Two price economies in continuous time

Abstract

Static and discrete time pricing operators for two price economies are reviewed and then generalized to the continuous time setting of an underlying Hunt process. The continuous time operators define nonlinear partial integro–differential equations that are solved numerically for the three valuations of bid, ask and expectation. The operators employ concave distortions by inducing a probability into the infinitesimal generator of a Hunt process. This probability is then distorted. Two nonlinear operators based on different approaches to truncating small jumps are developed and termed $QV$ for quadratic variation and $NL$ for normalized Lévy. Examples illustrate the resulting valuations. A sample book of derivatives on a single underlier is employed to display the gap between the bid and ask values for the book and the sum of comparable values for the components of the book.

Appendix

Proof that $c(A),$ (5) is a Choquet capacity.

To show that $c$ is a capacity we must show that

$$\begin{aligned} c(A\cup B)+c(A\cap B)\le c(A)+c(B) \end{aligned}$$

Defining

$$\begin{aligned} a&= Q^{*}(A-B) \\ b&= Q^{*}(B-A) \\ c&= Q^{*}(A\cap B) \end{aligned}$$

we must show that

$$\begin{aligned} \Psi (a+b+c)+\Psi (c)\le \Psi (a+c)+\Psi (b+c). \end{aligned}$$

Equivalently we have

$$\begin{aligned} \Psi (a+b+c)-\Psi (a+c)\le \Psi (b+c)-\Psi (c). \end{aligned}$$

Now

$$\begin{aligned} \Psi (a+b+c)-\Psi (a+c)&= \int \limits _{a+c}^{a+b+c}\Psi ^{\prime }(x)dx \\&= \int \limits _{c}^{b+c}\Psi ^{\prime }(a+x)dx \\&\le \int \limits _{c}^{b+c}\Psi ^{\prime }(x)dx, \text{ as } \Psi ^{\prime }\text{ is } \text{ decreasing. } \\&= \Psi (b+c)-\Psi (c). \end{aligned}$$

Proof that $\mathcal{G }_{QV}(u(.,t))\le \mathcal{L }(u(.,t))$

The operator $\mathcal{G }_{QV}(u(.,t))$ evaluates the distorted expectation of a random variable $Y_{x,t}$ with distribution function $F_{x,t}(v)$ that may be written as

$$\begin{aligned} -\int \limits _{-\infty }^{0}\Psi (F_{x,t}(v))dv+\int \limits _{0}^{\infty }\left( 1-\Psi (F_{x,t}(v)\right) dv. \end{aligned}$$

On the other hand the operator $\mathcal{L }$ evaluates the expectation that may be written as

$$\begin{aligned} -\int \limits _{-\infty }^{0}F_{x,t}(v)dv+\int \limits _{0}^{\infty }\left( 1-F_{x,t}(v)\right) dv. \end{aligned}$$

The results follows on noting that $\Psi (u)\ge u$.

Proof of Eq. (17)

We begin by writing

$$\begin{aligned} u\left( X_{t},T-t\right)&= u(X_{0},T)+\sum _{(t_{n})}\left( u(X_{t_{i+1}},T-t_{i+1}\right) -u(X_{t_{i}},T-t_{i})) \\&= u(X_{0},T)+(A)+(B) \end{aligned}$$

where $(t_{n})$ is a sequence of subdivisions of $[0,t],$ with mesh going to zero and

$$\begin{aligned} A&= \sum _{(t_{n})}\left\{ u\left( X_{t_{i+1}},T-t_{i+1}\right) -u(X_{t_{i}},T-t_{i+1})\right\} \\ B&= \sum _{(t_{n})}\left\{ u\left( X_{t_{i}},T-t_{i+1}\right) -u(X_{t_{i}},T-t_{i})\right\} \end{aligned}$$

For $B$ we note that if $u_{t}^{\prime }$ is jointly continuous

$$\begin{aligned} B=-\sum _{(t_{n})}\int \limits _{t_{i}}^{t_{i+1}}u_{t}^{\prime }(X_{t_{i}},T-s)ds \underset{N\rightarrow \infty }{\rightarrow }-\int \limits _{0}^{t}u_{t}^{\prime }(X_{s},T-s)ds. \end{aligned}$$

It remains to consider $A$.

$$\begin{aligned} A&= \sum _{(t_{n})}u\left( X_{t_{i+1}},T-t_{i+1}\right) -u(X_{t_{i}},T-t_{i+1}) \\&= \sum _{(t_{n})}\left\{ M_{t_{i+1}}^{(t_{i+1})}-M_{t_{i}}^{(t_{i+1})}\right\} +\sum _{(t_{n})}\int \limits _{t_{i}}^{t_{i+1}}\mathcal{L }(u(.,T-t_{i+1}))(X_{s})ds, \end{aligned}$$

where the notation $M$ with a superscript denotes a martingale. The second term converges to

$$\begin{aligned} \int \limits _{0}^{t}\mathcal{L }(u(.,T-s))(X_{s})ds \end{aligned}$$

and so the first converges to a limit we call $M_{t}$. We wish to show that $M_{t}$ is a martingale, but this follows from the tower property

$$\begin{aligned} M_{t}=L^{1}-\lim \left( \sum _{\left( t_{n}\right) \le t}\left( M_{t_{i+1}}^{(t_{i+1})}-M_{t_{i}}^{(t_{i+1})}\right) \right) \end{aligned}$$

Then for $s<t$

$$\begin{aligned} E\left[ M_{t}|\mathcal{F }_{s}\right]&= \sum _{\left( t_{n}\right) \le s}\left( M_{t_{i+1}}^{(t_{i+1})}-M_{t_{i}}^{(t_{i+1})}\right) +E\left[ \sum _{s\le t_{n}\le t}M_{t_{i+1}}^{(t_{i+1})}-M_{t_{i}}^{(t_{i+1})}| \mathcal{F }_{s}\right] \\&= M_{s}. \end{aligned}$$

Hence the result.