Model uncertainty, recalibration, and the emergence of delta–vega hedging

Abstract

We study option pricing and hedging with uncertainty about a Black–Scholes reference model which is dynamically recalibrated to the market price of a liquidly traded vanilla option. For dynamic trading in the underlying asset and this vanilla option, delta–vega hedging is asymptotically optimal in the limit for small uncertainty aversion. The corresponding indifference price corrections are determined by the disparity between the vegas, gammas, vannas and volgas of the non-traded and the liquidly traded options.

Appendix: Linearly constrained quadratic programming

Lagrangian duality. We first recall some basic Lagrange duality results from [8, Sect. 5.1.5]. Fix $n \in\mathbb{N}$ and functions $f,g,h:\mathbb{R}^{n} \to\mathbb{R}$. We refer to the problem

$$\begin{aligned} \text{minimise } f(\mathbf{z}) \qquad\text{subject to } \mathbf{z}\in\mathbb{R}^{n}, h(\mathbf{z}) = 0, g(\mathbf{z}) \leq0 \end{aligned}$$

(A.1)

as the primal problem and denote by

$$\begin{aligned} f^{*} &= \inf\lbrace f(\mathbf{z}) : \mathbf{z} \in\mathbb{R}^{n}, h(\mathbf {z}) = 0, g(\mathbf{z}) \leq0 \rbrace \end{aligned}$$

its optimal value. The corresponding Lagrangian is

$$\begin{aligned} L(\mathbf{z}, \lambda, \mu) &= f(\mathbf{z}) + \lambda h(\mathbf{z}) + \mu g(\mathbf{z}), \quad \mathbf{z} \in\mathbb{R}^{n}, \lambda,\mu\in\mathbb{R}, \end{aligned}$$

and a pair $(\mu^{*}, \lambda^{*})$ is called a Lagrange multiplier if

$$\begin{aligned} f^{*} &= \inf_{\mathbf{z} \in\mathbb{R}^{n}} L(\mathbf{z},\lambda^{*},\mu^{*}) \quad\text{and}\quad \mu^{*} \geq0. \end{aligned}$$

The dual problem for (A.1) is

$$\begin{aligned} \text{maximise } q(\lambda,\mu) \qquad\text{subject to } \lambda\in\mathbb{R}, \mu\geq0, \end{aligned}$$

where the dual function $q$ is

$$\begin{aligned} q(\lambda,\mu) &= \inf_{\mathbf{z} \in\mathbb{R}^{n}} L(\mathbf{z},\lambda,\mu), \quad \lambda,\mu\in\mathbb{R}. \end{aligned}$$

Finally, $q^{*} = \sup\lbrace q(\lambda,\mu):\lambda\in\mathbb{R}, \mu \geq0 \rbrace$ denotes the optimal value of the dual problem.

A quadratic programming problem with linear equality and inequality constraints. The following lemma provides the solution to a primal problem with a strictly convex quadratic cost function and specific linear equality and inequality constraints.

Lemma A.1

Fix $n \in\mathbb{N}$, a diagonal matrix $D = \operatorname {diag}(d_{1},\ldots,d_{n}) \in\mathbb{R}^{n\times n}$ with positive diagonal entries, and vectors $\mathbf{v} = (v_{1},\ldots,v_{n})^{\top}$ and $\mathbf{c} = (c_{1},\ldots, c_{n})^{\top}$ in $\mathbb{R}^{n}$ such that $c_{i} \neq0$ for some $i \in\lbrace1,\ldots,n-1\rbrace$. Moreover, set $\mathbf{1}_{A} = 1$ if $v_{n} - \frac{\mathbf{c}^{\top}D^{-1}\mathbf {v}}{\mathbf{c}^{\top}D^{-1}\mathbf{c}} c_{n} < 0$ and $\mathbf{1}_{A} = 0$ otherwise, and define

$$\begin{aligned} \lambda^{*} &= \frac{\mathbf{c}^{\top}D^{-1}\mathbf{v} - c_{n} v_{n} d_{n}^{-1} \mathbf {1}_{A}}{\mathbf{c}^{\top}D^{-1}\mathbf{c} - c_{n}^{2} d_{n}^{-1}\mathbf {1}_{A}}, \end{aligned}$$

(A.2)

$$\begin{aligned} \mu^{*} &= (v_{n} - \lambda^{*} c_{n})^{-}, \end{aligned}$$

(A.3)

$$\begin{aligned} \mathbf{z}^{*} &= D^{-1}(\mathbf{v} - \lambda^{*} \mathbf{c} + \mu^{*} \mathbf{e}_{n}). \end{aligned}$$

(A.4)

• (a) $\mathbf{z}^{*}$ is the unique optimiser of the primal problem

  $$\begin{aligned} \textit{minimise }\frac{1}{2}\mathbf{z}^{\top}D \mathbf{z} - \mathbf {v}^{\top}\mathbf{z} \qquad\textit{ subject to } \mathbf{z}\in\mathbb{R}^{n}, \mathbf{c}^{\top}\mathbf{z} = 0, z_{n} \geq0 \end{aligned}$$

  (A.5)

  and satisfies the bound $\left\vert\mathbf{z}^{*}\right\vert \leq d_{\min}^{-1} \left\vert\mathbf{v}\right\vert$, where $d_{\min} = \min(d_{1},\ldots,d_{n})$.

• (b) $(\lambda^{*},\mu^{*})$ is the unique optimiser of the dual problem for (A.5), which can be written as

  $$\begin{aligned} &\textit{maximise } -\frac{1}{2}(\mathbf{v} - \lambda\mathbf{c} + \mu \mathbf{e}_{n})^{\top}D^{-1}(\mathbf{v} - \lambda\mathbf{c} + \mu \mathbf{e}_{n}) \\ & \textit{subject to } \lambda\in\mathbb{R}, \mu\geq0. \end{aligned}$$

  (A.6)

  The optimiser satisfies the bound

  $$\begin{aligned} \vert\lambda^{*} \mathbf{c} - \mu^{*} \mathbf{e}_{n}\vert &\leq\left(1 + \frac{d_{\max}}{d_{\min}}\right)\left\vert \mathbf{v}\right\vert, \end{aligned}$$

  (A.7)

  where $d_{\max} = \max(d_{1},\ldots,d_{n})$.

• (c) The optimal values of the primal and dual problems coincide (i.e., there is no duality gap) and equal

  $$\begin{aligned} -\frac{1}{2}\mathbf{v}^{\top}\mathbf{z}^{*}. \end{aligned}$$

• (d) The triplet $(\mathbf{z}^{*},\mu^{*},\lambda^{*})$ satisfies the optimality conditions

  $$\begin{aligned} \mathbf{z}^{*} = \arg\min_{\mathbf{z}\in\mathbb{R}^{n}} L(\mathbf{z},\lambda ^{*},\mu ^{*}),\quad \mathbf{c}^{\top}\mathbf{z}^{*} = 0,\quad z^{*}_{n} \geq0,\quad \mu^{*} \geq0,\quad \mu^{*}z_{n}^{*} = 0, \end{aligned}$$

  (A.8)

  where $L$ is the Lagrangian corresponding to the primal problem. Moreover, the pair $(\lambda^{*},\mu^{*})$ is a Lagrange multiplier for the primal problem.

Proof

First of all, note that the Lagrangian

$$\begin{aligned} L(\mathbf{z},\lambda,\mu) &= \frac{1}{2}\mathbf{z}^{\top}D \mathbf{z} - \mathbf{v}^{\top}\mathbf{z} + \lambda\mathbf{c}^{\top}\mathbf{z} - \mu \mathbf{e}_{n}^{\top}\mathbf{z} \end{aligned}$$

corresponding to the primal problem is strictly convex over $\mathbf{z} \in\mathbb{R}^{n}$. Hence, the dual function $q(\lambda,\mu)= \inf _{\mathbf{z}\in\mathbb{R}^{n}} L(\mathbf{z},\lambda,\mu)$ can be computed explicitly by substituting the solution $\mathbf{z}'$ to the first-order condition $\mathbf{0} = \operatorname{D}_{\mathbf{z}} L(\mathbf{z}',\lambda,\mu) = D \mathbf{z}' - \mathbf{v} + \lambda \mathbf{c} - \mu \mathbf{e}_{n}$ back into $L(\mathbf{z}',\mu ,\lambda)$. This yields

$$\begin{aligned} q(\lambda,\mu) &= -\frac{1}{2}(\mathbf{v} - \lambda\mathbf{c} + \mu \mathbf {e}_{n})^{\top}D^{-1}(\mathbf{v} - \lambda\mathbf{c} + \mu \mathbf{e}_{n}), \end{aligned}$$

and thus the dual problem takes the form (A.6).

The crucial part of the proof is to show that the triplet $(\mathbf {z}^{*},\lambda^{*},\mu^{*})$ satisfies the optimality conditions (A.8). As $L$ is strictly convex over $\mathbf{z} \in\mathbb{R}^{n}$, the optimality conditions are equivalent to

$$\begin{aligned} D\mathbf{z}^{*} - \mathbf{v} + \lambda^{*} \mathbf{c} - \mu^{*} \mathbf {e}_{n} = 0,\quad \mathbf{c}^{\top}\mathbf{z}^{*} = 0,\quad z^{*}_{n} \geq0,\quad \mu^{*} \geq0,\quad \mu^{*} z^{*}_{n} = 0. \end{aligned}$$

(A.9)

Recall the definitions of $\lambda^{*}$, $\mu^{*}$ and $\mathbf{z}^{*}$ in (A.2)–(A.4) and note that the assumption that $c_{i} \neq0$ for some $i\in\lbrace 1,\ldots ,n-1\rbrace$ together with the positive definiteness of $D^{-1}$ ensures that $\lambda^{*}$ is well defined. The stationarity condition $D\mathbf{z}^{*} - \mathbf{v} + \lambda^{*} \mathbf{c} - \mu^{*} \mathbf {e}_{n} = 0$ holds by definition of $\mathbf{z}^{*}$. For the other conditions, we distinguish two cases. First, suppose that $\mathbf{1}_{A} = 0$, i.e., $v_{n} - \frac{\mathbf {c}^{\top}D^{-1}\mathbf{v}}{\mathbf{c}^{\top}D^{-1} \mathbf{c}} c_{n} \geq 0$. Then $\lambda^{*} = \frac{\mathbf{c}^{\top}D^{-1}\mathbf {v}}{\mathbf {c}^{\top}D^{-1} \mathbf{c}}$, $\mu^{*} = 0$ and $z^{*}_{n} = d_{n}^{-1}(v_{n} - \lambda^{*} c_{n}) \geq0$. Moreover, $\mathbf{c}^{\top}\mathbf{z}^{*} = \mathbf{c}^{\top}D^{-1} \mathbf{v} - \lambda^{*} \mathbf{c}^{\top}D^{-1} \mathbf{c} = 0$. Second, suppose that $\mathbf{1}_{A} = 1$, i.e., $v_{n} - \frac{\mathbf {c}^{\top}D^{-1}\mathbf{v}}{\mathbf{c}^{\top}D^{-1} \mathbf{c}} c_{n} < 0$, or, equivalently (multiply by $\mathbf{c}^{\top}D^{-1}\mathbf{c}$, add and subtract $c_{n}^{2} v_{n} d_{n}^{-1}$, and then divide by $\mathbf{c}^{\top}D^{-1} \mathbf{c} - c_{n}^{2} d_{n}^{-1} > 0$), $v_{n} -\lambda^{*} c_{n} < 0$. Then $\mu^{*} > 0$ and $z^{*}_{n} = 0$ by the definition of $\mu^{*}$ and $\mathbf{z}^{*}$. Finally, setting $c = \mathbf{c}^{\top}D^{-1} \mathbf {c}$ and $v = \mathbf{c}^{\top}D^{-1} \mathbf{v}$ for brevity, we obtain

$$\begin{aligned} \mathbf{c}^{\top}\mathbf{z}^{*} &= \mathbf{c}^{\top}D^{-1}(\mathbf{v} - \lambda^{*}\mathbf{c}+ \mu ^{*}\mathbf{e}_{n}) = v -\lambda^{*} c +\mu^{*} c_{n} d_{n}^{-1}\\ &= v -\lambda^{*} c - (v_{n} - \lambda^{*} c_{n}) c_{n} d_{n}^{-1} = v - \lambda^{*}(c - c_{n}^{2} d_{n}^{-1}) - c_{n} v_{n} d_{n}^{-1}\\ &= v - (v - c_{n} v_{n} d_{n}^{-1}) - c_{n} v_{n} d_{n}^{-1} = 0. \end{aligned}$$

So, (A.9) holds in both cases. By the characterisation of primal optimal solutions [8, Proposition 5.1.5], this implies that $\mathbf{z}^{*}$ is an optimiser for the primal problem, that $(\lambda^{*},\mu^{*})$ is a Lagrange multiplier, and that there is no duality gap. Moreover, $(\lambda^{*},\mu^{*})$ is an optimiser for the dual problem by a corollary [8, Proposition 5.1.4 (a)] of the weak duality theorem [8, Proposition 5.1.3]. As the primal and dual problems are strictly convex and strictly concave, respectively, the optimisers are unique.

Plugging the optimiser $(\lambda^{*},\mu^{*})$ of the dual problem into the cost function of the dual problem (A.6) and using the definition of $\mathbf{z}^{*}$, the optimal value $q^{*}$ (of both the primal and the dual problem) reads

$$\begin{aligned} q^{*}&:=-\frac{1}{2}(\mathbf{v} - \lambda^{*} \mathbf{c} + \mu^{*} \mathbf{e}_{n})^{\top}D^{-1}(\mathbf{v} - \lambda^{*} \mathbf{c} + \mu \mathbf{e}_{n}) = -\frac{1}{2} (\mathbf{v} - \lambda^{*} \mathbf{c} + \mu^{*} \mathbf {e}_{n})^{\top}\mathbf{z}^{*}. \end{aligned}$$

Now, note that $\mathbf{c}^{\top}\mathbf{z}^{*} = 0$ and $\mu^{*} \mathbf{e}_{n}^{\top}\mathbf{z}^{*} = \mu^{*} z_{n}^{*} = 0$ by (A.9). Hence, $q^{*} = -\frac{1}{2}\mathbf {v}^{\top}\mathbf{z}^{*}$.

Next, using that $\mathbf{z}^{*}$ achieves the optimal value $-\frac {1}{2}\mathbf{v}^{\top}\mathbf{z}^{*}$ for the primal problem and applying the Cauchy–Schwarz inequality, we obtain

$$\begin{aligned} \frac{1}{2}d_{\min} \vert\mathbf{z}^{*}\vert^{2} &\leq\frac{1}{2}(\mathbf{z}^{*})^{\top}D \mathbf{z}^{*} = \frac{1}{2}\mathbf{v}^{\top}\mathbf{z}^{*} \leq\frac{1}{2} \vert\mathbf{v}\vert \vert\mathbf{z}^{*}\vert. \end{aligned}$$

This yields the last claim of part (a). Finally, using (A.4), the triangle inequality and the bound $\left\vert\mathbf{z}^{*}\right\vert \leq d_{\min}^{-1}\left\vert \mathbf{v}\right\vert$ which we just proved, we obtain

$$\begin{aligned} \vert\lambda^{*} \mathbf{c} - \mu^{*} \mathbf{e}_{n}\vert &= \vert D\mathbf{z}^{*} - \mathbf{v}\vert \leq\vert D\mathbf{z}^{*}\vert + \vert\mathbf{v}\vert \leq d_{\max} \vert\mathbf{z}^{*}\vert + \vert\mathbf{v}\vert \leq\bigg(1+\frac{d_{\max}}{d_{\min}}\bigg) \vert\mathbf {v}\vert. \end{aligned}$$

This proves the last claim of part (b) and thereby concludes the proof. □