Game-theoretic derivation of upper hedging prices of multivariate contingent claims and submodularity

Abstract

We investigate upper and lower hedging prices of multivariate contingent claims from the viewpoint of game-theoretic probability and submodularity. By considering a game between “Market” and “Investor” in discrete time, the pricing problem is reduced to a backward induction of an optimization over simplexes. For European options with payoff functions satisfying a combinatorial property called submodularity or supermodularity, this optimization is solved in closed form by using the Lovász extension and the upper and lower hedging prices can be calculated efficiently. This class includes the options on the maximum or the minimum of several assets. We also study the asymptotic behavior as the number of game rounds goes to infinity. The upper and lower hedging prices of European options converge to the solutions of the Black–Scholes–Barenblatt equations. For European options with submodular or supermodular payoff functions, the Black–Scholes–Barenblatt equation is reduced to the linear Black–Scholes equation and it is solved in closed form. Numerical results show the validity of the theoretical results.

Appendices

Number of candidates ${\widetilde{\chi }}$ in three dimension

Here, we calculate the number of candidates ${\widetilde{\chi }}$ when $d=3$. Let $\mathrm{int} (A)$ be the interior of a set A and $A^c$ be the complement of A.

Lemma A.1

Let

$$\begin{aligned} T_1= & {} \mathrm{conv} \{ (0,0,0),(0,1,1),(1,1,0),(1,0,1) \}, \\ T_2= & {} \mathrm{conv} \{ (0,0,1),(0,1,0),(1,0,0),(1,1,1) \} \end{aligned}$$

be the regular tetrahedra in the cube $[0,1]^3$. Consider a point $x \in [0,1]^3$ and let $N(x) = \{ \{ z_0,z_1,z_2,z_3 \} \mid z_k \in \{ 0,1 \}^3, x \in \mathrm{conv} \{z_0,z_1,z_2,z_3\}, \dim \mathrm{conv}\{ z_0, z_1, z_2,$$z_3\}=3 \}$ be the set of tetrahedra containing x.

• If $x \in \mathrm{int} (T_1 \cap T_2)$, then $|N (x)| = 14$.

• If $x \in \mathrm{int} (T_1 \cap T^c_2)$ or $x \in \mathrm{int} (T^c_1 \cap T_2)$, then $|N (x)| = 11$.

• If $x \in \mathrm{int} (T^c_1 \cup T^c_2)$, then $|N (x)| = 8$.

Fig. 8

Cube and cutting planes

Proof

We only give a sketch of a proof, because we used computer to count the number of tetrahedra containing a point x.

The cube $[0,1]^3$ is denoted as the left picture of Fig. 8. There are 14 planes containing three or four vertices of the cube, which cut into the cube. There are 6 Type 1 planes containing four vertices with the equations

$$\begin{aligned} x=y, \ y=z,\ x=z, \ x+y=1, \ y+z=1, \ x+z=1, \end{aligned}$$

(38)

and there are 8 Type 2 planes containing three vertices with the equations

$$\begin{aligned}&x+y+z=1, \ x+y+z=2, \ x-y+z=0, \ x-y+z=1, \nonumber \\&x+y-z=0, \ x+y-z=1, \ -x+y+z=1, \ -x+y+z=0. \end{aligned}$$

(39)

There are 58 tetrahedra (simplexes) in 4 types as in Fig. 9. There are 8 Type 1 tetrahedra, which are corner simplexes. There are 2 Type 2 regular tetrahedra denoted as $T_1, T_2$ in the lemma. There are 24 Type 3 tetrahedra and there are 24 Type 4 tetrahedra. We keep the list of these 58 tetrahedra in a computer program.

On the other hand, it is easy to visualize the 14 planes in (38) and (39) by fixing z and drawing the sections of the cube as in Fig. 10. The 14 planes appear as 14 lines inside the unit squares in Fig. 10. Note that we only need to consider $z<1/2$ by symmetry: $z\leftrightarrow 1-z$. For each region of the sections of Fig. 10 we count the number of tetrahedra containing the region. Then we obtain the lemma.

$\square$

Fig. 9

Four types of tetrahedra

Fig. 10

Sections of the cube

Note that Lemma A.1 applies only for generic x. For x on the boundary of a tetrahedron, the number |N(x)| may be larger. For example $x=(1/2,1/2,1/2)$ is contained in $|N(x)|=50$ tetrahedra of Types 2–4.

Lower bound on the number of candidate ${\widetilde{\chi }}$

Here, we provide a lower bound on the number of candidates ${\widetilde{\chi }}$ for general d.

Lemma B.1

Assume $d \ge 2$. Consider a point x in the d-dimensional hypercube $[0,1]^d$ and define $N (x) = \{ \{ z_0,\ldots ,z_d \} \mid z_k \in \{ 0,1 \}^d, x \in \mathrm{conv} \{z_0,\ldots ,z_d\}, \dim \mathrm{conv}\{z_1,$$\ldots , z_d\}=d\}$. Then,

$$\begin{aligned} |N (x)| \ge 2^{d-2}. \end{aligned}$$

Proof

First, we consider the case where x is a generic point, Without loss of generality, we assume

$$\begin{aligned} 0< x_1< x_2< \cdots< x_d < 1/2. \end{aligned}$$

(40)

Let

$$\begin{aligned} c = \min \{x_1, x_2 - x_1, x_3- x_2, \ldots , x_d - x_{d-1}\} > 0, \end{aligned}$$

and $i_{*}$ be an index attaining the minimum:

$$\begin{aligned} c = x_{i_{*}} - x_{i_{*}-1}. \end{aligned}$$

(41)

Note that $c<1/4$ since $d \ge 2$. In the following, we focus on simplexes that have the zero vector as one vertex. Note that such simplex has nonempty interior if and only if the other d vertices $y_1,\ldots ,y_d$ are linearly independent.

Let $e_i \in \{ 0,1 \}^d$, $i=1,\ldots ,d+1$ be a 0-1 vector defined by

$$\begin{aligned} (e_i)_j={\left\{ \begin{array}{ll} 0 &{} \quad (j < i) \\ 1 &{}\quad (j \ge i) \end{array}\right. }. \end{aligned}$$

(42)

In particular, $e_{d+1}$ is the zero vector. Then, the vector x is decomposed as

$$\begin{aligned} x = \sum _{i=1}^{d+1} (x_i-x_{i-1}) e_i, \end{aligned}$$

(43)

where we define $x_0=0$ and $x_{d+1}=1$. Therefore, from (40),

$$\begin{aligned} x \in \mathrm{conv}\{ e_1,\ldots ,e_{d+1} \}. \end{aligned}$$

(44)

Now, we construct simplexes containing x by changing the vertex $e_{i_{*}}$ in (44). Note that every 0-1 vector $z \in \{ 0,1 \}^d$ is uniquely expressed as

$$\begin{aligned} z = \sum _{i=1}^d c_i e_i, \end{aligned}$$

where $c_i \in \{ -\,1,0,1 \}$ satisfies

$$\begin{aligned} \sum _{j=1}^i c_j \in \{ 0,1 \} \quad (i=1,\ldots ,d). \end{aligned}$$

Under this correspondence, there are $2^{d-2}$ vectors $\epsilon$ with $c_{i_{*}-1} = 0$ and $c_{i_{*}} = 1$. These vectors are given by

$$\begin{aligned} \epsilon = \cdots - e_k + e_{i_{*}} - e_l + \cdots \end{aligned}$$

(45)

for some $k < i_{*}$ and $l > i_{*}$. Thus,

$$\begin{aligned} \epsilon - \cdots + e_k +e_l - \cdots = e_{i_{*}}, \end{aligned}$$

(46)

where the sum of coefficients in the left hand side is 1 or 2. By substituting (46) into (43),

$$\begin{aligned} x&= \sum _{i \ne i_{*}} (x_i-x_{i-1}) e_i + c (\epsilon - \cdots + e_k +e_l - \cdots ) \\&= \sum _{i \ne i_{*}} b_i e_i + c \epsilon , \end{aligned}$$

where $b_i \ge 0$ and $\sum _{i \ne i_{*}} b_i + c=1$. Therefore,

$$\begin{aligned} x \in \mathrm{conv} \{ e_1,\ldots ,e_{i_{*}-1},\epsilon ,e_{i_{*}+1},\ldots ,e_{d+1} \}. \end{aligned}$$

Since there are $2^{d-2}$ choices of $\epsilon$, we have $|N (x)| \ge 2^{d-2}$.

Next, we consider the case where x is not a generic point. Without loss of generality, we assume

$$\begin{aligned} 0 \le x_1 \le x_2 \le \cdots \le x_d \le 1/2, \end{aligned}$$

where at least one inequality holds with equality. Then, we can take a sequence of generic points $y_1,y_2,\ldots$ satisfying (40) and having the same $i_*$ that converges to x. Let $N=\bigcap N (y_k)$. Then, from the above argument, $|N (y_k)| \ge 2^{d-2}$ for each k and the $2^{d-2}$ simplexes containing $y_k$ are common. In particular $|N| \ge 2^{d-2}$. Also, since the simplex is closed, x belongs to each simplex in N: $N \subset N(x)$. Therefore, $|N (x)| \ge 2^{d-2}$. $\square$