2.1. Semi-static variance-optimal hedging

Let $S\in{\mathcal{H}}^2$ be strictly positive and assume that S models the price process of some stock. This is a crucial assumption on the probability measure ${\mathrm{P}}$ : We require that ${\mathrm{P}}$ is already an equivalent martingale measure and we do not consider the problem under the historical measure, say ${\mathrm{Q}}$ , with respect to which S is, in general, a non-negative semimartingale. This latter case has been considered, e.g., in [Reference Schweizer22]. See also [Reference Schweizer, Jouini, Cvitanic and Musiela23] and the references therein.

A fixed basket of contingent claims ${\pmb{\eta}}\,:\!=({\eta}^1, \dots, {\eta}^d)^{\top}$ , ${\eta}^{\kern1ptj}\in L^2({\mathrm{P}})$ , $j=1,\ldots,d$ , is given. We stress that we regard ${\mathbb{R}}^d$ as a space of column vectors.

(2.1) \begin{equation} {{\varepsilon }^{2}}=\overset{\min }{\mathop{_{v\in {{\mathbb{R}}^{d}},\vartheta \in {{\text{L}}^{2}}(S),c\in \mathbb{R}}}}\,\text{E}[{{(c-\text{E}[{{v}^{\top }}\eta ]+\int_{0}^{T}{{{\vartheta }_{s}}}\text{d}{{S}_{s}}-({{\eta }^{0}}-{{v}^{\top }}\text{ }\eta ))}^{2}}]. \label{eqn1} \end{equation}

In other words, a semi-static strategy is a strategy which, beyond the dynamic (i.e. continuously rebalanced) position in S, also allows a static (i.e. buy-and-hold) position in the fixed basket $\pmb{\eta}=({\eta}^1, \dots, {\eta}^d)^{\top}$ of contingent claims.

Clearly, (2.1) is a generalization of the classic variance-optimal hedging problem (see, e.g., [Reference Föllmer, Sondermann, Hildenbrand, Mas-Colell and North-Holland12] or [Reference Schweizer, Jouini, Cvitanic and Musiela23]), and semi-static variance-optimal strategies typically allow for a reduction of the quadratic hedging error in comparison with classic variance-optimal strategies.

Following [Reference Di Tella, Haubold and Keller-Ressel9], the semi-static hedging problem can be split into an inner and an outer optimization problem. That is, it can be written as an inner problem,

(2.2) \begin{equation} {{\varepsilon }^{2}}(v)=\overset{\min }{\mathop{_{\vartheta \in {{L}^{2}}(S),c\in \mathbb{R}}}}\,\text{E}[{{(c-\text{E}[{{v}^{\top }}\eta ]+\int\limits_{0}^{T}{{{\vartheta }_{s}}}\text{d}{{S}_{s}}-({{\eta }^{0}}-{{v}^{\top }}\text{ }\eta ))}^{2}}], \label{eqn2} \end{equation}

and an outer problem,

(2.3) \begin{equation} \displaystyle\varepsilon^2 = \min_{\pmb v \in {\mathbb{R}}^d}{\varepsilon^2(\pmb v)}. \label{eqn3} \end{equation}

Now let $(\vartheta^{\kern1ptj},L^{\kern1ptj})$ be the GKW decomposition of ${\eta}^{\kern1ptj}$ with respect to S, $j=0,\ldots,d$ . Since (2.2) is a classic variance-optimal hedging problem, the optimal solution $(\vartheta^{\pmb v,\ast},c^{\ast})$ is given by the integrand $\vartheta^{\pmb v,\ast}\in L^2(S)$ appearing in the GKW decomposition of ${\eta}^0-\pmb v^{\top}\pmb{\eta}$ with respect to S, that is, $\vartheta^{\pmb v,\ast}=\vartheta^0-\sum_{j=1}^dv_j\vartheta^{\kern1ptj}$ , and $c^{\ast}={\mathrm{E}}[{\eta}^0]$ .

The outer problem (2.3) is a finite-dimensional quadratic optimization problem, since $\varepsilon^2(\pmb v)$ is of the form

$${{\varepsilon }^{2}}(v)=A-2\text{ }{{v}^{\top }}B+\text{ }{{v}^{\top }}Cv,$$

where the coefficients A, B, and C are given by

(2.4) \begin{align} A:=\text{E}[{{\left\langle {{L}^{0}},{{L}^{0}} \right\rangle }_{T}}],\ \ {{B}^{j}}:=\text{E}[{{\left\langle {{L}^{0}},{{L}^{j}} \right\rangle }_{T}}],\ \ {{C}^{ij}}:=\text{E}[{{\left\langle {{L}^{i}},{{L}^{j}} \right\rangle }_{T}}],\quad i,j=1,\ldots ,d. \label{eqn4} \end{align}

If C is invertible, then the optimal strategy $(\vartheta^{\ast},\pmb v^{\ast}, c^{\ast})$ for the semi-static variance-optimal hedging problem (2.1) is given by

\[\pmb v^* = C^{-1} B, \quad \vartheta^{\ast} = \vartheta^0 - \sum_{j=1}^dv_j^{\ast} \vartheta^{\kern1ptj},\quad c^{\ast}={\mathrm{E}}[{\eta}^0].\]

Hence, to solve the semi-static variance-optimal hedging problem, it is necessary to compute the predictable covariations of the residuals in the GKW decomposition of ${\eta}^{\kern1ptj}$ with respect to S, $j=1,\ldots,d$ , and then their expectation. This extends beyond the classic variance-optimal hedging problem, where it is enough to determine A in order to compute the squared hedging error and the optimal strategy.

3.1. Martingale representation

We consider a quasi-left continuous locally square-integrable semimartingale $X=(X^1,\ldots,X^n)^{\top}$ with state space $(E,{\mathcal{E}})\,:\!=({\mathbb{R}}^ n,{\mathcal{B}}({\mathbb{R}}^n))$ (see [16, Definition II.2.27]). Then, for $j=1,\ldots,n$ , the unidimensional semimartingale $X^{\kern1ptj}$ admits the unique decomposition

(3.1) \begin{equation} X^{\kern1ptj}=X^{\kern1ptj}_0+M^{\kern1ptj}+A^{\kern1ptj}, \label{eqn5} \end{equation}

where $A^{\kern1ptj}$ is a continuous adapted process of finite variation and $M^{\kern1ptj}\in{\mathcal{H}}^2_\mathrm{loc}$ is quasi-left continuous, $j=1,\ldots,n$ . We denote by $X^{j,c}$ the continuous martingale part of $X^{\kern1ptj}$ . Let $Y\in{\mathcal{H}}^2_{\mathrm{loc}}$ be of the form

(3.2) \begin{equation} Y_t = f(t,X_t),\quad f\in C^{1,2}({\mathbb{R}}_+\times{\mathbb{R}}^n). \label{eqn6} \end{equation}

We are going to show that Y has the representation property relative to X (see [16, Definition III.4.22]) and also obtain an explicit form of the integrand. To tackle the problem, we pursue the following classical idea: We consider $Y\in{\mathcal{H}}^2_\mathrm{loc}$ , which is given by a function $f\in C^{1,2}({\mathbb{R}}_+\times{\mathbb{R}}^n)$ of time and X. Then, applying Itô’s formula, we show that Y has the representation property relative to X.

Such kinds of martingale representation results have already been considered in the literature, especially for pricing of European options: In [8, Section 3] this result has been shown in a Brownian context, and in [11, Section 3] for the homogeneous Poisson case. In [5, Section 2] the case of a continuous semimartingale is considered. We add a new element to this picture by considering locally square-integrable quasi-left continuous semimartingales.

We denote by $\mu$ be the jump measure of X, by $\nu$ its predictable compensator, and we put $\overline\mu\,:\!=\mu-\nu$ . By $\mathcal{G}_{\text{loc}}^{2}\left(\mu\right)$ we denote the space of predictable integrands W on $[0,T]\times{\Omega}\times E$ such that the stochastic integral $Z=\int_0^{\cdot}\int_E W(s,x)\overline\mu({\mathrm{d}} s,{\mathrm{d}} x)$ is the unique (up to indistinguishability) purely discontinuous element of ${\mathcal{H}}^2_\mathrm{loc}$ such that $Z_0=0$ and $\Delta Z=W(\cdot,\Delta X)1_{\{\Delta X\neq0\}}$ . For details, see [16, Section II.1d].

For $Y\in{\mathcal{H}}^2_\mathrm{loc}$ , we denote by $Y^d\in{\mathcal{H}}^2_\mathrm{loc}$ the purely discontinuous part of Y (see [16, Theorem I.4.18]).

(3.3) \begin{equation} Y=Y_0+\sum_{j=1}^n \int_0^{\cdot} \partial_{x_j}\,f(s,X_{s-}){\,\mathrm{d}} X^{j,c}_s+\int_0^{\cdot}\int_E [\:f(s,x+X_{s-})-f(s,X_{s-})]\,\overline\mu({\mathrm{d}} s,{\mathrm{d}} x). \label{eqn7} \end{equation}

Proof. By Itô’s formula and (3.1), we deduce that

(3.4) \begin{equation} \begin{split} Y-{{Y}_{0}}&=\sum\limits_{j=1}^{n}{\int_{0}^{\cdot }{{{\partial }_{{{x}_{j}}}}}}f(s,{{X}_{s-}})\text{d}M_{s}^{j}+\int_{0}^{\cdot }{{{\partial }_{t}}}f(s,{{X}_{s-}})\text{d}s\\ &+\sum\limits_{j=1}^{n}{\int_{0}^{\cdot }{{{\partial }_{{{x}_{j}}}}}}f(s,{{X}_{s-}})\text{d}A_{s}^{j}+\frac{1}{2}\sum\limits_{j,k=1}^{n}{\int_{0}^{\cdot }{\partial _{{{x}_{j}}{{x}_{k}}}^{2}}}f(s,{{X}_{s-}})\text{d}{{\left\langle {{X}^{j,c}}{{X}^{k,c}} \right\rangle }_{s}}\\ &+\sum\limits_{s\le \cdot }{[}\Delta {{Y}_{s}}-\sum\limits_{j=1}^{n}{{{\partial }_{{{x}_{j}}}}}f(s,{{X}_{s-}})\Delta X_{s}^{j}], \end{split} \label{eqn8} \end{equation}

where $\partial^2_{\cdot\cdot}$ denotes the second partial derivative. We denote by N the first term and by G the sum of the second, the third, and the fourth terms on the right-hand side of (3.4). Finally, we set $ U(t,x)\,:\!=f(t,x+X_{t-})-f(t,X_{t-})-\sum_{j=1}^n\partial_{x_j}\, f(t,X_{t-}) x_j $ . The process U belongs to ${\mathcal{G}}^2_\mathrm{loc}(\mu)$ . Indeed, because of (3.4) and the continuity of G, $\Delta (Y-N)=U(\cdot,\Delta X)$ . Hence, $\sum_{s\leq\cdot}U_s^2\leq[Y-N,Y-N]$ and $[Y-N,Y-N]$ is locally integrable because $Y-N\in{\mathcal{H}}^2_\mathrm{loc}$ . Thus, $Z\,:\!=\int_0^{\cdot}\int_E U(s,x)\,\overline\mu({\mathrm{d}} s,{\mathrm{d}} x)\in{\mathcal{H}}^2_\mathrm{loc}$ . Furthermore, because of (3.4), we have $J\,:\!=\sum_{s\leq \cdot}U(s,\Delta X_s)+G\in{\mathcal{H}}^2_\mathrm{loc}$ and J is of finite variation. Hence, J is a purely discontinuous local martingale and $\Delta Z=\Delta J$ . From [16, Corollary I.4.19], Z and J are indistinguishable and (3.4) becomes

\[ Y=Y_0+\sum_{j=1}^n\int_0^{\cdot}\partial_{x_j}\, f(s,X_{s-}){\,\mathrm{d}} M^{\kern1ptj}_s+\int_0^{\cdot}\int_E U(s,x)\,\overline\mu({\mathrm{d}} s,{\mathrm{d}} x). \]

Since, for $j=1,\ldots,n$ , we have $M^{\kern1ptj}=X^{ j,c}+(M^{j})^d$ and ${{({{M}^{j}})}^{d}}={{x}_{j}}*\bar{\mu}$ , we get the desired result by [16, Proposition II.1.30 (b)] and by linearity.

We now assume that $S\in{\mathcal{H}}^2_\mathrm{loc}$ is of the form $S={\mathrm{e}}^{X^1}$ . Denoting by ${\skew3\check S}$ the stochastic logarithm of S, we have

(3.5) \begin{equation} S=S_0+\int_0^{\cdot} S_{s-}{\,\mathrm{d}} {\skew3\check S}_s. \label{eqn9} \end{equation}

Applying Theorem 3.1 to $f(t,x)={\mathrm{e}}^{x_1}$ yields

(3.6) \begin{equation} {\skew3\check S}\,:\!={\skew3\check S}_0+(X^{1, c}+{\skew3\check S}^d),\quad{\skew3\check S}^d\,:\!=\int_0^{\cdot}\int_E({\mathrm{e}}^{x_1}-1)\,\overline\mu({\mathrm{d}} s,{\mathrm{d}} x) \label{eqn10} \end{equation}

and, since $x\mapsto({\mathrm{e}}^{x_1}-1)\in{\mathcal{G}}^2_\mathrm{loc}(\mu)$ , we have ${\skew3\check S} \in{\mathcal{H}}^2_\mathrm{loc}$ . If Y is as in (3.2), then Theorem 3.1, (3.5), (3.6), and [16, Theorem II.1.33(a)] imply that

(3.7) \begin{align} \left\langle S,Y \right\rangle &=\sum\limits_{j=1}^{n}{\int_{0}^{\cdot }{{{S}_{s-}}}}{{\partial }_{{{x}_{j}}}}f(s,{{X}_{s-}})\text{d}{{\left\langle {{X}^{j,c}},{{X}^{1,c}} \right\rangle }_{s}}\\ &+\int_{0}^{\cdot }{\int_{E}{({{\text{e}}^{{{x}_{1}}}}-1)}}{{S}_{s-}}W(s,x)\nu (\text{d}s,\text{d}x). \label{eqn11}\end{align}

(3.8) \begin{equation} \vartheta _{t}^{i}=\frac{1}{{{S}_{t-}}}(\sum\limits_{j=1}^{n}{{{\partial }_{{{x}_{j}}}}}{{f}^{i}}(t,{{X}_{t-}})\frac{\text{d}{{\left\langle {{X}^{1,c}},{{X}^{j,c}} \right\rangle }_{t}}}{\text{d}{{\left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle }_{t}}}+\frac{\text{d}{{\left\langle {{\overset{}{\skew3\check S}\,}^{d}},{{Y}^{i,d}} \right\rangle }_{t}}}{\text{d}{{\left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle }_{t}}}),\quad i=1,2. \label{eqn12} \end{equation}

Moreover, setting

\[{{Q}^{jk}}:=\left\langle {{X}^{j,c}},{{X}^{k,c}} \right\rangle -\int_{0}^{\cdot }{\frac{d\left\langle {{X}^{1,c}},{{X}^{j,c}} \right\rangle s}{d{{\left\langle \overset{{\skew3\check S}}\,,\overset{{\skew3\check S}}\, \right\rangle }_{s}}}}\text{d}{{\left\langle {{X}^{1,c}},{{X}^{k,c}} \right\rangle }_{s,}}\]

we have

(3.9) \begin{equation} \begin{split} \text{d }{{\left\langle {{L}^{1}},{{L}^{2}} \right\rangle }_{t}}&=\sum\limits_{j=1}^{n}{\sum\limits_{k=1}^{n}{{{\partial }_{{{x}_{j}}}}}}{{f}^{1}}(t,{{X}_{t-}}){{\partial }_{{{x}_{k}}}}{{f}^{2}}(t,{{X}_{t-}})\text{d}Q_{t}^{jk}\\ &-\sum\limits_{j=1}^{n}{{{\partial }_{{{x}_{j}}}}}{{f}^{1}}(t,{{X}_{t-}})\frac{{{\left\langle \text{d}{{\overset{}{\skew3\check S}\,}^{d}},{{Y}^{2,d}} \right\rangle }_{t}}}{\text{d}{{\left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle }_{t}}}\text{d}{{\left\langle {{X}^{1,c}},{{X}^{j,c}} \right\rangle }_{t}}\\ &-\sum\limits_{j=1}^{n}{{{\partial }_{{{x}_{j}}}}}{{f}^{2}}(t,{{X}_{t-}})\frac{\text{d}{{\left\langle {{\overset{}{\skew3\check S}\,}^{d}}{{Y}^{1,d}} \right\rangle }_{t}}}{\text{d}{{\left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle }_{t}}}\text{d}{{\left\langle {{X}^{1,c}},{{X}^{j,c}} \right\rangle }_{t}}\\ &+\text{d}{{\left\langle {{Y}^{1,d}},{{Y}^{2,d}} \right\rangle }_{t}}-\frac{\text{d}{{\left\langle {{\overset{}{\skew3\check S}\,}^{d}},{{Y}^{1,d}} \right\rangle }_{t}}}{\text{d}{{\left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle }_{t}}}\text{d}{{\left\langle {{\overset{}{\skew3\check S}\,}^{d}}{{Y}^{2,d}} \right\rangle }_{t,}} \end{split} \label{eqn13} \end{equation}

where, denoting $W^i(t,x)\,:\!=f^i(t,x+X_{t-})-f^i(t,X_{t-})$ , $i=1,2$ , we have

(3.10) \begin{eqnarray} \left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle =\left\langle {{X}^{1,c}},{{X}^{1,c}} \right\rangle +\int_{0}^{\cdot }{\int_{E}{{{({{\text{e}}^{{{x}_{1}}}}-1)}^{2}}}}\nu (\text{d}s,\text{d}x), \label{eqn14} \end{eqnarray}

(3.11) \begin{eqnarray} \left\langle {{\overset{}{\skew3\check S}\,}^{d}},{{\overset{}{\skew3\check S}\,}^{d}} \right\rangle =\int_{0}^{\cdot }{\int_{E}{{{({{\text{e}}^{{{x}_{1}}}}-1)}^{2}}}}\nu (\text{d}s,\text{d}x), \label{eqn15}\end{eqnarray}

(3.12) \begin{eqnarray} \left\langle {{\overset{}{\skew3\check S}\,}^{d}},{{Y}^{i,d}} \right\rangle =\int_{0}^{\cdot }{\int_{E}{({{\text{e}}^{{{x}_{1}}}}-1)}}{{W}^{i}}(s,x)\nu (\text{d}s,\text{d}x), \label{eqn16}\end{eqnarray}

(3.13) \begin{eqnarray} \left\langle {{Y}^{1,d}},{{Y}^{2,d}} \right\rangle =\int_{0}^{\cdot }{\int_{E}{{{W}^{1}}}}(s,x){{W}^{2}}(s,x)\nu (\text{d}s,\text{d}x). \label{eqn17}\end{eqnarray}

Proof. Clearly, (3.10) and (3.11) follow from (3.6); (3.12) follows from (3.7); and (3.13) is a consequence of Theorem 3.1. Formula (3.8) follows by Theorem 3.1 and (3.6), because of ${{\vartheta }^{i}}=\text{d}\left\langle S,{{Y}^{i}} \right\rangle /\text{d}\left\langle S,S \right\rangle$ , $i=1,2$ , and of $\left\langle S,S \right\rangle =S_{-}^{2}\cdot \left\langle \overset{}{\skew3\check S}\,,\overset{}{\skew3\check S}\, \right\rangle$ . To show (3.9) we consider the identity $ \left\langle {{L}^{1}},{{L}^{2}} \right\rangle =\left\langle {{Y}^{1}},{{Y}^{2}} \right\rangle -\int_{0}^{\cdot }{\vartheta _{s}^{1}}\vartheta _{s}^{2}\text{d}{{\left\langle S,S \right\rangle }_{s}} $ , compute $\left\langle {{Y}^{1}},{{Y}^{2}} \right\rangle$ from Theorem 3.1, and use (3.8) and (3.10). The proof is complete.

Let $(B,C,\nu)$ denote the semimartingale characteristics of X (cf. [16, II.§2a]) relative to the truncation function $h=(h^1,\ldots,h^n)^{\top}$ , where $h^{\kern1ptj}(x)=x_j1_{\{|x_j|\leq1\}}$ , $j=1,\ldots,n$ . Let us assume that $(B,C,\nu)$ are absolutely continuous with respect to the Lebesgue measure on [0,T] (which we denote by ${\mathrm{Leb.}}$ ): By (b, c, F) we denote the predictable differential characteristics of X. We stress that $F_t({\omega},{\,\mathrm{d}} x)$ is a measure on $({\mathbb{R}}^n,{\mathcal{B}}({\mathbb{R}}^n))$ and $C^{\,jk}=\int_0^{\cdot} c^{\,jk}_s{\,\mathrm{d}} s$ , $j,k=1,\ldots,n$ . We get the following corollary to Theorem 3.2.

(i) Setting $\xi_t\,:\!=c_t^{11}+\int_E({\mathrm{e}}^{x_1}-1)^2F_t({\mathrm{d}} x)$ , $t\in[0,T]$ , we get, ${\mathrm{Leb.}}$ -a.e.,

(3.14) \begin{equation} \vartheta_t^i=\frac{1}{S_{t-}\xi_t}\bigg(\sum_{j=1}^n\partial_{x_j}\,f^i(t,X_{t-})c^{1j}_t+\int_E({\mathrm{e}}^{x_1}-1)W^i(t,x)F_t({\mathrm{d}} x)\bigg), \quad i=1,2, \label{eqn18} \end{equation}

and (3.9) becomes

(3.15) \begin{equation} \begin{split} \frac{\text{d }{{\left\langle {{L}^{1}},{{L}^{2}} \right\rangle }_{t}}}{\text{d}t}& =\sum_{j=1}^n\sum_{k=1}^n\partial_{x_j}\,f^1(t,X_{t-})\partial_{x_k}f^2(t,X_{t-})\bigg(c^{\,jk}_t-\frac{c^{\,1j}_t}{\xi_t}c^{\,1k}_t\bigg) \\&- \frac1{\xi_t}\sum_{j=1}^n\partial_{x_j}\,f^1(t,X_{t-})c^{\,j1}_t\int_E({\mathrm{e}}^{x_1}-1)W^2(t,x)F_t({\mathrm{d}} x) \\&- \frac1{\xi_t}\sum_{j=1}^n\partial_{x_j}\,f^2(t,X_{t-})c^{\,j1}_t\int_E({\mathrm{e}}^{x_1}-1)W^1(t,x)F_t({\mathrm{d}} x) \\&+ \int_EW^1(t,x)W^2(t,x)F_t({\mathrm{d}} x) \\&-\frac1{\xi_t}\bigg(\!\int_E({\mathrm{e}}^{x_1}-1)W^1(t,x)F_t({\mathrm{d}} x)\bigg)\!\bigg(\!\int_E({\mathrm{e}}^{x_1}-1)W^2(t,x)F_t({\mathrm{d}} x)\bigg). \end{split} \label{eqn19} \end{equation}

(ii) If, furthermore, X is continuous, then

\begin{equation*}\label{eq:var.opt.str.abs.con.con} \vartheta_t^i=\frac{1}{S_{t}c^{11}_t}\sum_{j=1}^n\partial_{x_j}\,f^i(t,X_{t})c^{1j}_t,\quad i=1,2, \end{equation*}

and

(3.16) \begin{equation} \frac{\text{d}{{\left\langle {{L}^{1}},{{L}^{2}} \right\rangle }_{t}}}{\text{d}t} =\sum_{j=2}^n\sum_{k=2}^n\partial_{x_j}\,f^1(t,X_{t})\partial_{x_k}f^2(t,X_{t})\bigg(c^{\,jk}_t-\frac{c^{\,1j}_t}{c^{\,11}_t}c^{\,1k}_t\bigg). \label{eqn20} \end{equation}

We remark that in the case of absolutely continuous semimartingale characteristics as described in Corollary 3.1, by (3.5) and (3.6) we get $\left\langle S,S \right\rangle =\int_{0}^{\cdot }{S_{s-}^{2}}{{\xi }_{s}}\text{d}s$ .

4.1. Complex-valued processes depending on a parameter

We denote by ${\mathcal{S}}\subseteq{\mathbb{C}}^n$ the space of parameters and consider the Borel ${\sigma}$ -algebra ${\mathcal{B}}({\mathcal{S}})$ . On $({\mathcal{S}},{\mathcal{B}}({\mathcal{S}}))$ a finite complex-valued measure ${\zeta}$ is given. We recall that with ${\zeta}$ we can associate the positive measure $|{\zeta}|$ , called the total variation of ${\zeta}$ . Moreover, ${\mathrm{d}}{\zeta}=h\,{\mathrm{d}}|{\zeta}|$ , where h is a complex-valued function such that $|h|=1$ . So, $L^1({\zeta})=L^1(|{\zeta}|)$ .

We set $\widetilde{\Omega}\,:\!={\Omega}\times[0,T]\times{\mathcal{S}}$ . If ${\mathcal{O}}$ (resp. ${\mathcal{P}}$ ) is the optional (resp. predictable) ${\sigma}$ -algebra, then we set $\widetilde {\mathcal{O}}\,:\!={\mathcal{O}}\otimes{\mathcal{B}}({\mathcal{S}})$ (resp. $\widetilde {\mathcal{P}}\,:\!={\mathcal{P}}\otimes{\mathcal{B}}({\mathcal{S}})$ ).

We denote by $\skew3\widetilde U$ a function on $\widetilde{\Omega}$ and we say that $(t,\omega ,z)\mapsto \text{ }\widetilde{U}{{(\omega ,z)}_{t}}$ is an optional (resp. a predictable) function if it is $\widetilde{\mathcal{O}}$ -measurable ($\widetilde{\mathcal{P}}$ -measurable). Sometimes, to specify the ${\sigma}$ -algebra ${\mathcal{B}}({\mathcal{S}})$ , we say that $\skew3\widetilde U$ is an optional (resp. predictable) function with respect to ${\mathcal{S}}$ . We denote by ${\mathrm{Leb.}}$ the Lebesgue measure on [0,T].

Convention. We assume that all quantities explicitly depending on $z\in{\mathcal{S}}$ are complex-valued.

\[ {\mathrm{E}}\left[\!\int_{\mathcal{S}}|\skew3\widetilde U(z)_t|^2|{\zeta}|({\mathrm{d}} z)\right]\leq |{\zeta}|({\mathcal{S}})\sup_{z\in{\mathcal{S}}} {\mathrm{E}}[|{ \skew3\widetilde U(z)_t}|^2]<+\infty,\quad t\in[0,T]. \]

So, $z\mapsto \skew3\widetilde U({\omega},z)_t\in L^2({\zeta})$ a.s., $t\in[0,T]$ , and $\skew3\widehat U$ is adapted, $(\omega ,z)\mapsto \widetilde{U}{{(\omega ,z)}_{t}}$ being ${\mathcal{F}}_t\otimes{\mathcal{B}}({\mathcal{S}})$ -measurable, for every $t\in[0,T]$ . Fubini’s theorem now yields that $\skew3\widehat U$ is a complex-valued square-integrable martingale (non-necessarily càdlàg). If $t\mapsto \skew3\widetilde U({\omega},z)_t$ is càdlàg, then $\left( s,\omega ,z \right)\mapsto {{\sup }_{t\in \left[ 0,s \right]}}|\widetilde{U}{{\left( \omega ,z \right)}_{t}}|$ is optional and Tonelli’s theorem and Doob’s inequality yield ${\mathrm{E}}[\!\int_{\mathcal{S}} \sup_{t\in[0,T]}|\skew3\widetilde U(z)_t|^2|{\zeta}|({\mathrm{d}} z)]<+\infty$ . Therefore, we have $|\skew3\widetilde U(z)_t|\leq\sup_{t\in[0,T]} |\skew3\widetilde U(z)_t|\in L^2({\zeta})$ a.s. Hence, by dominated convergence, $\skew3\widehat U$ is a.s. càdlàg. The proof is complete.

To compute the point brackets of square-integrable martingales with a representation as in Lemma 4.1, we shall use the next proposition.

(i) Then the function $ (t,\omega ,z)\mapsto \text{Tv}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{t}}(\omega) $ is predictable and

(4.1) \begin{equation} \text{E}\left[ \int_{\mathcal{S}}{\text{Tv}}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{T}}|\zeta |(\text{d}z) \right]<+\infty. \label{eqn21} \end{equation}

Moreover, the process $D=(D_t)_{t\in[0,T]}$ , $ {{D}_{t}}:={{\int_{\mathcal{S}}{\left\langle Z,\tilde{V}(z) \right\rangle }}_{t}}\zeta (\text{d}z) $ , is well defined for all $t\in[0,T]$ a.s. Furthermore, D is càdlàg, predictable, of finite variation, and $D_0=0$ .

(ii) Let $\skew3\widetilde U$ be an optional function and $\skew3\widetilde U(z)$ be a (non-necessarily càdlàg) version of $\widetilde V(z)$ for every $z\in{\mathcal{S}}$ . Then $\sup_{z\in{\mathcal{S}}}{\mathrm{E}}[|\skew3\widetilde U(z)_t|^2]<+\infty$ , $t\in[0,T]$ , and the process $\skew3\widehat U=(\skew3\widehat U_t)_{t\in[0,T]}$ , $\skew3\widehat U_t\,:\!=\int_{\mathcal{S}}\skew3\widetilde U(z)_t{\zeta}({\mathrm{d}} z)$ , is a complex-valued square-integrable martingale. Let $U^{\ast}$ be a càdlàg version of $\skew3\widehat U$ . Then

(4.2) \begin{equation} \left\langle Z,{{U}^{*}} \right\rangle =\int_{\mathcal{S}}{\left\langle Z\text{,}\tilde{V}(z) \right\rangle }\zeta (\text{d}z). \label{eqn22} \end{equation}

(iii) If $\widetilde K$ is a predictable function such that $\int_0^T|\widetilde K({\omega},z)_s|{\,\mathrm{d}} s<+\infty$ , $ (\omega ,z)\in \Omega \times \mathcal{S} $ , and

${{\left\langle Z,\tilde{V}(z) \right\rangle }_{t}}(\omega )=\int_{0}^{t}{\tilde{K}{{(\omega ,z)}_{s}}\text{d}s,}$

then $ (t,\omega ,z)\mapsto {{\left\langle Z,\tilde{V}(z) \right\rangle }_{t}}(\omega ) $ is a predictable function, $\widetilde K\in L^1(\widetilde {\mathcal{P}},{\mathrm{P}}\otimes{\mathrm{Leb.}}\otimes{\zeta})$ , and

(4.3) \begin{equation} \left\langle Z,{{U}^{*}} \right\rangle =\int_{\mathcal{S}}{\int_{0}^{\cdot }{\text{ }}}\tilde{K}{{(z)}_{s}}\text{d}s\zeta (\text{d}z)=\int_{0}^{\cdot }{\int_{\mathcal{S}}{\tilde{K}{{(z)}_{s}}\zeta (\text{d}z)\text{d}s}} \label{eqn23} \end{equation}

holds. Furthermore, $\text{E}[\left\langle Z,{{U}^{*}} \right\rangle ]=\int_{0}^{\cdot }{\int_{\mathcal{S}}{\text{E}}}[\tilde{K}{{(z)}_{s}}]\zeta (\text{d}z)\text{d}s$ .

Proof. We first prove (i). The predictability of $ (t,\omega ,z)\mapsto \text{Tv}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{t}}(\omega ) $ follows from the predictability of $ (t,\omega ,z)\mapsto {{\left\langle Z,\tilde{V}(z) \right\rangle }_{t}}(\omega ) $ and from

$ \text{Tv}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{t}}=\underset{n\to +\infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{|}{{\left\langle Z,\tilde{V}(z) \right\rangle }_{tk/n}}-{{\left\langle Z,\tilde{V}(z) \right\rangle }_{t(k-1)/n}}|. $

To prove (4.1), we observe that Tonelli’s theorem, the Kunita–Watanabe inequality, and the relation $ \text{E}[{{\left\langle X,\bar{X} \right\rangle }_{t}}]\le \text{E}[|{{X}_{t}}{{|}^{2}}] $ , $t\geq0$ , valid for every $X\in{\mathcal{H}}^2_{\mathbb{C}}$ ($\overline X$ denoting the complex conjugate of X), imply that

$ \text{E}[\int_{\mathcal{S}}{\text{Tv}}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{T}}|\zeta |(\text{d}z)]\le 2\text{E}{{[|{{Z}_{T}}{{|}^{2}}]}^{1/2}}{{(\underset{z\in \mathcal{S}}{\mathop{\sup }}\,\text{E}[|\tilde{V}{{(z)}_{T}}{{|}^{2}}])}^{1/2}}|\zeta |(\mathcal{S})<+\infty. $

This proves (4.1). From (4.1), since $\rm su{{p}_{t\in [0,T]}}|{{\left\langle Z,\tilde{V}(z) \right\rangle }_{t}}|\le \text{Tv}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{T}} $ , we get $ z\mapsto {{\left\langle Z,\tilde{V}(z) \right\rangle }_{t}}(\omega )\in {{L}^{1}}(\zeta ) $ for all $t\in[0,T]$ a.s. So, the process D is defined for every $t\in[0,T]$ a.s., and D is predictable. Since $ (t,\omega ,z)\mapsto \text{Tv}{{(\left\langle Z,\tilde{V}(z) \right\rangle )}_{t}}(\omega )\in {{L}^{1}}(\zeta ) $ a.s., by dominated convergence we see that D is a.s. càdlàg. From the definition of the total variation for $t\mapsto D_t({\omega})$ , by monotone convergence we obtain that D is of finite variation. Since $D_0=0$ holds, this concludes the proof of (i). We now verify (ii). Since $\skew3\widetilde U(z)$ is a version of $\widetilde V(z)$ for every z, we have ${\mathrm{E}}[|\skew3\widetilde U(z)_t|^2]={\mathrm{E}}[|\widetilde V(z)_t|^2]$ for every $t\in[0,T]$ and every $z\in{\mathcal{S}}$ . Hence, by the assumptions on $\widetilde V$ , we get $\sup_{z\in{\mathcal{S}}}{\mathrm{E}}[|\skew3\widetilde U(z)_t|^2]<+\infty$ for every $t\in[0,T]$ . By Lemma 4.1, the process $\skew3\widehat U=\int_{\mathcal{S}} \skew3\widetilde U(z){\zeta}({\mathrm{d}} z)$ is a complex-valued square-integrable martingale. We denote by $U^{\ast}$ a càdlàg version of $\skew3\widehat U$ . Hence, $U^{\ast}_t=\int_{\mathcal{S}} \skew3\widetilde U(z)_t{\zeta}({\mathrm{d}} z)$ a.s. for every $t\in[0,T]$ . Since $\skew3\widetilde U(z)$ is a version of $\widetilde V(z)$ , we get that $ Z\widetilde{U}(z)-\left\langle Z,\widetilde{V}(z) \right\rangle $ is a martingale. Furthermore, by Hölder’s inequality, for every $t\in[0,T]$ we have

\[ {\mathrm{E}}\left[\!\int_{\mathcal{S}}|Z_t\skew3\widetilde U(z)_t||{\zeta}|({\mathrm{d}} z)\right]\leq {\mathrm{E}}[|Z_t|^2]^{1/2}\sup_{z\in{\mathcal{S}}}{\mathrm{E}}[|\skew3\widetilde U(z)_t|^2]^{1/2}|{\zeta}|({\mathcal{S}})<+\infty. \]

To see (4.2), because of this latter estimate and (4.1), we can apply Fubini’s theorem to get that $ZU^{\ast}-D$ is a martingale. Indeed, a.s. and for every $s\leq t$ , we have

\[ \begin{split} \text{E}[{{Z}_{t}}U_{t}^{*}-{{D}_{t}}|{{\mathcal{F}}_{s}}]=\int_{\mathcal{S}}{\text{E}}[{{Z}_{t}}\widetilde{U}{{(z)}_{t}}-{{\left\langle Z,\widetilde{V}(z) \right\rangle }_{t}}|{{\mathcal{F}}_{s}}]\zeta (\text{d}z)={{Z}_{s}}U_{s}^{*}-{{D}_{s}}. \end{split} \]

So, (4.2) holds and the proof of (ii) is complete. Finally, we show (iii). The predictability of the function $\left( t,\omega ,z \right)\mapsto \int_{0}^{t}{\widetilde{K}}{{\left( \omega ,z \right)}_{s}}\text{d}s$ can be verified in a standard way. By (ii) we then deduce that $ \left\langle Z,{{U}^{*}} \right\rangle =\int_{\mathcal{S}}{\int_{0}^{\cdot }{\widetilde{K}{{(z)}_{s}}\text{d}s\zeta (\text{d}z)}} $ , which shows the first equality on the right-hand side of (4.3). From Tonelli’s theorem, we get

(4.4) \begin{matrix} \int_{0}^{T}{\int_{S}{|\widetilde{K}{{(z)}_{s}}||\zeta |(dz)}}\text{d}s=\int_{S}{\int_{0}^{T}{|\widetilde{K}{{(z)}_{s}}}}|ds|\zeta |(dz)\\ =\int_{S}{\text{Tv(}\left\langle Z,\widetilde{V}(z) \right\rangle \text{)}T\text{ }\!\!|\!\!\text{ }}\zeta |(dz)<+\infty ,\,\,\,\,\,\,\,\rm{a.s.}, \\ \end{matrix}

where, in the last estimate, we used the properties of the variation process and (4.1). Hence, taking the expectation in (4.4) and using (4.1) yields $\widetilde K\in L^1(\widetilde {\mathcal{P}},{\mathrm{P}}\otimes{\mathrm{Leb.}}~\otimes~{\zeta})$ . So, Fubini’s theorem implies that $z\mapsto \widetilde K({\omega},z)_t$ belongs to $L^1({\zeta})$ ${\mathrm{P}}\otimes{\mathrm{Leb.}}$ -a.e. and that $(t,{\omega})\mapsto\int_{\mathcal{S}} \widetilde K({\omega},z)_t{\zeta}({\mathrm{d}} z)$ is predictable, ${\mathrm{P}}\otimes{\mathrm{Leb.}}$ -integrable, and, finally, for every $t\in[0,T]$ , ${\mathrm{E}}[\!\int_{\mathcal{S}}\int_0^t \widetilde K(z)_s{\,\mathrm{d}} s{\zeta}({\mathrm{d}} z)]=\int_0^t\int_{\mathcal{S}} {\mathrm{E}}[\widetilde K(z)_s]{\zeta}({\mathrm{d}} z){\,\mathrm{d}} s$ . From (4.4), we also get $\int_{\mathcal{S}}\int_0^t\widetilde K(z)_s{\,\mathrm{d}} s{\zeta}({\mathrm{d}} z)=\int_0^t\int_{\mathcal{S}} \widetilde K(z)_s{\zeta}({\mathrm{d}} z){\,\mathrm{d}} s$ , $t\in[0,T]$ , a.s. The proof is complete.

4.2. Fourier representation of the GKW decomposition

Let X be a factor process taking values in ${\mathbb{R}}^n$ , and let $R\in{\mathbb{R}}^n$ be such that ${\mathrm{E}}[\!\exp(2R^{\top} X_T)]<~\infty$ . We define ${\mathcal{S}}\,:\!=\{z\in{\mathbb{C}}^n\,:\, {\mathrm{Re}}(z)=R\}$ . A European option ${\eta}\in L^2({\mathrm{P}})$ is given and its payoff is ${\eta}=h({X_T})$ for some real-valued functionh with domain in ${\mathbb{R}}^n$ . We remark that ${\eta}$ is a real-valued random variable. We assume that the two-sided Laplace transform $\hat{h}$ of h exists in R and that it is integrable on ${\mathcal{S}}$ . Then, h has the following representation:

(4.5) \begin{equation} h(x)=\frac1{(2\pi\,{\mathrm{i}})^n}\,\int_{R-{\mathrm{i}}\,\infty}^{R+{\mathrm{i}}\,\infty}\exp(z^{\top} x)\hat h(z){\,\mathrm{d}} z=\int_{\mathcal{S}} \exp(z^{\top} x){\zeta}({\mathrm{d}} z), \label{eqn24} \end{equation}

where ${\zeta}$ is the finite complex-valued absolutely continuous measure on ${\mathcal{S}}$ defined by

(4.6) \begin{equation} {\zeta}({\mathrm{d}} z)\,:\!=\frac1{(2\pi\,{\mathrm{i}})^n}\,\hat h(z)\,{\mathrm{d}} z. \label{eqn25} \end{equation}

Let $H\in{\mathcal{H}}^2$ be the unique (up to indistinguishability) càdlàg real-valued martingale satisfying $H_t={\mathrm{E}}[{\eta}\mid{\mathcal{F}}_t]$ a.s., $t\in[0,T]$ .

Let $S\in{\mathcal{H}}^2$ describe the price process of some traded asset. We stress that S is a real-valued process. Theorem 4.1 below shows that the Fourier representation (4.5) of ${\eta}$ and the GKW decomposition of ${\eta}$ with respect to S can be interchanged.

(i) We have $\widetilde{\vartheta }\in {{L}^{2}}(\widetilde{\mathcal{P}},\text{P}\otimes \left\langle S,S \right\rangle \otimes \zeta)$ , $\int_{\mathcal{S}}\widetilde\vartheta(z){\zeta}({\mathrm{d}} z)\in L^2(S)$ , and there exists an optional function $ (t,\omega ,z)\mapsto \widetilde{Y}{{(\omega ,z)}_{t}} $ such that $\widetilde Y(z)$ is a version of $\int_0^{\cdot} \widetilde\vartheta(z)_s{\,\mathrm{d}} S_s$ (and hence of $\int_0^{\cdot} \vartheta(z)_s{\,\mathrm{d}} S_s$ ) for every $z\in{\mathcal{S}}$ and

(4.7) \begin{equation} \int_{0}^{t}{\int_{\mathcal{S}}{\widetilde{\vartheta }{{(z)}_{s}}\zeta (\text{d}z)\text{d}{{S}_{s}}}}=\int_{\mathcal{S}}{\widetilde{Y}{{(z)}_{t}}\zeta (\text{d}z),}\quad a.s.,for all\ t\in [0,T]. \label{eqn26} \end{equation}

(ii) There exists an optional function $ (t,\omega ,z)\mapsto \text{ }\widetilde{L}{{(\omega ,z)}_{t}} $ such that $\widetilde L(z)$ is a version of L(z), for every $z\in{\mathcal{S}}$ . Furthermore, $\sup_{z\in{\mathcal{S}}}{\mathrm{E}}[|\widetilde L(z)_t|^2]<+\infty$ , $t\in[0,T]$ , and

(4.8) \begin{equation} {{L}_{t}}=\int_{\mathcal{S}}{\text{ }}\widetilde{L}{{(z)}_{t}}\zeta (\text{d}z),\quad \text{a}\text{.s}\text{.},\ for all\ t\in [0,T]. \label{eqn27} \end{equation}

(iii) The GKW decomposition of $H\in{\mathcal{H}}^2$ , $H_t={\mathrm{E}}[{\eta}\mid{\mathcal{F}}_t]$ a.s., $t\in[0,T]$ , is

(4.9) \begin{equation} H=H_0+\int_0^{\cdot}\int_{\mathcal{S}}\widetilde\vartheta(z)_s\zeta({\mathrm{d}} z){\,\mathrm{d}} S_s+L , \label{eqn28} \end{equation}

and $\vartheta$ is given by

(4.10) \begin{equation} \vartheta=\int_{\mathcal{S}}\widetilde\vartheta(z){\zeta}({\mathrm{d}} z). \label{eqn29} \end{equation}

Proof. First, we observe that, for every $z\in{\mathcal{S}}$ , the identity

\[ {\mathrm{E}}[|\skew3\widetilde H(z)_t|^2]={\mathrm{E}}\left[\Big|\int_0^{\cdot}\vartheta_s(z){\,\mathrm{d}} S_s\Big|^2+|L(z)_t|^2+|\skew3\widetilde H(z)_0|^2\right] \]

holds. Hence, Proposition 4.2 implies that

(4.11) \begin{equation} \sup_{z\in{\mathcal{S}}}{\mathrm{E}}\left[\Big|\int_0^t\vartheta_s(z){\,\mathrm{d}} S_s\Big|^2\right]<+\infty\,,\qquad \sup_{z\in{\mathcal{S}}}{\mathrm{E}}\Big[|L(z)_t|^2\Big]<+\infty, \quad t\in[0,T]. \label{eqn30} \end{equation}

We now verify (i). The stochastic integrals $\int_0^{\cdot}\vartheta_s(z){\,\mathrm{d}} S_s$ and $\int_0^{\cdot}\widetilde\vartheta_s(z){\,\mathrm{d}} S_s$ are indistinguishable, for every $z\in{\mathcal{S}}$ . So, from (4.11), we get $\sup_{z\in{\mathcal{S}}}{\mathrm{E}}[|\int_0^ t\widetilde\vartheta_s(z){\,\mathrm{d}} S_s|^2]<+\infty$ , $t\in[0,T]$ . From Tonelli’s theorem and Itô’s isometry, we then have

\[ \begin{split} \text{E}\left[ \int_{0}^{T}{\int_{\mathcal{S}}{|}}\widetilde{\vartheta }{{(z)}_{t}}{{|}^{2}}|\zeta |(\text{d}z)\text{d}{{\left\langle S,S \right\rangle }_{t}} \right]&=\int_{\mathcal{S}}{\text{E}}\left[ \int_{0}^{T}{|}\widetilde{\vartheta }{{(z)}_{t}}{{|}^{2}}\text{d}{{\left\langle S,S \right\rangle }_{t}} \right]|\zeta |(\text{d}z)\\ &\le |\zeta |(\mathcal{S})\underset{z\in \mathcal{S}}{\mathop{\sup }}\,\text{E}\left[ |\int_{0}^{T}{\widetilde{\vartheta }(z)\text{d}{{S}_{s}}{{|}^{2}}} \right]<+\infty , \end{split} \]

which shows that $ \widetilde{\vartheta }\in {{L}^{2}}(\widetilde{\mathcal{P}},\text{P}\otimes \left\langle S,S \right\rangle \otimes \zeta) $ . Therefore, applying Fubini’s theorem yields that $z\mapsto\widetilde\vartheta({\omega},z)_t\in L^2({\zeta})$ $ \text{P}\otimes \left\langle S,S \right\rangle $ -a.e., and that $(t,{\omega})\mapsto\int_{\mathcal{S}}|\widetilde\vartheta({\omega},z)_t|^2|{\zeta}|({\mathrm{d}} z)$ is predictable and belongs to $ {{L}^{1}}(\mathcal{P},\text{P}\otimes \left\langle S,S \right\rangle) $ . So, $\int_{\mathcal{S}}\widetilde\vartheta(z){\zeta}({\mathrm{d}} z)\in L^2(S)$ . To see (4.7), we apply the stochastic Fubini theorem as in [20, Theorem 3(a) and Corollary 2]: There exists an optional function $\left( t,\omega ,z \right)\mapsto \widetilde{Y}{{\left( \omega ,z \right)}_{t}}$ such that $\widetilde Y(z)$ is a version of $\int_0^{\cdot}\widetilde\vartheta(z)_s {\,\mathrm{d}} S_s$ , for every $z\in{\mathcal{S}}$ , and we have the identity (4.7). The proof of (i) is complete. We now verify (ii) and (iii). We define $\widetilde L({\omega},z)_t\,:\!=\skew3\widetilde H({\omega},z)_t-\skew3\widetilde H(z)_0-\widetilde Y({\omega},z)_t$ , for every $(t,\omega ,z)\in \widetilde{\Omega}$ . Since $\widetilde L(z)_t=\skew3\widetilde H(z)_t-\skew3\widetilde H(z)_0-\int_0^{\cdot}\widetilde \vartheta(z)_s{\,\mathrm{d}} S_s$ , $\widetilde Y(z)$ being a version of $\int_0^{\cdot}\widetilde \vartheta(z)_s{\,\mathrm{d}} S_s$ for every $z\in{\mathcal{S}}$ , we obtain that $\widetilde L(z)$ is a version of L(z) for every $z\in{\mathcal{S}}$ . Hence, (4.11) yields the estimate $\sup_{z\in{\mathcal{S}}}{\mathrm{E}}[|\widetilde L(z)_t|]<+\infty$ for every $t\in[0,T]$ . Therefore, since $\widetilde L$ is optional, by Lemma 4.1, $\widehat L_t=\int_{\mathcal{S}} \widetilde L(z)_t{\zeta}({\mathrm{d}} z)$ is a square-integrable martingale. We denote by $L^{\ast}$ a càdlàg version of $\widehat L$ . We can now compute a.s., and for every $t\in[0,T]$ ,

(4.12) \begin{align} & {{H}_{t}}=\int_{\mathcal{S}}{\widetilde{H}{{(z)}_{t}}\zeta (dz)}={{H}_{0}}+\int_{\mathcal{S}}{\widetilde{Y}{{(z)}_{t}}\zeta (dz)}+{{\widehat{L}}_{t}} \\ &={{H}_{0}}+\int_{0}^{t}{\int_{\mathcal{S}}{\widetilde{\vartheta }{{(z)}_{s}}\zeta (dz)}}\,\text{d}{{S}_{s}}+L_{t}^{*}, \\ \end{align}

where in the first equality we used Proposition 4.2. But the left-hand side and the right-hand side of the last equality in (4.12) are càdlàg. Hence, they are indistinguishable. To prove (4.9), it remains to show that $L^{\ast}$ is orthogonal to S. This yields $L=L^{\ast}$ . By Tonelli’s theorem and Hölder’s inequality we obtain, for every $t\in[0,T]$ , $ {\mathrm{E}}[\!\int_{\mathcal{S}} |S_t\widetilde L(z)_t||{\zeta}|({\mathrm{d}} z)]<+\infty $ . Since SL(z) is a martingale for every $z\in{\mathcal{S}}$ , $S\widetilde L(z)$ is a martingale for every $z\in{\mathcal{S}}$ too, $\widetilde L(z)$ being a version of L(z) for every $z\in{\mathcal{S}}$ . By Fubini’s theorem, we get, a.s., for every $0\leq s\leq t\leq T$ ,

\[ \begin{split} {\mathrm{E}}[S_t L^{\ast}_t\mid{\mathcal{F}}_s]&={\mathrm{E}}[S_t \widehat L_t\mid{\mathcal{F}}_s]=\int_{\mathcal{S}}{\mathrm{E}}[S_t \widetilde L(z)_t\mid{\mathcal{F}}_s]{\zeta}({\mathrm{d}} z)=S_sL^{\ast}_s. \end{split} \]

So, $L^{\ast}$ is orthogonal to S, and (4.9) and (4.8) hold. From (4.9) we get (4.10) because of $ \vartheta =\text{d}\left\langle S,H \right\rangle /\text{d}\left\langle S,S \right\rangle =\int_{\mathcal{S}}{\widetilde{\vartheta }(z)\zeta (\text{d}z)} $ . The proof is complete.

Let ${\eta}^{\kern1ptj}$ be a real-valued square-integrable European option with GKW decomposition $(\vartheta^{\kern1ptj}, L^{\kern1ptj})$ with respect to S, $j=1,2$ . In the following theorem we compute $\left\langle {{L}^{1}},{{L}^{2}} \right\rangle$ assuming that ${\eta}^{\kern1ptj}=h^{\kern1ptj}(X_T)$ and $h^{\kern1ptj}$ has the representation given in (4.5) and (4.6) on ${\mathcal{S}}^{\kern1ptj}\,:\!=\{z\in{\mathbb{C}}^n\,:\, {\mathrm{Re}}(z)=R^{\kern1ptj}\}$ with ${\mathrm{E}}[\!\exp(2(R^{\kern1ptj})^{\top} X_T)]<+\infty$ , $j=1,2$ . We use the notation ${\mathcal{S}}\,:\!={\mathcal{S}}^1\cup{\mathcal{S}}^2$ . We also introduce $\overline{\Omega}\,:\!=[0,T]\times{\Omega}\times{\mathcal{S}}^1\times{\mathcal{S}}^2$ and $\overline{\mathcal{P}}\,:\!={\mathcal{P}}\otimes{\mathcal{B}}({\mathcal{S}}^1)\otimes{\mathcal{B}}({\mathcal{S}}^2)$ .

(i) Then there exists an optional mapping $\widetilde L$ on $[0,T]\times{\Omega}\times{\mathcal{S}}$ such that the restriction of $\widetilde L$ to $[0,T]\times{\Omega}\times{\mathcal{S}}^{\kern1ptj}$ is optional with respect to ${\mathcal{S}}^{\kern1ptj}$ , $\widetilde L(z)$ is a version of L(z), for every $z\in{\mathcal{S}}$ , and $L^ j_t=\int_{{\mathcal{S}}^{\kern1ptj}}\widetilde L(z)_t{\zeta}^{\kern1ptj}({\mathrm{d}} z)$ a.s. for every $t\in[0,T]$ , $j=1,2$ .

(ii) If the function $ (t,\omega ,{{z}_{1}},{{z}_{2}})\mapsto {{\left\langle L({{z}_{1}})L({{z}_{2}}) \right\rangle }_{t}}(\omega)$ on $\overline{\Omega}$ is $\overline{\mathcal{P}}$ -measurable then so is $ (t,\omega ,{{z}_{1}},{{z}_{2}})\mapsto \text{Tv}{{(\left\langle L({{z}_{1}}),L({{z}_{2}}) \right\rangle )}_{t}}(\omega ) $ , and

(4.13) \begin{equation}\begin{split} \text{E}[{{\int }_{{{\mathcal{S}}^{1}}}}{{\int }_{{{\mathcal{S}}^{2}}}}\text{Tv}{{(\left\langle L({{z}_{1}}),L({{z}_{2}}) \right\rangle )}_{T}}|{{\zeta }^{2}}|(\text{d}{{z}_{2}})|{{\zeta }^{1}}|(\text{d}{{z}_{1}})]<+\infty. \end{split} \label{eqn31} \end{equation}

Moreover, $ \int_{{{\mathcal{S}}^{1}}}{\int_{{{\mathcal{S}}^{2}}}{\left\langle L({{z}_{1}})L({{z}_{2}}) \right\rangle }}{{\zeta }^{2}}(\text{d}{{z}_{2}}){{\zeta }^{1}}(\text{d}{{z}_{1}}) $ is a predictable process and

(4.14) \begin{equation} \left\langle {{L}^{1}},{{L}^{2}} \right\rangle =\int_{{{\mathcal{S}}^{1}}}{\int_{{{\mathcal{S}}^{2}}}{\left\langle L({{z}_{1}}),L({{z}_{2}}) \right\rangle }}{{\zeta }^{2}}(\text{d}{{z}_{2}}){{\zeta }^{1}}(\text{d}{{z}_{1}}). \label{eqn32} \end{equation}

(iii) If $\widetilde K$ is a $\overline{\mathcal{P}}$ -measurable function on $\overline{\Omega}$ such that $\int_0^T|\widetilde K({\omega},z_1,z_2)_s|{\,\mathrm{d}} s<+\infty$ , for $ (\omega ,{{z}_{1}},{{z}_{2}})\in \Omega \times {{\mathcal{S}}^{1}}\times {{\mathcal{S}}^{2}} $ , and

$$ {{\left\langle L({{z}_{1}})L({{z}_{2}}) \right\rangle }_{t}}(\omega )=\int_{0}^{t}{\text{ }}\widetilde{K}{{(\omega ,{{z}_{1}},{{z}_{2}})}_{s}}\text{d}s,$$

then $\widetilde K$ is $\overline{\mathcal{P}}$ -measurable, $\widetilde K\in L^1(\overline{\mathcal{P}},{\mathrm{P}}\otimes{\mathrm{Leb.}}\otimes{\zeta}^1\otimes{\zeta}^2)$ ,

$$ \left\langle {{L}^{1}},{{L}^{2}} \right\rangle =\int_{0}^{\cdot }{\int_{{{\mathcal{S}}^{1}}}{\int_{{{\mathcal{S}}^{2}}}{\widetilde{K}{{({{z}_{1}},{{z}_{2}})}_{s}}{{\zeta }^{2}}(\text{d}{{z}_{2}}){{\zeta }^{1}}(\text{d}{{z}_{1}})\text{d}s}}}$$

and $ \text{E}[{{\left\langle {{L}^{1}},{{L}^{2}} \right\rangle }_{T}}]=\int_{0}^{T}{\int_{{{\mathcal{S}}^{1}}}{\int_{{{\mathcal{S}}^{2}}}{\text{E}}}}[\text{ }\widetilde{K}{{({{z}_{1}},{{z}_{2}})}_{s}}]{{\zeta }^{2}}(\text{d}{{z}_{2}}){{\zeta }^{1}}(\text{d}{{z}_{1}})\text{d}s $ .

Notice that the process $\left\langle {{L}^{1}},{{L}^{2}} \right\rangle$ in Theorem 4.2 is real valued, while $\left\langle L({{z}_{1}}),({{z}_{2}}) \right\rangle$ is complex valued, $z_j\in{\mathcal{S}}^{\kern1ptj}$ , $j=1,2$ .

We now combine the results of this section with those of Section 3 for semimartingales with absolutely continuous characteristics. Let us consider the factor process $X=(X^1,\ldots,X^n)^{\top}$ and let $R^{\kern1ptj}\in{\mathbb{R}}$ be such that ${\mathrm{E}}[\!\exp(2R^{\kern1ptj} X_T^ 1)]<+\infty$ . We define ${\mathcal{S}}^{\kern1ptj}\,:\!=\{z\in{\mathbb{C}}\,:\, {\mathrm{Re}}(z)=R^{\kern1ptj}\}$ , $j=1,\ldots,d$ , and set ${\mathcal{S}}\,:\!=\bigcup_{j=1}^d{\mathcal{S}}^{\kern1ptj}$ .

(i) Let $f\,:\,[0,T]\times{\mathbb{R}}^n\times{\mathcal{S}} \rightarrow{\mathbb{C}} $ be such that the restriction of f to $[0,T]\times{\mathbb{R}}^n\times{\mathcal{S}}^{\kern1ptj}$ is ${\mathcal{B}}([0,T])\otimes{\mathcal{B}}({\mathbb{R}}^n)\otimes{\mathcal{B}}({\mathcal{S}}^{\kern1ptj})$ -measurable, $j=1,\ldots,d$ , $f(\cdot,\cdot, z)\in C^{1,2}([0,T]\times{\mathbb{R}}^n)$ , for $z\in{\mathcal{S}}$ , and $\skew3\widetilde H({\omega},z)_t=f(t,X_t({\omega}),z)$ , $\skew3\widetilde H(z)_t={\mathrm{E}}[{\mathrm{e}}^{zX^1_T}\mid{\mathcal{F}}_t]$ , a.s., $t\in[0,T]$ , $z\in{\mathcal{S}}$ .

(ii) We define the complex-valued function $W(t,{\omega},x,z)\,:\!=f(t,x+X_{t-}({\omega}),z)-f(t,$ $X_{t-}({\omega}),z)$ , for $z\in{\mathcal{S}}$ , and assume that $\int_E|W(t,{\omega},x,z)|^2 F_t({\omega},{\mathrm{d}} x)<+\infty$ , for every $(t,{\omega},z)\in[0,T]\times{\Omega}\times{\mathcal{S}}$ , and that $\int_E({\mathrm{e}}^{ x_1}-1)^2F_t({\omega},{\mathrm{d}} x)<+\infty$ , $\xi_t({\omega})>0$ for every $(t,{\omega})\in[0,T]\times{\Omega}$ (see Corollary 3.1(i) for the definition of the process $\xi$ ).

(iii) Let $(\vartheta(z),L(z))$ denote the GKW decomposition of $\skew3\widetilde H(z)$ with respect to S, for $z\in{\mathcal{S}}$ , and let $(t,{\omega},z)\mapsto\widetilde\vartheta({\omega},z)_t$ be the function defined on the right-hand side of (3.14) in terms of $f(t,X_{t-}({\omega}),z)$ and $W(t,{\omega},x,z)$ (see (ii) above), that is,

\[ \widetilde\vartheta(z)_t=\frac{1}{S_{t-}\xi_t}\bigg(\sum_{k=1}^n\partial_{x_k}f(t,X_{t-},z)c^{1k}_t+\int_E({\mathrm{e}}^{x_1}-1)W(t,x,z)F_t({\mathrm{d}} x)\bigg),\quad z\in{\mathcal{S}}. \]

Claim 1. The mapping $ (t,\omega )\mapsto \text{ }\widetilde{H}{{(\omega ,z)}_{t}} $ is càdlàg, for $z\in{\mathcal{S}}$ , and the restriction of $\skew3\widetilde H$ to ${\mathcal{S}}^{\kern1ptj}$ is optional with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . Furthermore, $\widetilde\vartheta(z)=\vartheta(z)$ $ \text{P}\otimes \text{d}\left\langle S,S \right\rangle $ -a.e., for $z\in{\mathcal{S}}$ , holds, and the restriction of $\widetilde\vartheta$ to ${\mathcal{S}}^{\kern1ptj}$ is a predictable function with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . In particular, the assumptions of Theorem 4.1 are satisfied with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ .

(iv) We denote by $f^{\kern1ptj}$ and $W^{\kern1ptj}$ the complex-valued functions obtained taking the restriction of f and W to ${\mathcal{S}}^{\kern1ptj}$ , respectively, so that $\skew3\widetilde H({\omega},z)_t=f^{\kern1ptj}(t,X_t({\omega}),z)$ , $z\in{\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . Let $\widetilde K({\omega},z_i,z_j)$ denote the density of $ \left\langle L({{z}_{i}}),L({{z}_{j}}) \right\rangle $ with respect to the Lebesgue measure given in (3.15) (now defined with respect to $f^i$ , $f^ j$ , $W^i$ , and $W^{\kern1ptj}$ , $i,j=1,\ldots,d$ ) and let $\int_0^T|\widetilde K({\omega},z_i,z_j)_t|{\,\mathrm{d}} t<+\infty$ for every $ (t,\omega ,{{z}_{i}},{{z}_{j}})\in [0,T]\times \Omega \times {{\mathcal{S}}^{i}}\times {{\mathcal{S}}^{j}} $ , for every $i,j=1,\ldots,d$ .

Claim 2. The function $ (t,\omega ,{{z}_{i}},{{z}_{j}})\mapsto \int_{0}^{t}{\widetilde{K}{{(\omega ,{{z}_{i}},{{z}_{j}})}_{s}}\text{d}s} $ is predictable with respect to ${\mathcal{S}}^ i\times{\mathcal{S}}^{\kern1ptj}$ and we have $\text{E}[{{\left\langle {{L}^{i}},{{L}^{j}} \right\rangle }_{T}}]=\int_{0}^{T}{\int_{{{\mathcal{S}}^{i}}}{\int_{{{\mathcal{S}}^{j}}}{\text{E}}}}[\widetilde{K}{{(\omega ,{{z}_{i}},{{z}_{j}})}_{t}}]{{\zeta }^{j}}(\text{d}{{z}_{j}}){{\zeta }^{i}}(\text{d}{{z}_{i}})\text{d}t$ , for $i,j=1,\ldots,d$ .

5.1. The Heston model

The Heston model for ${{(X,V)}^{\top }}$ (see [Reference Heston14]) is given by

\begin{equation*}\label{eq:Heston} {\mathrm{d}} X_{\,t}=-\frac12 V_{\,t}{\,\mathrm{d}} t+\sqrt{V_{\,t}} {\,\mathrm{d}} W_{\,t}^{\,1},\quad {\mathrm{d}} V_{\,t}=-\lambda(V_{\,t}-\kappa ){\,\mathrm{d}} t+\sigma\sqrt{V_{\,t}} {\,\mathrm{d}} W_{\,t}^{\,2}, \end{equation*}

where $W^1$ and $W^2$ are correlated Brownian motions with $ {{\left\langle {{W}^{1}},{{W}^{2}} \right\rangle }_{t}}=\rho t$ , $\rho\in[\!-1,1]$ and $\lambda,\sigma,\kappa>0$ , $2{\lambda}\kappa\geq{\sigma}^2$ so that $V_t>0$ . Notice that ${{(X,V)}^{\top }}$ is affine and

(5.1) \begin{equation} {\mathrm{E}}[\!\exp( u_1X_T+u_2V_T)\mid{\mathcal{F}}_t]=\exp(\phi_{T-t}(u_1,u_2)+u_1 X_t+\psi_{T-t}(u_1,u_2)V_t)=\!:\,\skew3\widetilde H(z)_t, \label{eqn33} \end{equation}

where $(t,u_1,u_2)\mapsto\phi_t(u_1,u_2)$ and $(t,u_1,u_2)\mapsto\psi_t(u_1,u_2)$ are complex-valued functions that can be expressed in terms of elementary functions. See [Reference Heston14], [2, Section 4.2.1], [Reference Albrecher, Mayer, Schoutens and Tistaert1,Reference Keller-Ressel and Mayerhofer19] etc. for concrete expressions, domain of validity, and numerical issues.

For $z\in{\mathcal{S}}^{\kern1ptj}$ , we have $\skew3\widetilde H(z)\in{\mathcal{H}}^2_{\mathbb{C}}$ , $j=1,\ldots,d$ . Defining $r^0_{z_1,z_2}\,:\!=\phi_t(z_1,0)+\phi_t(z_2,0)$ and $r^ 1_{z_1,z_2}\,:\!=\psi_t(z_1,0)+\psi_t(z_2,0)$ for $z_1\in{\mathcal{S}}^i$ , $z_2\in{\mathcal{S}}^{\kern1ptj}$ , from [18, Equation (5.18)], for $t\in[0,T]$ , the random variable $V_t\skew3\widetilde H(z_1)_t\skew3\widetilde H(z_2)_t$ is integrable and

(5.2) \begin{equation} \begin{split} {\mathrm{E}}[V_t\skew3\widetilde H(z_1)_t\skew3\widetilde H(z_2)_t]=&\ S_0^{z_1+z_2}{\mathrm{e}}^{r^0_{z_1,z_2}}\times\exp(\phi_t(z_1+z_2,r^1_{z_1,z_2})\\&+\psi_t(z_1+z_2,r^ 1_{z_1,z_2})V_0)\times\partial_{u_2}[\phi_t(z_1+z_2,u_2)\\ &+\psi_t(z_1+z_2,u_2)V_0]|_{u_2=r^1_{z_1,z_2}}. \end{split} \label{eqn34} \end{equation}

5.1.1. Semi-static variance-optimal hedging

We now compute the quantities A, B, and C defined in (2.4) for the variance swap ${\eta}^0=\int_0^T V_s{\,\mathrm{d}} s-k$ in the Heston model. We start by computing A.

Proposition 5.1. Setting ${\alpha}(t)\,:\!={\lambda}^{-1}(1-{\mathrm{e}}^{-{\lambda}(T-t)})$ , we have

(5.3) \begin{equation} A={\sigma}^2(1-\varrho^2)\int_0^T{\alpha}^2(t)[(V_0-\kappa){\mathrm{e}}^{-{\lambda} t}+\kappa]\,{\mathrm{d}} t. \label{eqn35} \end{equation}

(5.4) \begin{equation} {\mathrm{E}}[V_s\mid{\mathcal{F}}_t]=\kappa(1-{\mathrm{e}}^{-{\lambda} (s-t)})+{\mathrm{e}}^{-{\lambda} (s-t)}V_t,\quad s\geq t. \label{eqn36} \end{equation}

We consider $H^0\in{\mathcal{H}}^2$ such that $H^0_t={\mathrm{E}}[\!\int_0^T V_s{\,\mathrm{d}} s|{\mathcal{F}}_t]$ a.s., $t\in[0,T]$ . By Fubini’s theorem we obtain $H^0_t=f^0(t,X_t,V_t,[X,X]_t)$ , where $f^0(t,x,v,w)=\tilde{\alpha}(t)+{\alpha}(t)v+w$ , with $\tilde{\alpha}(t)\,:\!=\kappa{\lambda}^{-1}({\lambda}(T-t)-1+{\mathrm{e}}^{-{\lambda}(T-t)})$ and ${\alpha}(t)$ as above, $t\in[0,T]$ . We can now apply Corollary 3.1(ii) with $Y^i=H^0$ , $f^i=f^0$ , $i=1,2$ , and $ {{({{X}^{1}},{{X}^{2}},{{X}^{3}})}^{\top }}={{(X,V,[X,X])}^{\top }} $ . The only relevant derivative is $\partial_v f^0={\alpha}(t)$ . Taking the expectation in (3.16), and applying Fubini’s theorem and (5.4) with $t=0$ , we get (5.1). The proof is complete.

We stress that according to Corollary 3.1(ii) we have $\vartheta^0_t=\rho{\sigma}{\alpha}(t)S_t^{-1}$ . As a next step, we compute the vector B in (2.4).

Proposition 5.2. For $B=(B^1,\ldots,B^d)^{\top}$ we have

\[ B^{\kern1ptj}={\sigma}^2(1-\varrho^2)\int_0^T\int_{{\mathcal{S}}^{\kern1ptj}}{\alpha}(t)E(t,z)\psi_{T-t}(z,0){\zeta}^{\kern1ptj}({\mathrm{d}} z){\,\mathrm{d}} t, \]

where ${\alpha}(t)={\lambda}^{-1}(1-{\mathrm{e}}^{-{\lambda}(T-t)})$ and the explicit expression of $E(t,z)\,:\!={\mathrm{E}}[\skew3\widetilde H(z)_tV_t]$ is given by (5.2) with $z_1=z\in{\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ , and $z_2=0$ .

Because of (5.1), $\widetilde K(z)_t\,:\!={\sigma}^2(1-\rho^2){\alpha}(t)\psi_{T-t}(z,0)\skew3\widetilde H(z)_tV_t$ is a predictable function with respect to each ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . Furthermore, it is continuous in t. Because of (5.1) and Theorem 4.3, the assumptions of Theorem 4.1 are satisfied with respect to ${\mathcal{S}}^{\kern1ptj}$ , for each $j=1,\ldots,d$ , so we have $L^ j_t=\int_{{\mathcal{S}}^{\kern1ptj}}\widetilde L(z){\zeta}^{\kern1ptj}({\mathrm{d}} z)$ , a.s., $t\in[0,T]$ . Since $\sup_{z\in{\mathcal{S}}^{\kern1ptj}}{\mathrm{E}}[|\widetilde L(z)_t|^2]<+\infty$ , $t\in[0,T]$ , $j=1,\ldots,d$ , from Proposition 4.1(iii) we get

$${{B}^{j}}=\text{E}[{{\left\langle {{L}^{0}},{{L}^{j}} \right\rangle }_{T}}]={{\sigma }^{2}}(1-{{\rho }^{2}})\int_{0}^{T}{\int_{{{\mathcal{S}}^{j}}}{\alpha }}(t){{\psi }_{T-t}}(z,0)\text{E}[\widetilde{H}{{(z)}_{t}}{{V}_{t}}]{{\zeta }^{j}}(\text{d}z)\text{d}t.$$

Finally, from (5.2) we get the explicit expression of ${\mathrm{E}}[\skew3\widetilde H(z)_tV_t]$ .

Proposition 5.3. For $C=(C^{ij})_{i,j=1,\ldots,d}$ , we have

\[\begin{split} C^{ij}={\sigma}^2(1-\varrho^2)\int_0^T\int_{{\mathcal{S}}^i}\int_{{\mathcal{S}}^{\kern1ptj}}\psi_{T-t}(z_1,0)\psi_{T-t}(z_2,0)J(t,z_1,z_2){\zeta}^{\kern1ptj}({\mathrm{d}} z_2){\zeta}^i({\mathrm{d}} z_1){\,\mathrm{d}} t, \end{split}\]

where $J(t,z_1,z_2)\,:\!={\mathrm{E}}[\skew3\widetilde H(z_1)_t\widetilde H(z_2)_tV_t]$ is given by (5.2), $i,j=1,\ldots,d$ .

where, $(t,{\omega},z_1,z_2)\mapsto\widetilde K({\omega},z_1,z_2)_t$ is ${\mathcal{P}}\otimes{\mathcal{B}}({\mathcal{S}}^i)\otimes{\mathcal{B}}({\mathcal{S}}^{\kern1ptj})$ -measurable and continuous in t. Since we have $V_t>0$ , the claim follows from Theorem 4.3.

5.2. The 3/2 model

The 3/2 model for ${{(X,V)}^{\top }}$ is given by

(5.5) \begin{equation} {\mathrm{d}} X_t=-\frac{V_t}{2}{\,\mathrm{d}} t+\sqrt{V_t}{\,\mathrm{d}} W_t^1,\qquad {\mathrm{d}} V_t=V_t({\lambda}-\kappa V_t){\,\mathrm{d}} t+{\sigma} V_t^{3/2}{\,\mathrm{d}} W^ 2_t,\quad V_0>0, \label{eqn37} \end{equation}

where $W^{1}$ and $W^{2}$ are Brownian motions with ${{\left\langle {{W}^{1}},{{W}^{2}} \right\rangle }_{t}}=\rho t$ , $\rho\in[\!-1,1]$ . We assume $\kappa>0$ , which ensures both non-explosion of V (see [10, Equation (3)]) and $V_t>0$ , $t\in[0,T]$ . The two-dimensional semimartingale ${{(X,V)}^{\top }}$ given by (5.5) is not an affine process. For the computation of the conditional moment-generating function we mainly refer to [Reference Grasselli13], that is, we regard the 3/2 model as a special case of the so-called 4/2 model. More precisely, setting $a=0$ , $b=1$ , and $r=0$ in [13, Equation (2.1)], we get the dynamics of the price process $S={\mathrm{e}}^X$ (and hence of $X=\log S$ ) under the local martingale measure in the 3/2 model. We stress that, denoting by $\kappa_G, \theta_G, {\sigma}_G$ the parameters in [13, Equation (2.2)], the relation between ${\lambda}$ , $\kappa$ , and ${\sigma}$ in (5.5) and $\kappa_G, \theta_G, {\sigma}_G$ is

(5.6) \begin{equation} \kappa=\kappa_G\theta_G-{\sigma}^2_G,\quad {\lambda}=\kappa_G,\quad {\sigma}=-{\sigma}_G. \label{eqn38} \end{equation}

In [17, Proposition 6.3.2.1] the transition density of the process $R=(R_t)_{t\in[0,T]}$ , $R_t\,:\!=1/V_t$ is given. From this, we obtain the density function $f_{V_t}(v)$ of $V_t$ : $f_{V_t}(v)=0$ for $v\leq0$ , and for $v>0$

(5.7) \begin{equation} \begin{split} f_{V_t}(v)=&\frac{1}{2v^2}\,\frac{4{\lambda}}{{\sigma}^2({\mathrm{e}}^{{\lambda} t}-1)}\bigg(\frac{V_0}v\bigg)^{q/2}\\&\quad\times\exp\bigg({\lambda}\Big(1+\frac{ q}2\Big)t-\frac{2{\lambda}}{{\sigma}^2({\mathrm{e}}^{{\lambda} t}-1)}\Big(\frac{1}{V_0}+\frac{{\mathrm{e}}^{{\lambda} t}}{v}\Big)\bigg) I_q\bigg(\frac{4{\lambda}{\mathrm{e}}^{\frac{{\lambda}}{2}t}}{{\sigma}^2({\mathrm{e}}^{{\lambda} t}-1)\sqrt{V_0v}}\bigg), \end{split} \label{eqn39} \end{equation}

where $q\,:\!=2({\kappa}+{\sigma}^2)/{\sigma}^2-1$ and $I_p(\cdot)$ denotes the modified Bessel function of the first kind.

From [13, p. 17], denoting by $M(a,b,\cdot)$ the confluent hypergeometric function, we have

\begin{equation*}\label{eq:mom.v_t} \begin{split} \text{E}[V_{t}^{\eta }]={{\left( \frac{2\lambda {{\text{e}}^{\lambda t}}}{{{\sigma }^{2}}({{\text{e}}^{\lambda t}}-1)} \right)}^{\eta }}\exp \left( -\frac{2\lambda }{{{\sigma }^{2}}({{\text{e}}^{\lambda t}}-1){{V}_{0}}} \right)\frac{\Gamma (q+1-\eta )}{\Gamma (q+1)}\times M\left( q+1-\eta ,q+1,\frac{2\lambda }{{{\sigma }^{2}}({{\text{e}}^{\lambda t}}-1){{V}_{0}}} \right) \end{split} \end{equation*}

for ${\eta}\in{\mathbb{R}}$ such that $q+1-{\eta}>0$ , which always holds for ${\eta}=2+\delta$ , for some fixed $\delta>0$ : In the 3/2 model, contrary to the Heston model, $V_t$ does not have finite moments of every order. According to [13, Proposition 3.1], for $z\in{\mathbb{C}}$ we have $\exp(zX_T)\in L^1({\mathrm{P}})$ if ${\mathrm{Re}}(z)\in{\mathcal{A}}_{0,+\infty}$ , where the set ${\mathcal{A}}_{0,+\infty}$ is defined in [13, Equation (3.7)].

Let $\exp(zX_T)$ be integrable, $z\in{\mathbb{C}}$ . By the relation $M(x,y,z)={\mathrm{e}}^z\,M(y-x,y,-z)$ , $z\in{\mathbb{C}}$ , for the confluent hypergeometric function, from [13, Equation (3.3)] with $a=0$ and $b=1$ , we obtain $\skew3\widetilde H(z)_t={\mathrm{E}}[\!\exp(zX_T)\mid{\mathcal{F}}_t]$ , a.s., $t\geq0$ , where

(5.8) \begin{equation} \skew3\widetilde H(z)_t={\mathrm{e}}^{zX_t}g(t,V_t,z), \label{eqn40} \end{equation}

and the function g is given by

(5.9) \begin{equation} \begin{split} g(t,v,z)\,:\!=\frac{\Gamma({\beta}_z-{\alpha}_z)}{\Gamma({\beta}_z)}&\left(\frac{2{\lambda}}{v{\sigma}^2({\mathrm{e}}^{{\lambda}(T-t)}-1)}\right)^{{\alpha}_z}\\&\times M\left({\alpha}_z,{\beta}_z,-\frac{2{\lambda}}{v{\sigma}^2({\mathrm{e}}^{{\lambda}(T-t)}-1)}\right) \end{split} \label{eqn41} \end{equation}

and ${\alpha}_z=-\frac12-\frac{\tilde\kappa_z}{{\sigma}^2}+c_z$ , ${\beta}_z=1+2c_z$ , $\tilde\kappa_z=k-z\varrho{\sigma}$ , $c_z\,:\!=\{(\frac12+\frac{\tilde\kappa_z}{{\sigma}^2})^2+\frac{z-z^2}{{\sigma}^2}\}^{1/2}$ .

In the proof of [7, Theorem 3, Section 8], it was shown that $v\mapsto g(t,v,z)$ (therein $\ell$ ) solves an ordinary differential equation of the second order and $g(T,V_T,z)=1$ . So, $t\mapsto\skew3\widetilde H({\omega},z)_t$ is continuous. Taking $m_z$ as in [13, Equation (3.5)] with $a=0$ and $b=1$ , then $c_z=\frac{m_z}2$ , by the properties of the confluent hypergeometric function M (x, y, z), and we get:

(5.10) \begin{equation} \begin{split} \partial_{v} g(t,v,z)=&\frac{{\alpha}_z}{v}(\delta(t,v))^{{\alpha}_z}\frac{\Gamma({\beta}_z-{\alpha}_z)}{\Gamma({\beta}_z)}\\&\times\bigg[\frac{\delta(t,v)}{{\beta}_z}M({\alpha}_z+1,{\beta}_z+1,-\delta(t,v))-M({\alpha}_z,{\beta}_z,-\delta(t,v))\bigg], \end{split} \label{eqn42} \end{equation}

where $\delta(t,v)\,:\!=\frac{2{\lambda}}{{\sigma}^2({\mathrm{e}}^{{\lambda}(T-t)}-1)v}$ . We require $\kappa-\varrho{\sigma}\geq-\frac{{\sigma}^2}{2}$ (i.e., Feller’s condition), which ensures that $S={\mathrm{e}}^X$ is a true martingale (see [13, Remark 3.3] with the identification of parameters in (5.6)). To get $S\in{\mathcal{H}}^2$ , we require $2\in{\mathcal{A}}_{0,+\infty}$ .

We observe that the variance swap ${\eta}^0$ is square integrable in this model. This follows because, analyzing the explicit expression of the Laplace transform of ${\eta}^0$ , which was calculated in [7, Theorem 3] (or [13, Appendix A]), we recognize that it is defined in an open neighborhood of zero.

5.2.1. Variance-optimal semi-static hedging

We now compute the quantities A, B, and C defined in (2.4) for the variance swap ${\eta}^0=\int_0^T V_s{\,\mathrm{d}} s-k$ in the 3/2 model.

We start by computing A. First, we introduce the function

(5.11) \begin{equation} h^{\prime}(y)=\frac{{\sigma}^2}{2}\Gamma(q){\mathrm{e}}^{-{2/{\sigma}^2y}}y^{-1}{\gamma}^{\ast}\Big(q,-\frac{2}{{\sigma}^2y}\Big), \label{eqn43} \end{equation}

where $\gamma^{\ast}(a,z)\,:\!=\frac1{\Gamma(a)}\int_0^1u^{a-1}{\mathrm{e}}^{-zu}{\,\mathrm{d}} u$ , such that $\gamma^{\ast}(a,z)=\frac{z^{-a}}{\Gamma(a)}\,\gamma(a,z)$ , $\gamma$ denoting the lower incomplete Gamma function. We then put $h(y)\,:\!=\int_0^ yh^{\prime}(x){\,\mathrm{d}} x$ . The function h was introduced in [7, Theorem 4] and this is just a more compact way to write h.

Proposition 5.4. Setting ${\beta}(t)\,:\!={\lambda}^{-1}({\mathrm{e}}^{{\lambda}(T-t)}-1)$ , we have

\[ \begin{split} A&={\sigma}^2(1-\rho^2)\int_0^T\int_0^{\infty}{\beta}(t)^2[h^{\prime}({\alpha}(t)v)]^2v^3f_{V_t}(v) {\,\mathrm{d}} v\,{\,\mathrm{d}} t, \end{split} \]

where $f_{V_t}(v)$ is as in (5.7) and $h^{\prime}$ is from (5.11).

Proof. The process $(X,V,[X,X])^{\top}$ is a locally square-integrable semimartingale with absolutely continuous characteristics. The second differential characteristic c is given by $c_t^{11}=V_t$ , $c_t^{12}=\rho{\sigma} V_t^2$ , $c_t^{22}={\sigma}^2V_t^3$ , and $c_t^{i3}=0$ , $i=1,2,3$ . Since $[X,X]=\int_0^{\cdot} V_s{\,\mathrm{d}} s$ , we consider $H^0\in{\mathcal{H}}^2$ such that $H_t^0={\mathrm{E}}[\!\int_0^TV_s{\,\mathrm{d}} s\mid{\mathcal{F}}_t]-k$ a.s., $t\in[0,T]$ . By the Markov property of V and [7, Theorem 4], we get $H^0_t+k=f^0(t,X_t,V_t,[X,X]_t)$ , with $f^0(t,x,v,w)\,:\!=h({\beta}(t)v)+w$ , and moreover $h(y)\,:\!=\int_0^yh^{\prime}(x){\,\mathrm{d}} x$ . The function $f^0$ is twice continuously differentiable (see [7, Equation (81)]) and $\partial_{v}f^0(t,v,w)={\beta}(t)h^{\prime}({\beta}(t)v)$ . Therefore, by Corollary 3.1(ii), we obtain

\[ \begin{split} {{\left\langle {{L}^{0}},{{L}^{0}} \right\rangle }_{T={{\sigma }^{2}}(1-{{\rho }^{2}})}}\int_{0}^{T}{\beta }{{(t)}^{2}}{{[{{h}^{\prime }}(\beta (t){{V}_{t}})]}^{2}}V_{t}^{3}\text{d}t. \end{split} \]

Taking the expectation and using Fubini’s theorem yields the result.

We stress that according to Corollary 3.1(ii) we have $\vartheta^0_t=\rho{\sigma}{\beta}(t)h^{\prime}({\beta}(t)V_t)V_tS_t^{-1}$ .

As the next step, we compute the vector B. The explicit expression of ${\mathrm{E}}[{\mathrm{e}}^{zX_T}\mid V_t]$ can be obtained from [13, Proposition 4.1], with $a=0$ and $b=1$ (observe that $R_t=V_t^{-1}$ , where ${{({{R}_{t}})}_{t\in [0,T]}}$ denotes the volatility process in [13, Equation (2.2)]), and reads

(5.12) \begin{equation} \begin{split} {\mathrm{E}}[{\mathrm{e}}^{zX_t}\mid V_t]=\exp\bigg(zX_0-z\frac{{\lambda}\rho}{{\sigma}}\,t\bigg)\bigg(\frac{V_t}{V_0}\bigg)^{\frac{z\rho}{{\sigma}}}\;\frac{I_{\frac{2}{{\sigma}^2}\,\sqrt{B_z}}\left(\frac{4{\lambda}{\mathrm{e}}^{\frac{{\lambda}}{2}t}}{{\sigma}^2({\mathrm{e}}^{{\lambda} t}-1)\sqrt{V_0V_t}}\right)}{I_q\left(\frac{4{\lambda}{\mathrm{e}}^{\frac{{\lambda}}{2}t}}{{\sigma}^2({\mathrm{e}}^{{\lambda} t}-1)\sqrt{V_0V_t}}\right)}, \end{split} \label{eqn44} \end{equation}

where $ B_z\,:\!=(\kappa+\frac{{\sigma}^2}{2})^2+2{\sigma}^2[z(\frac{\rho}{{\sigma}}(\kappa+\frac{{\sigma}^2}{2})-\frac12)+\frac{z^2}{2}(1-\rho^2)]. $

Proposition 5.5. For $B=(B^1,\ldots,B^d)^{\top}$ , setting ${\beta}(t)={\lambda}^{-1}({\mathrm{e}}^{{\lambda}(T-t)}-1)$ ,

\[ B^{\kern1ptj}={\sigma}^2(1-\rho^2)\int_0^T\int_{{\mathcal{S}}^{\kern1ptj}} \int_0^{\infty} {\beta}(t)h^{\prime}({\beta}(t)v)\partial_vg(t,v,z)G(t,v,z)v^3f_{V_t}(v){\,\mathrm{d}} v\,{\zeta}^{\kern1ptj}({\mathrm{d}} z){\,\mathrm{d}} t, \]

where $G(t,v,z)\,:\!={\mathrm{E}}[{\mathrm{e}}^{zX_T}\mid V_t=v]$ is given by (5.12) with $V_t\rightarrow v$ , $\partial_vg(t,v,z)$ by (5.10), $f_{V_t}(v)$ by (5.7), and $z\in{\mathcal{S}}^{\kern1ptj}$ .

Proof. Let $(\vartheta^{\kern1ptj},L^{\kern1ptj})$ and $(\vartheta(z),L(z))$ be the GKW decompositions with respect to S of ${\eta}^{\kern1ptj}$ and $\skew3\widetilde H(z)$ , respectively, $j=0,\ldots,d$ , $z\in{\mathcal{S}}\,:\!=\bigcup_{j=1}^d{\mathcal{S}}^{\kern1ptj}$ . We now compute $\left\langle {{L}^{0}},L(z) \right\rangle$ , $z\in{\mathcal{S}}$ . Let $H^0$ and $f^0$ be as in the proof of Proposition 5.4. According to Corollary 3.1(ii) with $f^1=f^0$ and $f^2=f$ , $f(t,x,v,w,z)={\mathrm{e}}^{zx}g(t,v,z)$ , and g from (5.9), $\vartheta(z)=(zH(z)_t+{\sigma}\rho{\mathrm{e}}^{zX_t}\partial_v g(t,V_t)V_t)S_t^{-1}=\!:\,\widetilde\vartheta(z)$ $\text{P}\otimes \text{d}\left\langle S,S \right\rangle$ -a.e., $z\in{\mathcal{S}}$ . Thus, the restriction of $\widetilde\vartheta$ to ${\mathcal{S}}^{\kern1ptj}$ is a predictable function with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . Furthermore,

$$\text{d}{{\left\langle {{L}^{0}},L(z) \right\rangle }_{t}}={{\sigma }^{2}}(1-{{\rho }^{2}}){{h}^{\prime }}(\beta (t){{V}_{t}})\beta (t){{\partial }_{v}}g(t,{{V}_{t}},z)V_{t}^{3}{{\text{e}}^{z{{X}_{t}}}}\text{d}t=:\widetilde{K}{{(z)}_{t}}\text{d}t.$$

Since h is twice continuously differentiable and g is twice continuously differentiable in the second variable, $t\mapsto \widetilde K({\omega},z)_t$ is a continuous function. Because of (5.8) and $V_t>0$ , by Theorem 4.3, the assumptions of Theorem 4.1 are satisfied with respect to ${\mathcal{S}}^{\kern1ptj}$ , for each $j=1,\ldots,d$ , so we have $L^{\kern1ptj}_t=\int_{{\mathcal{S}}^{\kern1ptj}}\widetilde L(z){\zeta}^{\kern1ptj}({\mathrm{d}} z)$ , a.s., $t\in[0,T]$ . Since $\sup_{z\in{\mathcal{S}}^{\kern1ptj}}{\mathrm{E}}[|\widetilde L(z)_t|^2]<+\infty$ , $t\in[0,T]$ , $j=1,\ldots,d$ , from Proposition 4.1(iii), the identity ${\mathrm{E}}[\,\cdot\,]={\mathrm{E}}[{\mathrm{E}}[\,\cdot\mid V_t]]$ , and (5.12), we obtain the claim.

Proposition 5.6. For $C=(C^{ij})_{i,j=1,\ldots,d}$ we have

\[ \begin{split} C^{ij}= {\sigma}^2(1-\rho^2)\int_0^T\int_{{\mathcal{S}}^i}\int_{{\mathcal{S}}^{\kern1ptj}}\int_0^{\infty}\Big\{&\partial_vg(t,v,z_1)\partial_vg(t,v,z_2)\times G(t,v,z_1+z_2)v^3f_{V_t}(v)\Big\}\\&\times{\,\mathrm{d}} v\,{\zeta}^{\kern1ptj}({\mathrm{d}} z_2)\,{\zeta}^i({\mathrm{d}} z_1)\,{\,\mathrm{d}} t, \end{split} \]

where $\partial_vg$ is as in (5.10), G(t, v, z) as in Proposition 5.5, and $f_{V_t}(v)$ is from (5.7).

Proof. Let $(\vartheta^{\kern1ptj}, L^{\kern1ptj})$ and $(\vartheta(z), L(z))$ be the GKW decompositions with respect to S of ${\eta}^{\kern1ptj}$ and of H(z), $j=0,\ldots,d$ , $z\in{\mathcal{S}}\,:\!=\bigcup_{j=1}^d{\mathcal{S}}^{\kern1ptj}$ . By Corollary 3.1(ii) with $f^1(t,x,v,w,z)=f(t,x,v,w,z_1)$ , $z\in{\mathcal{S}}^ i$ , and $f^2(t,x,v,z)=f(t,x,v,w,z_2)$ , $z\in{\mathcal{S}}^{\kern1ptj}$ , $i,j=1,\ldots,d$ , $f(t,x,v,z)={\mathrm{e}}^{zx}g(t,v,z)$ , we get

$$ \text{d}{{\left\langle L({{z}_{1}}),L({{z}_{2}}) \right\rangle }_{t}}={{\sigma }^{2}}(1-{{\varrho }^{2}}){{\partial }_{v}}g(t,{{V}_{t}},{{z}_{1}}){{\partial }_{v}}g(t,{{V}_{t}},{{z}_{2}}){{\text{e}}^{({{z}_{1}}+{{z}_{2}}){{X}_{t}}}}V_{t}^{3}\text{d}t=:\widetilde{K}({{z}_{1}},{{z}_{2}})\text{d}t.$$

Proof. Let $(\vartheta^{\kern1ptj}, L^{\kern1ptj})$ and $(\vartheta(z), L(z))$ be the GKW decomposition with respect to S of ${\eta}^{\kern1ptj}$ and $\skew3\widetilde H(z)$ , respectively, $z\in{\mathcal{S}}\,:\!=\bigcup_{j=1}^d{\mathcal{S}}^{\kern1ptj}$ , $j=0,\ldots,d$ . We now compute $\left\langle {{L}^{0}},L(z) \right\rangle$ , $z\in{\mathcal{S}}$ . Let $H^0$ and $f^0$ be as in the proof of Proposition 5.1. According to Corollary 3.1 (applied now to the complex-valued square-integrable martingale $\skew3\widetilde H(z)$ and to $H^0$ ) with $f^1=f^0$ and $f^2=f$ , $f(t,x,v,w,z)=\exp(\phi_t(z,0)+\psi_t(z,0)v+zx)$ , we deduce that $\vartheta(z)_t=\frac{\skew3\widetilde H(z)_t}{S_t}(z+{\sigma}\rho\psi_{T-t}(z,0))=\!:\,\widetilde\vartheta(z)_t$ $\text{P}\otimes \text{d}\left\langle S,S \right\rangle$ -a.e., $z\in{\mathcal{S}}$ . Thus, the restriction of $\widetilde\vartheta$ to ${\mathcal{S}}^{\kern1ptj}$ is a predictable function with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . Moreover,

$$\text{d}{{\left\langle {{L}^{0}},L(z) \right\rangle }_{t}}={{\sigma }^{2}}(1-{{\rho }^{2}})\alpha (t){{\psi }_{T-t}}(z,0)\widetilde{H}{{(z)}_{t}}{{V}_{t}}\text{d}t.$$

Proof. We first show claim 1. By assumption (i) of the theorem, $(t,{\omega},z)\mapsto \skew3\widetilde H({\omega},z)_t$ is optional with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ , and $(t,\omega )\mapsto \text{ }\widetilde{H}{{(\omega ,z)}_{t}}$ is càdlàg, $z\in{\mathcal{S}}$ . By assumptions (i) and (ii), the function $(t,{\omega},z)\mapsto\int_E({\mathrm{e}}^{x_1}-1)W(t,{\omega},x,z)F_t({\omega},{\mathrm{d}} x)$ is predictable with respect to ${\mathcal{S}}^{\kern1ptj}$ , for every $j=1,\ldots,d$ , the process $\xi$ is predictable, and hence the restriction of $\widetilde\vartheta$ to ${\mathcal{S}}^{\kern1ptj}$ is predictable with respect to ${\mathcal{S}}^{\kern1ptj}$ , $j=1,\ldots,d$ . Now let $(\vartheta(z), L(z))$ be the GKW decomposition of $\skew3\widetilde H(z)$ with respect to S, where $z\in{\mathcal{S}}$ . From Corollary 3.1, we have $\vartheta(z)=\widetilde\vartheta(z)$ $\text{P}\otimes \text{d}\left\langle S,S \right\rangle$ -a.e. for every $z\in{\mathcal{S}}$ . This shows claim 1. To see claim 2, since ${{\left\langle L({{z}_{i}}),L({{z}_{j}}) \right\rangle }_{t}}=\int_{0}^{t}{\widetilde{K}{{(\omega ,{{z}_{i}},{{z}_{j}})}_{s}}\text{d}s}$ by Corollary 3.1, we use claim 1 and Theorem 4.2(iii) applied to ${\mathcal{S}}^i$ and ${\mathcal{S}}^{\kern1ptj}$ .

Proof. The process ${{(X,V,[X,X])}^{\top }}$ is a locally square-integrable semimartingale whose characteristics are absolutely continuous with respect to the Lebesgue measure. The second differential characteristic c is given by $c^{11}=V$ , $c^{12}=c^{21}=\rho{\sigma} V$ , $c^{22}={\sigma}^2V$ , and $c_t^{i3}=0$ , $i=1,2,3$ . Since V is a Cox–Ingersoll–Ross process, we have

Proof. Let $(\vartheta^{\kern1ptj}, L^{\kern1ptj})$ and $(\vartheta(z), L(z))$ be the GKW decomposition with respect to S of ${\eta}^{\kern1ptj}$ and of $\skew3\widetilde H(z)$ , respectively, $j=1,\ldots,d$ , $z\in{\mathcal{S}}\,:\!=\bigcup_{j=1}^d{\mathcal{S}}^{\kern1ptj}$ . By Corollary 3.1(ii) with $f^ 1(t,x,v,w,z)=f(t,x,v,w,z_1)$ , $z\in{\mathcal{S}}^i$ , and $f^2(t,x,v,w,z)=f(t,x,v,w,z_2)$ , $z\in{\mathcal{S}}^{\kern1ptj}$ , $f(t,x,v,w,z)=\exp(\phi_t(z,0)+\psi_t(z,0)v+zx)$ , we get

$$ \text{d}{{\left\langle L({{z}_{1}}),L({{z}_{2}}) \right\rangle }_{t}}={{\sigma }^{2}}(1-{{\varrho }^{2}}){{\partial }_{v}}g(t,{{V}_{t}},{{z}_{1}}){{\partial }_{v}}g(t,{{V}_{t}},{{z}_{2}}){{\text{e}}^{({{z}_{1}}+{{z}_{2}}){{X}_{t}}}}V_{t}^{3}\text{d}t=:\widetilde{K}({{z}_{1}},{{z}_{2}})\text{d}t.$$

Since $\widetilde K$ is ${\mathcal{P}}\otimes{\mathcal{B}}({\mathcal{S}}^i)\otimes{\mathcal{B}}({\mathcal{S}}^{\kern1ptj})$ -measurable, $t\mapsto\widetilde K({\omega},z_1,z_2)_t$ is continuous, and $V_t>0$ , the claim follows from Theorem 4.3 and (5.12).

5.3. An example with jumps

We now consider a special exponential Lévy model with unit jump sizes and compute the quantities A, B, and C for the variance swap.

Let $N^1$ and $N^2$ be independent Poisson processes. Let ${\lambda}_i$ be the parameter of $N^i$ , that is, ${{({{\bar{N}}^{i}})}_{t\in [0,T]}}$ is a martingale, $\overline N^i_t\,:\!=N^i_t-{\lambda}_it$ , $i=1,2$ . We now introduce the processes $M^1_t\,:\!=N^1_t-{\lambda}_1 t({\mathrm{e}}-1)$ and $M^2_t\,:\!=N^2_t+{\lambda}_2 t({\mathrm{e}}^{-1}-1)$ and consider the semimartingale $X=(X^1,X^2)\,:\!=(X^1_0+M^1-M^2,N^2)$ . Since $\Delta N^1\Delta N^2=0$ , the jump measure $\mu$ of X is given on $E=\{(-1,1),(1,0)\}$ by

\[ \mu({\omega},{\mathrm{d}} t,{\mathrm{d}} x_1,{\mathrm{d}} x_2)={\mathrm{d}} N^1_t({\omega})\delta_{(1,0)}({\mathrm{d}} x_1,{\mathrm{d}} x_2)+{\mathrm{d}} N^2_t({\omega})\delta_{(-1,1)}({\mathrm{d}} x_1,{\,\mathrm{d}} x_2), \]

$\delta_a$ denoting the Dirac measure concentrated at the point a. Hence, the predictable compensator $\nu$ of $\mu$ is

\[ \nu({\omega},{\mathrm{d}} t,{\mathrm{d}} x_1,{\mathrm{d}} x_2)= {\lambda}_1\delta_{(1,0)}({\mathrm{d}} x_1,{\mathrm{d}} x_2){\,\mathrm{d}} t+{\lambda}_2\delta_{(-1,1)}({\mathrm{d}} x_1,{\mathrm{d}} x_2){\,\mathrm{d}} t. \]

The third differential semimartingale characteristic of X is

\[F_t({\omega},{\mathrm{d}} x)=F({\mathrm{d}} x)={\lambda}_1\delta_{(1,0)}({\mathrm{d}} x_1,{\mathrm{d}} x_2)+{\lambda}_2\delta_{(-1,1)}({\mathrm{d}} x_1,{\mathrm{d}} x_2).\]

We consider the market model $S={\mathrm{e}}^{M^1-M^2}$ . Then, S is a square-integrable martingale. We now compute the quantities A, B, and C combining Corollary 3.1(i) and Theorem 4.3. The quantity $\xi_t$ in Corollary 3.1(i) fulfills $\xi_t=\xi=({\mathrm{e}}-1)^2({\lambda}_1+\frac{{\lambda}_2}{{\mathrm{e}}^2})>0$ . For the variance swap we now have ${\eta}^0+k=[X^1,X^1]_T=N^1_T+N^2_T$ and $k=({\lambda}_1+{\lambda}_2)T$ . We now define $f^0(t,x)\,:\!=x^1+2x^2+[{\lambda}_1({\mathrm{e}}-2)+{\lambda}_2({\mathrm{e}}^{-1}-2)] t$ . Then $H^0_t={\mathrm{E}}[{\eta}^0\mid{\mathcal{F}}_t]=f^0(t,X)$ holds. Therefore, Corollary 3.1(i) with $L^i=L^0$ and $W^i=W^0$ , $i=1,2$ , where we define $W^0\,:\!=f^0(t,x+X_{t-})-f^0(t,X_{t-})=x_1+2x_2$ , yields

$$A={{\left\langle {{L}^{0}},{{L}^{0}} \right\rangle }_{T}}=\left[ {{\lambda }_{1}}+{{\lambda }_{2}}-\frac{{{({{\lambda }_{1}}-\frac{{{\lambda }_{2}}}{\text{e}})}^{2}}}{{{\lambda }^{1}}+\frac{{{\lambda }^{2}}}{{{\text{e}}^{2}}}} \right]T.$$

We now compute B. Let ${\mathcal{S}}\,:\!=\bigcup_{j=1}^d{\mathcal{S}}^{\kern1ptj}$ . We observe that $X^1$ has exponential moments of every order. Furthermore, since $X^1$ has independent increments with respect to the filtration ${\mathbb{F}}$ , setting $q(z)\,:\!={\lambda}_1({\mathrm{e}}^z-1-z({\mathrm{e}}-1))+{\lambda}_2({\mathrm{e}}^{-z}-1-z({\mathrm{e}}^{-1}-1)), $ we get ${\mathrm{E}}[{\mathrm{e}}^{zX^1_T}\mid{\mathcal{F}}_t]=f(t,X_t,z)=\!:\,\skew3\widetilde H(z)_t$ , where $f(t,x,z)\,:\!={\mathrm{e}}^{zx_1+q(z)(T-t)}$ , $z\in{\mathbb{C}}$ . Let us consider the GKW decomposition $(\vartheta(z),L(z))$ of $\skew3\widetilde H(z)$ with respect to S, $z\in{\mathcal{S}}$ . We define $W(t,x,z)\,:\!=H(z)_{t-}({\mathrm{e}}^{zx_1}-1)$ . Then, by Corollary 3.1(i) applied to $L^1=L^0$ and $L^2=L(z)$ , with $W^1=W^0$ and $W^2=W$ , we get $\text{d }{{\left\langle {{L}^{0}},L(z) \right\rangle }_{t}}=Q(z)\widetilde{H}{{(z)}_{t-}}\text{d}t$ , where we define

\[ Q(z)\,:\!=({\mathrm{e}}^z-1)\bigg({\lambda}_1-\frac{{\lambda}_2}{{\mathrm{e}}^ z}-\frac{({\lambda}_1-\frac{{\lambda}_2}{{\mathrm{e}}})({\lambda}_1+\frac{{\lambda}_2}{{\mathrm{e}}^{z+1}})}{{\lambda}_1+\frac{{\lambda}_2}{{\mathrm{e}}^2}}\bigg).\]

Assumptions (ii) of Theorem 4.3 are satisfied, and hence the assumptions of Theorem 4.1 are satisfied. Thus, $L^{\kern1ptj}_t=\int_{{\mathcal{S}}^{\kern1ptj}}\widetilde L(z)_t{\zeta}^{\kern1ptj}({\mathrm{d}} z)$ , a.s., $t\in[0,T]$ , $j=1,\ldots,d$ . Since $\skew3\widetilde H(z)_-$ is locally bounded, it is integrable with respect to ${\mathrm{Leb.}}$ on [0,T]. So, Proposition 4.1(iii) yields $B^{\kern1ptj}=T\int_{{\mathcal{S}}^{\kern1ptj}}\exp(zX_0^1+q(z)T)Q(z){\zeta}^{\kern1ptj}({\mathrm{d}} z)$ .

We now conclude by computing the covariance matrix C. We consider ${\mathcal{S}}^i$ and ${\mathcal{S}}^{\kern1ptj}$ and define $f^1(t,x,z)\,:\!=f(t,x,z_1)$ , $z_1\in{\mathcal{S}}^ i$ , $f^2(t,x,z)\,:\!=f(t,x,z_2)$ , $z_2\in{\mathcal{S}}^{\kern1ptj}$ , and

\[ P(z_1,z_2)\,:\!=({\mathrm{e}}^{z_1}-1)({\mathrm{e}}^ {z_2}-1)\left(\!{\lambda}_1\!+\!\frac{{\lambda}_2}{\exp(z_1+z_2)}-\frac{\left({\lambda}_1+\frac{{\lambda}_2}{\exp({z_1+1})}\right)\left({\lambda}_1+\frac{{\lambda}_2}{\exp({z_2+1})}\right)}{{\lambda}_1+\frac{{\lambda}_2}{{\mathrm{e}}^2}}\!\right). \]

Then Corollary 3.1(i) yields $\text{d}{{\left\langle L({{z}_{1}}),L({{z}_{2}}) \right\rangle }_{t}}=P({{z}_{1}},{{z}_{2}})H{{({{z}_{1}})}_{t-}}H{{({{z}_{2}})}_{t-}}\text{d}t$ , $z_1\in{\mathcal{S}}^i$ , $z_2\in{\mathcal{S}}^{\kern1ptj}$ , and the density is locally bounded. From Theorem 4.3, we get

\[ \begin{split} C^{ij}=\int_0^T\int_{{\mathcal{S}}^i}\int_{{\mathcal{S}}^{\kern1ptj}}\Big\{\!&\exp((z_1+z_2)X^1_0+q(z_1+z_2)t+(q(z_1)+q(z_2))(T-t))\\ &\,\times P(z_1,z_2)\Big\}{\zeta}^{\kern1ptj}({\mathrm{d}} z_2){\zeta}^i({\mathrm{d}} z_1){\,\mathrm{d}} t. \end{split} \]