Small-Time Asymptotics for the At-the-Money Implied Volatility in a Multi-dimensional Local Volatility Model

Abstract

We consider a basket or spread option based on a multi-dimensional local volatility model. Bayer and Laurence (Commun. Pure. Appl. Math., 67(10), 2014, [5]) derived highly accurate analytic formulas for prices and implied volatilities of such options when the options are not at the money. We now extend these results to the ATM case. Moreover, we also derive similar formulas for the local volatility of the basket.

To the memory of Peter Laurence, who passed away unexpectedly during the final stage of the preparation of this manuscript.

Appendix A: Proof of Lemma 3.7

We present a proof of Lemma 3.7. Recall that we want to compute the determinant of the Hessian Q of the map

$$\begin{aligned} \Phi (\mathbf {G}) := \frac{1}{2}d\left( \mathbf {F}_0, \left( \mathbf {G}, F_N(\mathbf {G}, K) \right) \right) ^2 \end{aligned}$$

evaluated at $\mathbf {G} = \left( F_{0,1}, \ldots , F_{0,n-1} \right)$. Let $\mathfrak {S}_i(x)$ denote the anti-derivative of $1/\sigma _i$ satisfying (for simplicity) $\mathfrak {S}_i(F_{0,i}) {=} 0$. Now consider the change of variables $\mathbf {F} \rightarrow \mathbf {y}$ with $y_i := \mathfrak {S}_i(F_i)$, $i=1, \ldots , n$. As verified in [5], this transformation turns the Riemannian geometry introduced above into an (almost) Euclidean geometry, with

$$\begin{aligned} d(\mathbf {F}_0, \mathbf {F})^2 = \mathbf {y}^T \rho ^{-1} \mathbf {y}. \end{aligned}$$

Of course, the constraint on $\mathbf {F}$ translates into a constraint on $\mathbf {y}$, which can be removed by eliminating one variable. Indeed, setting $\mathbf {x} := (y_1, \ldots , y_{n-1})$, we get

$$\begin{aligned} y_n(\mathbf {x}) = \mathfrak {S}_n\left( F_n \right) = \mathfrak {S}_n\left( \frac{1}{w_n} \left( K - \sum _{j=1}^{n-1} w_j \mathfrak {S}_j^{-1}(y_j) \right) \right) . \end{aligned}$$

This way, we understand $\Phi (\mathbf {G})$ as a function $\varphi (\mathbf {x})$ in the new (reduced) coordinates, and obtain for the Hessian

$$\begin{aligned} H_{\mathbf {G}} \Phi (\mathbf {G}) = J(\mathbf {G})^T H_{\mathbf {x}} \varphi (\mathbf {x}) J(\mathbf {G}), \end{aligned}$$

where $H_{\mathbf {G}}$ and $H_{\mathbf {x}}$ denote the Hessians in the $\mathbf {G}$- and $\mathbf {x}$-coordinates, respectively, and $J(\mathbf {G})$ denotes the Jacobian matrix of the change of coordinates $\mathbf {G} \rightarrow \mathbf {x}$. As $\mathfrak {S}_i^\prime =1/\sigma _i$, we have $J(\mathbf {G}) = {\text {diag}}(1/\sigma _1(F_1), \ldots , 1/\sigma _{n-1}(F_{n-1}))$. Regarding the matrix $H_{\mathbf {x}} \varphi$, an elementary calculation using the fact that $\mathbf {F} = \mathbf {F}_0$ corresponds to $\mathbf {y} = 0$, we obtain

$$\begin{aligned} H_{\mathbf {x}}\varphi (0) = \left( \rho ^{\textit{ij}} - \rho ^{in} \frac{w_j \sigma _j(F_{0,j})}{w_n \sigma _n(F_{0,n})} - \rho ^{\textit{jn}} \frac{w_i \sigma _i(F_{0,i})}{w_n \sigma _n(F_{0,n})} + \rho ^{nn} \frac{w_i\sigma _i(F_{0,i}) w_j\sigma _j(F_{0,j})}{w_n^2 \sigma _n(F_{0,n})^2} \right) _{i,j=1}^{n-1}. \end{aligned}$$

From the structure of the above expression and the expression in Lemma 3.7, we see that we may assume that $w_i = 1$, $i=1, \ldots , n$, and $\sigma _n(F_{0,n}) = 1$. In this case, we are left to prove that the determinant of the matrix

$$\begin{aligned} A := \left( \rho ^{\textit{ij}} - \rho ^{\textit{in}} s_j - \rho ^{\textit{jn}} s_i + \rho ^{\textit{nn}} s_is_j \right) _{i,j=1}^{n-1} \end{aligned}$$

is equal to the expression $a := \mathbf {s}^T \rho \mathbf {s}/\det \rho$, where we used the short-hand notation $s_i = \sigma _i(F_{0,i})$, $i=1, \ldots , n-1$, and $s_n = 1$, and $\mathbf {s} = (s_1, \ldots , s_n)$.

As both $\det A$ and a are polynomials in $s_1, \ldots , s_{n-1}$, we prove this equality by establishing that they have the same coefficients. Here, Cramer’s rule is the essential tool:

$$\begin{aligned} B^{-1} = \frac{1}{\det B} {{\mathrm{Adj}}}(B), \end{aligned}$$

where the adjugate matrix ${{\mathrm{Adj}}}\,B$ is the transpose of the matrix of co-factors, i.e.,

$$\begin{aligned} \left( {{\mathrm{Adj}}}\,B\right) _{ij} = (-1)^{i+j} \det B_{\hat{j}\hat{i}}, \end{aligned}$$

with $B_{\hat{j}\hat{i}}$ being obtained from B by removing the jth row and the ith column. By symmetry, we hence have

$$\begin{aligned} \frac{\rho _{\textit{ij}}}{\det \rho } = (-1)^{i+j} \det \rho ^{-1}_{\hat{i}\hat{j}}, \quad \forall (i,j) \in \{1, \ldots , n-1\}^2, \end{aligned}$$

(A.1)

where $\rho ^{-1}_{\hat{i}\hat{j}}$ is understood in the sense of $(\rho ^{-1})_{\hat{i}\hat{j}}$.

Let us also establish a few notations. Let $S_{n-1}$ be the set of all permutations of $\{1, \ldots , n-1\}$ and let, similarly, S(A; B) denote the set of all bijective maps from $A \subset \mathbb {N}$ to $B \subset \mathbb {N}$, with A, B having the same (finite) size. Moreover, the definition of the signature ${{\mathrm{sign}}}$ is extended to S(A; B) in the obvious way (as being ${\pm }1$ depending on the number of inversions being even or odd). Moreover, for a monomial x in the variables $s_1, \ldots , s_{n-1}$ we denote by $\pi _{x} p$ the coefficient of any polynomial p w.r.t. the monomial x. In order to establish Lemma 3.7, we need to prove that

$$\begin{aligned} \forall x \in \bigcup _{k=0}^{2(n-1)} \{s_1, \ldots , s_{n-1}\}^k: \ \pi _x \det A = \pi _x a. \end{aligned}$$

We distinguish different cases according to the degree.

Case 0. For $\deg x = 0$, i.e., $x = 1$, we have

$$\begin{aligned} \pi _1 \det A = \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) \prod _{i=1}^{n-1} \rho ^{i\sigma (i)} = \det \rho ^{-1}_{\hat{n}\hat{n}} = {{\mathrm{Adj}}}(\rho ^{-1})_{\textit{nn}} = \frac{\rho _{\textit{nn}}}{\det \rho } = \pi _1 a. \end{aligned}$$

Case 1. For some fixed $s_k$ we have

$$\begin{aligned} \pi _{s_k} \det A&= \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) (-1) \left[ \rho ^{\sigma ^{-1}(k)n} \prod _{i \in \{1, \ldots , n-1\} \setminus \{\sigma ^{-1}(k)\}} \rho ^{i \sigma (i)}\right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \,\,\left. +\, \rho ^{\sigma (k) n} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k\}} \rho ^{i\sigma (i)} \right] \\&= -2 \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) \rho ^{\sigma (k)n} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k\}} \rho ^{i\sigma (i)} \end{aligned}$$

by symmetry of $\rho ^{-1}$. There is a one-to-one correspondence between $S_{n-1}$ and $S(\{1, \ldots , n\}\setminus \{k\}; \{1, \ldots , n-1\})$ given by $\sigma \mapsto \tilde{\sigma }$ defined by

$$\begin{aligned} \tilde{\sigma }(i) = {\left\{ \begin{array}{ll} \sigma (i), &{} i \in \{1, \ldots , n-1\} \setminus \{k\},\\ \sigma (k), &{} i = n. \end{array}\right. } \end{aligned}$$

Moreover, one can see that ${{\mathrm{sign}}}(\tilde{\sigma }) = (-1)^{k+n-1} {{\mathrm{sign}}}(\sigma )$. Hence, we obtain

$$\begin{aligned} \pi _{s_k} \det A&= -2 \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) \rho ^{n\tilde{\sigma }(n)} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k\}} \rho ^{i\tilde{\sigma (i)}} \\&= 2 (-1)^{k+n} \sum _{\tilde{\sigma } \in S(\{1, \ldots , n\}\setminus \{k\}; \{1, \ldots , n-1\})} {{\mathrm{sign}}}(\tilde{\sigma }) \rho ^{n\tilde{\sigma }(n)} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k\}} \rho ^{i\tilde{\sigma (i)}} \\&= 2 (-1)^{k+n} \det \rho ^{-1}_{\hat{k}\hat{n}}\\&= 2 {{\mathrm{Adj}}}(\rho ^{-1})_{kn} = \frac{2 \rho _{kn}}{\det \rho } = \pi _{s_k} a. \end{aligned}$$

Case 2. We consider $x = s_k s_l$. For simplicity, we assume $k=l$ ($k \ne l$ works analogously). We have

$$\begin{aligned} \pi _{s_k^2} \det A&= \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) \left[ \mathbf {1}_{k=\sigma (k)} \rho ^{nn} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k\}} \rho ^{i\sigma (i)}\,+ \right. \\&\left. \qquad \qquad \qquad \qquad \quad +\,\mathbf {1}_{k \ne \sigma (k)} \rho ^{\sigma (k)n} \rho ^{\sigma ^{-1}(k)n} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k, \sigma ^{-1}(k)\}} \rho ^{i\sigma (i)} \right] . \end{aligned}$$

We construct a bijective map from $S_{n-1}$ to $S(\{1, \ldots , n\}\setminus \{k\}; \{1, \ldots , n\} \setminus \{k\})$ by mapping $\sigma \in S_{n-1}$ to $\tilde{\sigma }$ defined by

$$\begin{aligned} \tilde{\sigma }(i) = {\left\{ \begin{array}{ll} \sigma (i), &{} i \in \{1, \ldots , n-1\} \setminus \{k\}, \\ n, &{} i = n, \end{array}\right. } \end{aligned}$$

for the case $k = \sigma (k)$ and

$$\begin{aligned} \tilde{\sigma }(i) = {\left\{ \begin{array}{ll} \sigma (i), &{} i \in \{1, \ldots , n-1\} \setminus \{k, \sigma ^{-1}(k)\}, \\ n, &{} i = \sigma ^{-1}(k),\\ \sigma (k), &{} i = n, \end{array}\right. } \end{aligned}$$

else. Note that it is easy to see that ${{\mathrm{sign}}}(\sigma ) = {{\mathrm{sign}}}(\tilde{\sigma })$. Hence, we have

$$\begin{aligned} \pi _{s_k^2} \det A&= \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) \prod _{i \in \{1, \ldots , n\} \setminus \{k\}} \rho ^{i \tilde{\sigma }(i)} \\&= \sum _{\tilde{\sigma } \in S(\{1, \ldots , n\} \setminus \{k\}; \{1, \ldots , n\} \setminus \{k\})} {{\mathrm{sign}}}(\tilde{\sigma }) \prod _{i \in \{1, \ldots , n\} \setminus \{k \}} \rho ^{i\tilde{\sigma }(i)} \\&= \det \rho ^{-1}_{\hat{k}\hat{k}} = \pi _{s_k^2} a. \end{aligned}$$

Higher order terms. Regarding the higher order terms, we note that $\pi _x a = 0$ for any monomial of degree larger than two. Therefore, the same should be true for $\det A$, where it does not to seem to follow from an obvious argument. Note that we only need to consider polynomials where each individual variable $s_k$ appears at most two times, as any other monomial cannot appear in $\det A$ by the definition of A and of the determinant. But any coefficient of $\det A$ with respect to such monomials can be understood as the determinant of a matrix $\widetilde{\rho ^{-1}}$, which is obtained from $\rho ^{-1}$ by omitting one row and one column and by replacing some rows/columns by copies of other rows/columns. Of course, any such matrix $\tilde{\rho }$ has vanishing determinant, implying that $\pi _x \det A = 0$. For concreteness, we indicate this mechanism by appealing to two special cases. First, take $x = s_k^2 s_l$, $l \ne k$. Similarly to the case of $x = s_k$, one can show that

$$\begin{aligned} \pi _{s_k^2 s_l} \det A = -2&\sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) \Biggl [ \mathbf {1}_{k=\sigma (k)} \rho ^{\textit{nn}} \rho ^{\sigma ^{-1}(l)n} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k,l\}} \rho ^{i\sigma (i)}+ \\&\quad \, + \mathbf {1}_{k \ne \sigma (k)} \rho ^{\sigma (k)n} \rho ^{\sigma ^{-1}(k)n} \rho ^{\sigma ^{-1}(l)n} \prod _{i \in \{1, \ldots , n-1\} \setminus \{k, \sigma ^{-1}(k) \sigma ^{-1}(l)\}} \rho ^{i\sigma (i)}\Biggr ], \end{aligned}$$

which is (the multiple of) the determinant of $\widetilde{\rho ^{-1}}$, which is obtained from $\rho ^{-1}_{\hat{k}\hat{k}}$ by replacing the lth row by the last row. As the last row appears twice in $\widetilde{\rho ^{-1}}$, the determinant, and hence $\pi _{s_k^2 s_l} \det A$, vanishes.

The mechanism is even more transparent for the most extreme monomial $x = s_1^2 \cdots s_{n-1}^2$. In this case,

$$\begin{aligned} \pi _{s_1^2\cdots s_{n-1}^2} \det A = \sum _{\sigma \in S_{n-1}} {{\mathrm{sign}}}(\sigma ) (\rho ^{\textit{nn}})^{n-1} = 0, \end{aligned}$$

as the determinant of the $(n-1) \times (n-1)$ matrix with all entries being equal to $\rho ^{\textit{nn}}$.