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Local risk-minimization for Barndorff-Nielsen and Shephard models

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Abstract

We obtain explicit representations of locally risk-minimizing strategies for call and put options in Barndorff-Nielsen and Shephard models, which are Ornstein–Uhlenbeck-type stochastic volatility models. Using Malliavin calculus for Lévy processes, Arai and Suzuki (Int. J. Financ. Eng. 2:1550015, 2015) obtained a formula for locally risk-minimizing strategies for Lévy markets under many additional conditions. Supposing mild conditions, we make sure that the Barndorff-Nielsen and Shephard models satisfy all the conditions imposed in (Arai and Suzuki in Int. J. Financ. Eng. 2:1550015, 2015). Among others, we investigate the Malliavin differentiability of the density of the minimal martingale measure. Moreover, we introduce some numerical experiments for locally risk-minimizing strategies.

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Acknowledgements

Takuji Arai gratefully acknowledges the financial support of Ishii Memorial Securities Research Promotion Foundation and Scientific Research (C) No. 15K04936 from the Ministry of Education, Culture, Sports, Science and Technology of Japan.

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Correspondence to Takuji Arai.

Appendix

Appendix

1.1 A.1 Theorem 3.7 of [3]

Theorem 3.7 of [3], which provides an explicit representation formula for LRM strategies in Lévy markets, is frequently referred to in this paper. Therefore, we give its statement for BNS models under Assumption 2.2. Note that although Assumption 2.1 of [3] is imposed in Theorem 3.7 of [3], it is satisfied under Assumption 2.2. For more details, see Remark 2.3.

Theorem A.1

(Theorem 3.7 of [3])

Let \(F\) be an \(L^{2}({\mathbb {P}})\) random variable satisfying the following three conditions:

AS1: (Assumption 2.6 in [3]) \(Z_{T}F\) is in \(L^{2}({\mathbb {P}})\).

AS2: (Assumption 3.4 in [3]) Conditions (C1)(C6) for \(F\) are satisfied.

AS3: ((3.1) in [3]) We have

$$ {\mathbb {E}}\left[\int_{0}^{T}\left((h^{0}_{t})^{2}+\int_{0}^{\infty}(h^{1}_{t,z})^{2}\nu(dz)\right)dt\right ]< \infty, $$

where \(h_{t,z}^{1}={\mathbb {E}}_{{\mathbb {P}}^{*}}[F(H^{*}_{t,z}-1)+zH^{*}_{t,z}D_{t,z}F|{\mathcal {F}} _{t-}]\) and

$$\begin{aligned} h^{0}_{t} &= {\mathbb {E}}_{{\mathbb {P}}^{*}}\bigg[D_{t,0}F-F\bigg(\int_{0}^{T}D_{t,0}u_{s}dW^{{\mathbb {P}}^{*}}_{s} \\ & \phantom{=:{\mathbb {E}}_{{\mathbb {P}}^{*}}\Bigg[D_{t,0}F-F\bigg(}+\int_{0}^{T}\int_{0}^{\infty}\frac{D_{t,0}\theta_{s,x}}{1-\theta_{s,x}}\widetilde{N}^{{\mathbb {P}}^{*}}(ds,dx)\bigg)\bigg|{\mathcal {F}}_{t-}\bigg]. \end{aligned}$$

Then the LRM strategy \(\xi^{F}\) for the claim \(F\) is given by

$$ \xi^{F}_{t}=\frac{1}{S_{t-}(\sigma^{2}_{t}+C_{\rho})}\left(h^{0}_{t}\sigma_{t}+\int _{0}^{\infty}h^{1}_{t,z}(e^{\rho z}-1)\nu(dz)\right). $$

1.2 A.2 Properties of \(\sigma_{t}\) and related Malliavin derivatives

The squared volatility process \((\sigma_{t}^{2})\), given as a solution to the SDE (1.2), is represented as

$$ \sigma_{t}^{2}=e^{-\lambda t}\sigma_{0}^{2}+\int_{0}^{t}e^{-\lambda(t-s)}dJ_{s}. $$
(A.1)

Remark that we have

$$ \sigma_{t}^{2}\geq e^{-\lambda t}\sigma_{0}^{2}\geq e^{-\lambda T}\sigma_{0}^{2} $$
(A.2)

and

$$ \int_{0}^{t}\sigma_{s}^{2}ds = \frac{1}{\lambda}(J_{t}-\sigma_{t}^{2}+\sigma_{0}^{2}) \leq\frac{1}{\lambda}(J_{t}+\sigma_{0}^{2}). $$
(A.3)

Next, we calculate some related Malliavin derivatives.

Lemma A.2

For any \(s\in[0,T]\), we have \(\sigma_{s}^{2}\in{\mathbb {D}}^{1,2}\) and

$$ D_{t,z}\sigma_{s}^{2}=e^{-\lambda(s-t)}{\mathbf{1}}_{[0,s]\times(0,\infty)}(t,z) $$
(A.4)

for \((t,z)\in[0,T]\times[0,\infty)\).

Proof

We can rewrite (A.1) as

$$\begin{aligned} \sigma_{s}^{2} &= e^{-\lambda s}\sigma_{0}^{2}+\int_{0}^{s}\int_{0}^{\infty}e^{-\lambda(s-u)}x\nu (dx)du \\ & \phantom{=:}+\int_{[0,T]\times[0,\infty)}e^{-\lambda(s-u)}{\mathbf{1}}_{[0,s]\times(0,\infty)}(u,x)Q(du,dx). \end{aligned}$$

Moreover, we have \(\int_{[0,T]\times[0,\infty)}e^{-2\lambda(s-u)}{\mathbf{1}}_{[0,s]\times(0,\infty)}(u,x)q(du,dx)<\infty\). The lemma follows by Definition 2.8. □

Lemma A.3

For any \(s\in[0,T]\), we have \(\sigma_{s}\in{\mathbb {D}}^{1,2}\) and

$$ D_{t,z}\sigma_{s}=\frac{\sqrt{\sigma_{s}^{2}+ze^{-\lambda(s-t)}}-\sigma _{s}}{z}{\mathbf{1}}_{[0,s]\times(0,\infty)}(t,z) $$

for \((t,z)\in[0,T]\times[0,\infty)\). Furthermore, we have \(0\leq D_{t,z}\sigma_{s}\leq\frac{1}{\sqrt{z}}{\mathbf{1}}_{[0,s]}(t)\) for \(z>0\).

Proof

Taking a \(C^{1}\)-function \(f\) such that \(f^{\prime}\) is bounded and \(f(r) = \sqrt{r}\) for \(r\geq e^{-\lambda T}\sigma_{0}^{2}\), we have \(\sigma_{s}=f(\sigma_{s}^{2})\) by (A.2). Proposition 2.6 in [22] implies that we have \(\sigma_{s}\in{\mathbb {D}} ^{1,2}\), \(D_{t,0}\sigma_{s}=f^{\prime}(\sigma_{s}^{2})D_{t,0}\sigma_{s}^{2}=0\), and

$$ D_{t,z}\sigma_{s} = \frac{f(\sigma_{s}^{2}+zD_{t,z}\sigma_{s}^{2})-f(\sigma_{s}^{2})}{z} = \frac{\sqrt{\sigma_{s}^{2}+ze^{-\lambda(s-t)}}-\sigma_{s}}{z}{\mathbf{1}}_{[0,s]}(t) $$

for \(z>0\) since \(D_{t,z}\sigma_{s}^{2}\) is nonnegative by (A.4). In addition, we have

$$ D_{t,z}\sigma_{s}\leq\frac{\sqrt{ze^{-\lambda(s-t)}}}{z}{\mathbf{1}}_{[0,s]}(t) \leq\frac{1}{\sqrt{z}}{\mathbf{1}}_{[0,s]}(t) $$

for \(z>0\). □

Lemma A.4

We have \(\int_{0}^{T}\sigma_{s}^{2}ds\in{\mathbb {D}}^{1,2}\) and

$$ D_{t,z}\int_{0}^{T}\sigma_{s}^{2}ds={\mathcal {B}}(T-t){\mathbf{1}}_{(0,\infty)}(z) $$

for \((t,z)\in[0,T]\times[0,\infty)\), where the functionis defined just before Assumption  2.2.

Proof

First, we have

$$\begin{aligned} \int_{0}^{T}\sigma_{s}^{2}ds &= \sigma_{0}^{2}\int_{0}^{T}e^{-\lambda s}ds+\int_{0}^{T}\int_{0}^{s}e^{-\lambda (s-u)}dJ_{u}ds \\ &= \sigma_{0}^{2}\frac{1-e^{-\lambda T}}{\lambda}+\int_{0}^{T}\int _{u}^{T}e^{-\lambda(s-u)}dsdJ_{u} \\ &= \sigma_{0}^{2}{\mathcal {B}}(T)+\int_{0}^{T}{\mathcal {B}}(T-u)dJ_{u}. \end{aligned}$$

From Definition 2.8 we obtain \(\int_{0}^{T}\sigma_{s}^{2}ds\in{\mathbb {D}}^{1,2}\) and

$$ D_{t,z}\int_{0}^{T}\sigma_{s}^{2}ds= {\mathcal {B}}(T-t){\mathbf{1}}_{(0,\infty)}(z) $$

for \((t,z)\in[0,T]\times(0,\infty)\). □

Lemma A.5

We have \(\int_{0}^{T}\sigma_{s}dW_{s}\in{\mathbb {D}}^{1,2}\) and

$$ D_{t,z}\int_{0}^{T}\sigma_{s}dW_{s}=\sigma_{t}{\mathbf{1}}_{\{0\}}(z)+\int_{t}^{T}\frac {\sqrt{\sigma_{s}^{2}+ze^{-\lambda(s-t)}}-\sigma_{s}}{z}dW_{s}{\mathbf{1}}_{(0,\infty)}(z) $$

for \((t,z)\in[0,T]\times[0,\infty)\).

Proof

To begin, we show that \(\sigma\in{\mathbb {L}}_{0}^{1,2}\). Lemma A.3 implies that \(\sigma_{s}\in{\mathbb {D}}^{1,2}\) for any \(s\in[0,T]\). We have \({\mathbb {E}}[\int_{0}^{T}\sigma_{s}^{2}ds]<\infty\) by (A.3) and the integrability of \(J_{T}\). Since \(|D_{t,z}\sigma_{s}|^{2}\le\frac{1}{z}\) by Lemma A.3, item (c) of the definition of \({\mathbb {L}}_{0}^{1,2}\) is satisfied. Hence, Lemma 3.3 in [10] provides \(\int_{0}^{T}\sigma_{s}dW_{s}\in{\mathbb {D}} ^{1,2}\) and

$$\begin{aligned} D_{t,z}\int_{0}^{T}\sigma_{s}dW_{s} &= D_{t,z}\int_{[0,T]\times[0,\infty)}\sigma_{s}{\mathbf{1}}_{\{0\} }(x)Q(ds,dx) \\ &= \sigma_{t}{\mathbf{1}}_{\{0\}}(z)+\int_{0}^{T}D_{t,z}\sigma_{s}dW_{s} \\ &= \sigma_{t}{\mathbf{1}}_{\{0\}}(z)+\int_{t}^{T}\frac{\sqrt{\sigma _{s}^{2}+ze^{-\lambda(s-t)}}-\sigma_{s}}{z}dW_{s}{\mathbf{1}}_{(0,\infty)}(z) \end{aligned}$$

for \((t,z)\in[0,T]\times[0,\infty)\) by Lemma A.3. □

Finally, we calculate \(D_{t,z}L_{T}\) as follows.

Proposition A.6

\(L_{T}\in{\mathbb {D}}^{1,2}\), and for \((t,z)\in[0,T]\times[0,\infty)\), we have

$$\begin{aligned} &D_{t,z}L_{T} \\ &=\sigma_{t}{\mathbf{1}}_{\{0\}}(z)+\bigg(-\frac{1}{2}{\mathcal {B}}(T-t)+\int_{t}^{T}\frac {\sqrt{\sigma_{s}^{2}+ze^{-\lambda(s-t)}}-\sigma_{s}}{z}dW_{s}+\rho\bigg){\mathbf{1}}_{(0,\infty)}(z). \end{aligned}$$

Proof

By (2.1) we have \(L_{T}=\mu T-\frac{1}{2}\int_{0}^{T}\sigma_{s}^{2}ds+\int _{0}^{T}\sigma_{s}dW_{s}+\rho J_{T}\). Since \(J_{T}\in{\mathbb {D}}^{1,2}\) and \(D_{t,z}J_{T}={\mathbf{1}}_{(0,\infty)}(z)\), we obtain the result by Lemmas A.4 and A.5. □

1.3 A.3 Properties of \(u_{s}\) and \(\theta_{s,x}\) and related Malliavin derivatives

We begin with recalling the two constants defined in (2.2):

$$ C_{u} := \max\bigg\{ \frac{|\alpha|e^{\frac{\lambda T}{2}}}{\sigma_{0}},\frac {|\alpha|}{C_{\rho}}\bigg\} \qquad\mbox{and}\qquad C_{\theta}:= \max\bigg\{ \frac{|\alpha|}{C_{\rho}},1\bigg\} . $$

The next lemma is cited often throughout the paper.

Lemma A.7

For any \(s\in[0,T]\) and any \(x\in(0,\infty)\), the following hold:

1. \(|u_{s}|\leq C_{u}\).

2. \(|\theta_{s,x}|\leq C_{\theta}\) and \(|\theta_{s,x}|\leq C_{\theta}(1-e^{\rho x})\leq C_{\theta}|\rho|x\).

3. \(\theta_{s,x}<1-e^{\rho x}\).

4. \(|\log(1-\theta_{s,x})|\leq C_{\theta}|\rho|x\).

5. \(\frac{1}{1-\theta_{s,x}}<\hat{C}_{\theta}\) for some \(\hat{C}_{\theta}>0\).

Proof

1. We have \(|u_{s}|\leq\frac{|\alpha|}{\sigma_{s}}\leq\frac{|\alpha|e^{\frac {\lambda T}{2}}}{\sigma_{0}}\) for any \(s\in[0,T]\) by (A.2).

2. \(|\theta_{s,x}|\leq\frac{|\alpha|}{C_{\rho}}(1-e^{\rho x})\leq C_{\theta}\) and \(1-e^{\rho x}\leq|\rho|x\) for any \(x>0\).

3. As seen in Remark 2.3, \(\frac{\alpha}{\sigma^{2}_{s}+C_{\rho}}>-1\) for any \(s\in[0,T]\). We have then \(\theta_{s,x}<1-e^{\rho x}\).

4. When \(\theta_{s,x}\geq0\), we have

$$ 0\geq\log(1-\theta_{s,x}) >\log\big(1-(1-e^{\rho x})\big)=\rho x\geq C_{\theta}\rho x. $$

On the other hand, if \(\theta_{s,x}< 0\), then \(0<\log(1-\theta _{s,x})\leq-\theta_{s,x}\leq C_{\theta}|\rho|x\).

5. If \(\theta_{s,x}\leq0\), then \(\frac{1}{1-\theta_{s,x}}\leq1\); otherwise, if \(\theta_{s,x}>0\), equivalently \(\alpha<0\), then

$$ 1-\theta_{s,x} = 1+\frac{\alpha}{\sigma^{2}_{s}+C_{\rho}}(1-e^{\rho x}) \geq 1+\frac{\alpha}{\sigma^{2}_{s}+C_{\rho}} \geq1+\frac{\alpha}{e^{-\lambda T}\sigma^{2}_{0}+C_{\rho}}>0 $$

by Assumption 2.2. This completes the proof. □

Next, we calculate some Malliavin derivatives related to \(u_{s}\) and \(\theta_{s,x}\).

Lemma A.8

For any \(s\in[0,T]\), we have \(u_{s}\in{\mathbb {D}}^{1,2}\) and

$$\begin{aligned} D_{t,z}u_{s} &= \frac{f_{u}(\sigma_{s}+zD_{t,z}\sigma_{s})-f_{u}(\sigma_{s})}{z}{\mathbf{1}}_{[0,s]\times(0,\infty)}(t,z) \\ &= \frac{f_{u}(\sqrt{\sigma^{2}_{s}+ze^{-\lambda(s-t)}})-f_{u}(\sigma _{s})}{z}{\mathbf{1}}_{[0,s]\times(0,\infty)}(t,z) \end{aligned}$$
(A.5)

for \((t,z)\in[0,T]\times[0,\infty)\), where \(f_{u}(r) := \frac{\alpha r}{r^{2}+C_{\rho}}\) for \(r\in{\mathbb {R}}\). Moreover, we have

$$ |D_{t,z}u_{s}|\leq\frac{C_{u}}{\sqrt{z}}{\mathbf{1}}_{[0,s]}(t) \quad\textit {and}\quad|D_{t,z}u_{s}|\leq\frac{C_{u}^{\prime}}{z}{\mathbf{1}}_{[0,s]}(t) $$

for some \(C_{u}^{\prime}>0\).

Proof

Note that \(f^{\prime}_{u}(r)=\alpha\frac{C_{\rho}-r^{2}}{(r^{2}+C_{\rho})^{2}}\) and \(|f^{\prime}_{u}(r)|\leq\frac{|\alpha|}{C_{\rho}}\leq C_{u}\). Since \(u_{s}=f_{u}(\sigma_{s})\) and \(\sigma_{s}\in{\mathbb {D}}^{1,2}\), Proposition 2.6 in [22], together with Lemma A.3, implies \(u_{s}\in {\mathbb {D}}^{1,2}\) and (A.5). In particular, we have \(D_{t,0}u_{s} = f^{\prime}_{u}(\sigma_{s})D_{t,0}\sigma _{s} = 0\). Further, Lemma A.3 again yields \(|D_{t,z}u_{s}|\leq\frac {1}{z}|zD_{t,z}\sigma_{s}|C_{u}\leq\frac{1}{\sqrt{z}}{\mathbf{1}}_{[0,s]}(t)C_{u}\). Moreover, since \(f_{u}(r)\) is bounded, we can find a \(C_{u}^{\prime}>0\) such that \(|D_{t,z}u_{s}|\leq\frac{C_{u}^{\prime}}{z}\). □

Lemma A.9

For any \((s,x)\in[0,T]\times(0,\infty)\), we have \(\theta_{s,x}\in{\mathbb {D}} ^{1,2}\) and

$$\begin{aligned} D_{t,z}\theta_{s,x} &= \frac{f_{\theta}(\sigma_{s}+zD_{t,z}\sigma_{s})-f_{\theta}(\sigma _{s})}{z}(e^{\rho x}-1){\mathbf{1}}_{[0,s]\times(0,\infty)}(t,z) \\ &= \frac{f_{\theta}(\sqrt{\sigma^{2}_{s}+ze^{-\lambda(s-t)}})-f_{\theta}(\sigma _{s})}{z}(e^{\rho x}-1){\mathbf{1}}_{[0,s]\times(0,\infty)}(t,z) \end{aligned}$$
(A.6)

for \((t,z)\in[0,T]\times[0,\infty)\), where \(f_{\theta}(r) := \frac{\alpha }{r^{2}+C_{\rho}}\) for \(r\in{\mathbb {R}}\). Moreover, we have

$$ |D_{t,z}\theta_{s,x}|\leq\frac{C^{\prime}_{\theta}}{\sqrt{z}}(1-e^{\rho x}){\mathbf{1}}_{[0,s]}(t) \ \ \textit{ and } \ \ |D_{t,z}\theta_{s,x}|\leq\frac{2C_{\theta}}{z}(1-e^{\rho x}){\mathbf{1}}_{[0,s]}(t) $$
(A.7)

for some \(C_{\theta}^{\prime}>0\).

Proof

Note that \(\theta_{s,x}=f_{\theta}(\sigma_{s})(e^{\rho x}-1)\) and \(f^{\prime}_{\theta}(r)=-\frac{2\alpha r}{(r^{2}+C_{\rho})^{2}}\). Hence, \(|f^{\prime}_{\theta}(r)|\) is bounded. Therefore, the same argument as for Lemma A.8 implies (A.6). In addition, (A.7) is given by the boundedness of \(f_{\theta}\) and \(f_{\theta}^{\prime}\). □

Lemma A.10

For any \((s,x)\in[0,T]\times(0,\infty)\), we have \(\log(1-\theta _{s,x})\in{\mathbb {D}}^{1,2}\) and

$$ D_{t,z}\log(1-\theta_{s,x})=\frac{\log(1-\theta_{s,x}-zD_{t,z}\theta _{s,x})-\log(1-\theta_{s,x})}{z}{\mathbf{1}}_{(0,\infty)}(z) $$

for \((t,z)\in[0,T]\times[0,\infty)\). Moreover, we have

$$ |D_{t,z}\log(1-\theta_{s,x})|\leq|D_{t,z}\theta_{s,x}|e^{-\rho x}. $$

Proof

For \(x>0\), we denote

$$\begin{aligned} g_{x}(r) := \left\{ \textstyle\begin{array}{ll} \log(1-r), &\quad r< 1-e^{\rho x}, \\ -e^{-\rho x}r+e^{-\rho x}-1+\rho x, &\quad r\geq1-e^{\rho x}. \end{array}\displaystyle \right. \end{aligned}$$

Note that \(g_{x}\) is a \(C^{1}\)-function satisfying \(|g^{\prime}_{x}(r)|\leq e^{-\rho x}\) for all \(r\in{\mathbb {R}}\). Because \(\theta_{s,x}\in{\mathbb {D}}^{1,2}\) and \(\log(1-\theta_{s,x})=g_{x}(\theta _{s,x})\) by item 3 of Lemma A.7, we have

$$ D_{t,z}\log(1-\theta_{s,x})=\frac{g_{x}(\theta_{s,x}+zD_{t,z}\theta _{s,x})-g_{x}(\theta_{s,x})}{z}{\mathbf{1}}_{(0,\infty)}(z). $$

Lemma A.9 implies, for \(t\in[0,s]\) and \(z\in(0,\infty)\),

$$\begin{aligned} \theta_{s,x}+zD_{t,z}\theta_{s,x} &= f_{\theta}\big(\sqrt{\sigma^{2}+ze^{-\lambda(s-t)}}\big)(e^{\rho x}-1) \\ &= \frac{\alpha(e^{\rho x}-1)}{\sigma^{2}_{s}+ze^{-\lambda(s-t)}+C_{\rho}} < 1-e^{\rho x}. \end{aligned}$$
(A.8)

We have then \(g_{x}(\theta_{s,x}+zD_{t,z}\theta_{s,x})=\log(1-\theta _{s,x}-zD_{t,z}\theta_{s,x})\). □

1.4 A.4 On \(D_{t,z}\log Z_{T}\)

We show that \(\log Z_{T}\in{\mathbb {D}}^{1,2}\) and calculate \(D_{t,z}\log Z_{T}\). Equality (2.5) implies that

$$\begin{aligned} \log Z_{T} &= -\int_{0}^{T}u_{s}dW_{s}-\frac{1}{2}\int_{0}^{T}u_{s}^{2}ds+\int_{0}^{T}\int_{0}^{\infty}\log (1-\theta_{s,x})\widetilde{N}(ds,dx) \\ & \phantom{=:}+\int_{0}^{T}\int_{0}^{\infty}\big(\log(1-\theta_{s,x})+\theta _{s,x}\big)\nu(dx)ds. \end{aligned}$$
(A.9)

We discuss each term of (A.9) separately. As seen in Sect. 4, we have \(u\in{\mathbb {L}}_{0}^{1,2}\). Therefore, Lemma 3.3 of [10] implies that

$$ D_{t,0}\int_{0}^{T}u_{s}dW_{s}=u_{t}+\int_{0}^{T}D_{t,0}u_{s}dW_{s}=u_{t} $$

and \(D_{t,z}\int_{0}^{T}u_{s}dW_{s}=\int_{0}^{T}D_{t,z}u_{s}dW_{s}\) for \(z>0\). Similarly, we have

$$ D_{t,0}\int_{0}^{T}\int_{0}^{\infty}\log(1-\theta_{s,x})\widetilde{N}(ds,dx)=0 $$

and

$$\begin{aligned} {D_{t,z}\int_{0}^{T}\int_{0}^{\infty}\log(1-\theta_{s,x})\widetilde{N}(ds,dx)} &= \frac{\log(1-\theta_{t,z})}{z}\\ &\phantom{=:}+\int_{0}^{T}\int_{0}^{\infty}D_{t,z}\log(1-\theta_{s,x})\widetilde{N}(ds,dx) \end{aligned}$$

for \(z>0\). As for \(D_{t,z}\int_{0}^{T}u_{s}^{2}ds\), because \(u^{2}\in{\mathbb {L}}_{0}^{1,2}\) by Sect. 4, Lemma 3.2 of [10] yields

$$ D_{t,z}\int_{0}^{T}u_{s}^{2}ds=\int_{0}^{T}D_{t,z}u_{s}^{2}ds=2\int _{0}^{T}u_{s}D_{t,z}u_{s}ds+z\int_{0}^{T}(D_{t,z}u_{s})^{2}ds $$

for \(z\geq0\). In particular, \(D_{t,0}\int_{0}^{T}u_{s}^{2}ds=0\). For the fourth term of (A.9), because \(\log(1-\theta)+\theta\in\widetilde{{\mathbb {L}} }_{1}^{1,2}\), Proposition 3.5 of [22] implies

$$\begin{aligned} &D_{t,z}\int_{0}^{T}\int_{0}^{\infty}\big(\log(1-\theta_{s,x})+\theta _{s,x}\big)\nu(dx)ds \\ &= \int_{0}^{T}\int_{0}^{\infty}\big(D_{t,z}\log(1-\theta_{s,x})+D_{t,z}\theta _{s,x}\big)\nu(dx)ds \end{aligned}$$

for \(z\geq0\). Collectively, we conclude the following:

Proposition A.11

We have \(\log Z_{T}\in{\mathbb {D}}^{1,2}\), \(D_{t,0}\log Z_{T}=u_{t}\), and

$$\begin{aligned} D_{t,z}\log Z_{T} &= -\int_{0}^{T}D_{t,z}u_{s}dW_{s}-\int_{0}^{T}u_{s}D_{t,z}u_{s}ds-\frac{z}{2}\int _{0}^{T}(D_{t,z}u_{s})^{2}ds \\ & \phantom{=:}+\int_{0}^{T}\int_{0}^{\infty}D_{t,z}\log(1-\theta_{s,x})\widetilde{N} (ds,dx) \\ & \phantom{=:}+\int_{0}^{T}\int_{0}^{\infty}\big(D_{t,z}\log(1-\theta _{s,x})+D_{t,z}\theta_{s,x}\big)\nu(dx)ds+\frac{\log(1-\theta_{t,z})}{z} \end{aligned}$$

for \(z>0\).

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Arai, T., Imai, Y. & Suzuki, R. Local risk-minimization for Barndorff-Nielsen and Shephard models. Finance Stoch 21, 551–592 (2017). https://doi.org/10.1007/s00780-017-0324-8

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