptyset$ and $A\cup B\ne \Omega$. The bets $1_A0, 1_B0\in \mathbb {R}^\Omega$ have matching extreme revenues. Hence, by assumption

$$\begin{aligned} \tilde{\pi }(1_A0+1_B0)=\nu (A\cap B)+\nu (A\cup B) \end{aligned}$$

which is equal to $\nu (A)+\nu (B)$. Hence, the result is

$$\begin{aligned} \nu (A\cup B)-\nu ( B)=\nu (A)-\nu (A\cap B). \end{aligned}$$

We denote $E=A \backslash A \cap B$, $F=A\cap B$ and $G=B$. We get Property A with $F\subset G$:

$$\begin{aligned} \nu (E\cup G)-\nu (G)=\nu (E\cup F)-\nu (F). \end{aligned}$$

Moreover, if we let $F_1,F_2\subset G$ then

$$\begin{aligned} \nu (E\cup F_1)-\nu (F_1)=\nu (E\cup F_2)-\nu (F_2). \end{aligned}$$

Now, we assume that the capacity is a GNAC then by the definition of a GNAC pricing rule, it is immediate that it is additive among payoffs with matching extreme revenues.□