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Utility maximization of the exponential Lévy switching models Yu. Donga, L. Vostrikovab a National University of Singapore, Institute of Operations Research and Analytics, Singapore b LAREMA, Département de Mathématiques, Université d'Angers, France
Англоязычная версия:
Theory of Probability and its Applications, 2024, Volume 69, Issue 1, Pages DOI: https://doi.org/10.1137/S0040585X97T991799 
Реферативные базы данных:
  1. IntroductionRegime switching processes are processes whose parameters depend on some Markov process, taking a finite number of values. Those processes have been applied successfully since more than 20 years to model the behavior of economic time series and financial time series. Hamilton and al. in [19] considered regime switching autoregressive conditional heteroskedastic models and proposed a way for their calibration. Cai in [2] considered regime switching ARCH models. So and al. in [29] considered the so-called Markov switching volatility models to capture the changing of the behavior of the volatility due to the economic factors. Lévy switching processes was widely used for pricing due to their flexibility to reflect the real data and their commodity as mathematical tool. For two state variance gamma model, Konikov and Madan in [27] have developed a method based on fast Fourier transform to price the vanilla options. Later, using the implementation of numerical methods of resolving PDE equations, Chourdakis [9] has priced the exotic contracts, like barrier, Bermuda and American options. Then, Elliott and Osakwe in [10] used multi-state pure jump processes to price vanilla derivatives. Later, Elliott and al. in [11] considered Heston Markov switching model to price variance swap and volatility swap with probabilistic and PDE approaches. Hainaut in [20] gave an overview of the tools related to Lévy switching models, like moments and the conditions under which such processes are martingales, the measure transforms preserving Lévy switching structure, and also the questions of option pricing and calibration. The question of optimal hedging in the discrete time methodology that minimizes the expected value of a given penalty function of the hedging error was studied by So and al. in [29]. Francois and al. in [15] considered the question of the portfolio optimization for affine models with Markov switching. Escobar and al. in [12] derived the optimal investment strategies for the expected utility maximization of terminal wealth. The dual method for utility maximization has been deeply studied by Goll and Ruschendorf [17] in general semi-martingale setting. This method permits to find a semi-explicit NA condition as well as a semi-explicit expression of the optimal (asymptotically optimal) strategy for utility maximization. This method was applied for pricing and hedging of exponential Lévy models by Miyahara [28], Fujiwara and Miyahara [16], Chouli and al. [7], [8], Essche and Schweizer[13], Hubalek and Sgarra [21], Jeanblanc and al. [25], Cawston and Vostrikova [4], [5]. This method was also efficient to determine the indifference prices in the papers by Ellanskaya and Vostrikova [14] and for the problem of utility maximization of change-point exponential Lévy models in the paper by Cawston and Vostrikova [6]. We notice that exponential Lévy switching models are the natural generalizations of exponential change point models. This is the reason to choose the dual method for utility maximization of the exponential Lévy switching models. To introduce a Lévy switching model we consider $N$ independent Lévy processes $X^{(j)}=(X_t^{(j)})_{0\leqslant t\leqslant T}$, $j=1,\dots,N$, on the interval $[0,T]$, with the values in $\mathbf R^d$ and starting from $0$. We assume that these processes are defined on a probability space $(\Omega_1,\mathcal G,P_1)$, where $(\Omega_1,\mathcal G)=(D(\mathbf{R}^{d}, [0,T]), \mathcal{D}(\mathbf{R}^{d}, [0,T]))$ is the Skorohod space of càdlàg functions equipped with the natural filtration $\mathbb G=(\mathcal G_t)_{0\leqslant t\leqslant T}$ satisfying usual conditions. As known, each Lévy process $X^{(j)}$ can be characterised by the characteristic triplet denoted by $(b^{(j)},c^{(j)},\nu^{(j)})$, where $b^{(j)}$ is the drift parameter, $c^{(j)}$ is the density of quadratic variation of its continuous martingale part with respect to Lebesgue measure and $\nu^{(j)}$ is a Lévy measure which satisfies the usual condition
$$
\begin{equation*}
\int_{\mathbf R^d}(\|x\|^2\land1)\,\nu^{(j)}(dx)<+\infty
\end{equation*}
\notag
$$
with $\|\,{\cdot}\,\|$ the Euclidean norm in $\mathbf{R}^d$. As known, the law of $X^{(j)}$ is entirely characterized by the characteristic function $\varphi_t^{(j)}$ of $X_t^{(j)}$ at $t>0$, which is, for $\lambda \in \mathbf R^d$, given by the formula
$$
\begin{equation*}
\varphi_t^{(j)}(\lambda)=\exp\{t\psi^{(j)}(\lambda)\},
\end{equation*}
\notag
$$
where $\psi^{(j)}$ is the characteristic exponent of $X^{(j)}$. We recall that $\psi^{(j)}$ is expressed via the Lévy–Khinchin formula as follows:
$$
\begin{equation*}
\psi^{(j)}(\lambda)=i\langle\lambda,b^{(j)}\rangle -\frac{1}{2}\langle\lambda,c^{(j)}\lambda\rangle+\int_{\mathbb R^d}\bigl(e^{i\langle\lambda,x\rangle}-1-i \langle\lambda,x\rangle \mathbf I_{\{\|x\|\leqslant 1\}} \bigr) \, \nu^{(j)}(dx),
\end{equation*}
\notag
$$
where $\mathbf I(\,{\cdot}\,)$ denotes the indicator function and $\langle \,{\cdot}\,, {\cdot}\, \rangle$ is the scalar product in $\mathbf{R}^d$. Let also $\alpha=(\alpha_t)_{{0\leqslant t\leqslant T}}$ be a homogeneous Markov process with the values in the set $\{1,\dots,N\}$, given on a probability space $(\Omega_2,\mathcal H,P_2)$, where $(\Omega_2,\mathcal H) = (D(\mathbf{R}, [0,T]), \mathcal{D}(\mathbf{R}, [0,T]))$ is the Skorohod space of càdlàg functions equipped with the natural filtration $\mathbb H=(\mathcal H_t)_{t \geqslant 0}$ satisfying usual properties. The process $\alpha$ is supposed to be independent of the Lévy processes $X^{(1)},\dots,X^{(N)}$. Then, on the product space $(\Omega,\mathcal F, P)=(\Omega_1\times\Omega_2,\mathcal G \otimes \mathcal H,P_1\times P_2)$ we can define two filtrations. One is the progressively enlarged filtration $\mathbb F=(\mathcal F_t)_{0\leqslant t\leqslant T}$, where for $t<T$
$$
\begin{equation*}
\mathcal F_t=\bigcap_{s>t}\mathcal G_s \otimes \mathcal H_s \quad\text{and}\quad \mathcal{F}_T=\mathcal G_T \otimes \mathcal H_T
\end{equation*}
\notag
$$
and the second one is the initially enlarged filtration $\widehat{\mathbb F}=(\widehat{\mathcal F}_t)_{0\leqslant t\leqslant T}$ defined for $t<T$ as
$$
\begin{equation*}
\widehat{\mathcal F}_t=\bigcap_{s>t}\mathcal G_s \otimes \mathcal H_T \quad\text{and}\quad \widehat{\mathcal{F}}_T=\mathcal G_T \otimes \mathcal H_T.
\end{equation*}
\notag
$$
For more information about these filtrations see [23], [1], [3], [18]. For technical reasons we will work in the initially enlarged filtration, and surprisingly at the first glance, we will obtain the results in the progressively enlarged filtration as for the minimal martingale measures also for the optimal strategies for utility maximization. Such phenomenon can be explained by the fact that firstly both filtrations coincide at the time $T$, and secondly, by the fact that the Lévy processes $X^{(j)}$, $j=1,\dots, N$, remain independent of the Markov process $\alpha$ under any $f$-divergence minimal martingale measure (cf. [5]). On the defined above product space we now introduce a Lévy switching process $X$ such that the increments of this process coincide with the increments of Lévy process $X^{(j)}$ when the process $\alpha$ states in the state $j$:
$$
\begin{equation}
dX_t=\sum_{j=1}^N dX_t^{(j)}\mathbf I_{\{\alpha_{t-}=j\}}\quad\text{or, equivalently},\quad X_t=\sum_{j=1}^N \int_0^t \mathbf I_{\{\alpha_{s-}=j\}}dX_s^{(j)}.
\end{equation}
\tag{1}
$$
More explicitly, if, for example, at $t=0$, $\alpha_0=i_0$ and $\tau_1$ is the first time change of the state $i_0$ into another state $i_1$ of the Markov process $\alpha$ and $\tau_k=\inf\{t>\tau_{k-1}\,|\, \alpha_t \neq i_{k-1}\}$ for $k \geqslant 2$, then
$$
\begin{equation*}
X_t=\begin{cases} X_t^{(i_0)}, &\text{for } t < \tau_1, \\ X_{\tau_1}^{(i_0)}+X_t^{(i_1)}-X_{\tau_1}^{(i_1)}, &\text{for } \tau_1 \leqslant t < \tau_2, \\ \dots \\ X_{\tau_n}^{(i_{n-1})}+X_t^{(i_n)}-X_{\tau_{n}}^{(i_n)}, &\text{for } \tau_n \leqslant t < \tau_{n+1}, \\ \dots\,. \end{cases}
\end{equation*}
\notag
$$
The characteristic function of $X$ can be easily found, since $X$ is a process with conditionally to the process $\alpha$ independent increments. Due to mutual independence of the Lévy processes $X^{(1)}, \dots, X^{(N)}$ and the Markov process $\alpha$, we have
$$
\begin{equation*}
\begin{aligned} \, \mathbf{E}_P\bigl(\exp(i\langle\lambda,X_t\rangle)\bigr)&= \mathbf{E}_P\bigl(\mathbf{E}_P\bigl(\exp(i\langle\lambda,X_t\rangle)\bigm|\alpha\bigr)\bigr) \\ &=\mathbf{E}_P\biggl(\prod_{j=1}^N\mathbf{E}_P\biggl(\exp\biggl(i \biggl\langle\lambda,\int_0^t \mathbf I_{\{\alpha_{s-}=j\}}\, dX_s^{(j)}\biggr\rangle\biggr)\biggm| \alpha\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
We remark that for any real-valued deterministic function $q=(q_s)_{s\geqslant0}$ such that $\int_0^tq_s\, dX_s^{(j)}$ exists, the characteristic function of $\int_0^tq_s\, dX_s^{(j)}$ is expressed as follows:
$$
\begin{equation*}
\begin{aligned} \, &\mathbf{E}_P\biggl(\exp\biggl(i\biggl\langle\lambda,\int_0^tq_s\, dX_s^{(j)}\biggr\rangle\biggr)\biggr) \\ &\qquad=\exp\biggl(i\langle\lambda,b^{(j)} \rangle\int_0^tq_s\, ds -\frac{1}{2}\langle \lambda, c^{(j)}\lambda \rangle\int_0^tq_s^2\, ds \\ &\qquad\qquad+\int_0^t\int_{\mathbf R^d}\bigl(e^{i\langle\lambda q_s,x\rangle}-1-i\langle\lambda q_s,x\rangle \mathbf I_{\{\|x\|\leqslant 1\}}\bigr) \nu^{(j)}(dx)\,ds\biggl). \end{aligned}
\end{equation*}
\notag
$$
If $q_s=\mathbf I_{\{\alpha_{s-}=j\}}$ for $s \geqslant 0$, then $q_s^2=q_s$. Moreover, considering two cases for the values of the integral with respect to the compensator $\nu^{(j)}$, namely, $q_s=\mathbf I_{\{\alpha_{s-}=j\}}=1$ and $q_s=\mathbf I_{\{\alpha_{s-}=j\}}=0$, we find finally that
$$
\begin{equation*}
\mathbf{E}_P\bigl(\exp(i\langle\lambda,X_t\rangle)\bigr) =\mathbf{E}_P\biggl(\exp\biggl( \sum_{j=1}^N \psi^{(j)}(\lambda)\int_0^t \mathbf I_{\{\alpha_{s-}=j\}}\, ds\biggr) \biggr).
\end{equation*}
\notag
$$
In this paper we consider the problem of utility maximization of the price process $S=(S_t)_{t\in [0,T]}$ such that the components of $S$ denoted by $S^{(k)}$, $1\leqslant k \leqslant d$, are Doléans–Dade exponentials of the corresponding components of $X$. More precisely, if we denote by $\chi^{(k)}$, $1\leqslant k\leqslant d$, the components of $X$, then for all $t\in [0,T]$ the components of $S^{(k)}$ satisfy where $\mathcal{E}(\,{\cdot}\,)$ is Doléans-Dade exponential. It is known that one can rewrite each component of the process $S$ as the usual exponential of an another Lévy switching process. The advantage of use of the stochastic exponent is that the set of all equivalent martingale measures of $X$ coincide with the one of the process $S$.As utility functions, we consider HARA utilities, which are logarithmic, power and exponential utilities, defined as
$$
\begin{equation*}
\begin{aligned} \, u(x) &=\ln x \text{ with } x >0, \\ u(x) &=\frac{x^p}{p}\text{ with } x>0 \text{ and } p \in (-\infty,0)\cup(0,1), \\ u(x) &=1-\exp(-x) \text{ with } x\in \mathbf R. \end{aligned}
\end{equation*}
\notag
$$
Let us denote by $\mathcal A$ the set of all self-financing admissible strategies. We recall that an admissible strategy is a predictable process $\Phi=(\eta,\phi)$ taking the values in $\mathbf R^{d+1}$, where $\eta$ represents the quantity invested in the risk-free asset $B$ and $\phi=(\phi^{(1)},\dots,\phi^{(d)})$ contains the quantities invested in the risky assets $S^{(1)},\dots,S^{(d)}$, respectively, and it holds: $\eta$ is $B$-integrable and there exists $ a \in \mathbf R^+$ such that, for $t\in[0,T]$,
$$
\begin{equation*}
\sum_{k=1}^d\int_0^t\phi_s^{(k)}\, dS_s^{(k)}\geqslant -a.
\end{equation*}
\notag
$$
In what follows, we suppose as usual, that the interest rate of risk-free asset $B$ is equal to zero. We recall that a strategy $\widehat \phi \in \mathcal A$, $\widehat \phi= (\widehat \phi^{(1)}, \dots, \widehat \phi^{(d)})$, is said to be $u$-optimal on $[0,T]$ if
$$
\begin{equation*}
\begin{aligned} \, &\mathbf{E}_{\mathbb P}\biggl(u\biggl(x_0+\sum_{k=1}^d\int_0^T\widehat{\phi}_s^{(k)}\, dS_s^{(k)}\biggr) \biggr) \\ &\qquad= \sup_{\phi \in \mathcal A} \mathbf{E}_{\mathbb P}\biggl(u\biggl(x_0+ \sum_{k=1}^d\int_0^T\phi_s^{(k)}\, dS_s^{(k)}\biggr)\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $x_0>0$ is an initial capital. A sequence of admissible strategies $(\widehat \phi^n)_{n\geqslant1}$ with $\widehat \phi^n = (\widehat \phi^{n,(1)}, \dots, \widehat \phi^{n,(d)})$ is said to be asymptotically $u$-optimal on $[0,T]$ if
$$
\begin{equation*}
\begin{aligned} \, &\lim_{n \to \infty}\mathbf{E}_{\mathbb P} \biggl(u\biggl(x_0+\sum_{k=1}^d \int_0^T\widehat{\phi}_s^{n,(k)} \, dS_s^{(k)}\biggr)\biggr) \\ &\qquad=\sup_{\phi \in \mathcal A} \mathbf{E}_{\mathbb P}\biggl(u\biggl(x_0+\sum_{k=1}^d\int_0^T\phi_s^{(k)}\, dS_s^{(k)}\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
The paper is organized in the following way. In Section 2 a description of all equivalent martingale measures for Lévy switching processes is given. In Section 3 $f$-divergence minimal martingale measures for Lévy switching processes and HARA utilities are derived (see Theorem 1). In Section 4 the known result about the optimal strategies is recalled (see Proposition 6). Then, in Propositions 7–9 of Section 4 the expressions of the optimal strategies in the progressively enlarged filtration are given and the values of the maximal expected utilities are calculated. In Section 5 we apply our result to the Brownian switching model. 2. Equivalent martingale measures of exponential Lévy switching processesSince the components $S^{(k)},1\leqslant k \leqslant d$, of the price process $S$ are expressed via the stochastic exponentials of the corresponding components $\chi^{(k)}$ of the process $X$, the set of all equivalent martingale measures of the process $(X, \alpha)$ coincides with the set of all equivalent martingale measures of the process $(S,\alpha)$. For this reason we will describe now the set of all equivalent martingale measures of the $(d+1)$-dimensional process $(X, \alpha)$. Let $\mathbb{P}_T$ be the law of $(X,\alpha)$ on the Skorohod space $(D(\mathbf{R}^{d+1}, [0,T]), \mathcal{D}(\mathbf{R}^{d+1}, [0,T]))$ and let $\mathbf{P}_T^X$ and $\mathbf{P}_T^{\alpha}$ be the laws of $X$ and $\alpha$, respectively, on the time interval $[0,T]$. We denote also by $ \mathbf{P}_T^X(\,{\cdot}\,|\, \alpha)$ the regular conditional law of $X$ given $\alpha$. Proposition 1. The law $\mathbb{Q}_T$ is absolutely continuous with respect to (w.r.t.) $\mathbb{P}_T$, i.e., $\mathbb{Q}_T\ll \mathbb{ P}_T$ if and only if there exists a regular conditional law of $X$, denoted by $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)$ such that $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)\ll \mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$ ($\mathbf{P}^{\alpha}_T$-a.s.), and the law of $\alpha$, denoted $\mathbf{Q}_T^{\alpha}$, such that $\mathbf{Q}^{\alpha}_T\ll \mathbf{P}^{\alpha}_T$. Moreover, $\mathbb{P}_T$-a.s. Proof. Let $\mathbb{Q}_T\ll \mathbb{ P}_T$. For all $A\in \mathcal{D}(\mathbf{R}^d,[0,T])$ and $B\in \mathcal{D}(\mathbf{R},[0,T])$,
$$
\begin{equation*}
\mathbb{P}_T(A\times B) = \int_B \mathbf{P}^X_T(A\,|\,\alpha=y)\,d\mathbf{P}_T^{\alpha}(y) = \int_{A\times B} d\mathbf{P}^X_T(x\,|\,\alpha=y)\,d\mathbf{P}_T^{\alpha}(y)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathbb{Q}_T(A\times B) = \int_B \mathbf{Q}^X_T(A\,|\,\alpha=y)\,d\mathbf{Q}_T^{\alpha}(y) = \int_{A\times B} d\mathbf{Q}^X_T(x\,|\,\alpha=y)\,d\mathbf{Q}_T^{\alpha}(y).
\end{equation*}
\notag
$$
At the same time, the condition $\mathbb{Q}_T\ll \mathbb{P}_T$ is equivalent to the existence of a density $f(x,y)$ such that
$$
\begin{equation*}
\mathbb{Q}_T(A\times B) = \int_{A\times B}f(x,y)\,d\mathbb{P}_T(x,y) = \int_{A\times B}f(x,y)\, d\mathbf{P}^X_T(x\,|\,\alpha=y)\,d\mathbf{P}_T^{\alpha}(y).
\end{equation*}
\notag
$$
Further, obviously $\mathbb{Q}_T\ll \mathbb{P}_T$ implies the condition $\mathbf{Q}^{\alpha}_T\ll \mathbf{P}^{\alpha}_T$ and
$$
\begin{equation*}
\mathbb{Q}_T(A\times B) = \int_{A\times B} d\mathbf{Q}^X_T(x\,|\,\alpha=y)\,\xi_T^{(\alpha)}(y)\,d\mathbf{P}_T^{\alpha}(y),
\end{equation*}
\notag
$$
where $\xi_T^{(\alpha)}=d\mathbf{Q}_T^{\alpha}/d\mathbf{P}_T^{\alpha}$. Since these relations hold for all $A\in \mathcal{D}(\mathbf{R}^d,[0,T])$ and all $B\in \mathcal{D}(\mathbf{R},[0,T])$ we get that $\mathbb{P}_T$-a.s.
$$
\begin{equation*}
\frac{d\mathbb{Q}_T}{d\mathbb{P}_T}(x,y)=f(x,y) = \frac{d\mathbf{Q}^X_T(x\,|\,\alpha=y)}{d\mathbf{P}^X_T(x\,|\,\alpha=y)}\,\xi_T^{(\alpha)}(y).
\end{equation*}
\notag
$$
Conversely, if $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)\ll \mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$ ($\mathbf{P}^{\alpha}_T$-a.s.) and $\mathbf{Q}^{\alpha}_T\ll \mathbf{P}^{\alpha}_T$, then, evidently, $\mathbb{P}_T(A\times B)=0$ implies $\mathbb{Q}_T(A\times B)=0$. Since the $\sigma$-algebra $\mathcal{D}(\mathbf{R}^{d+1},[0,T])$ can be generated by a countable set of the events of the type $A\times B$, it gives the claim. Proposition 1 is proved. Remark 1. The main message of Proposition 1 is that if $\mathbb{Q}_T\ll \mathbb{P}_T$, there exists a jointly measurable version of the density of regular probability distributions $d\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)/d\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$. This fact will be used later. From Proposition 1 we obtain the following result about absolute continuity. Corollary 1. The law $\mathbb{Q}_T$ is equivalent to $\mathbb{P}_T$, i.e., $\mathbb{Q}_T\sim \mathbb{P}_T$ if and only if $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)\sim \mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$ $\mathbf{P}^{\alpha}_T$-a.s. and $\mathbf{Q}^{\alpha}_T\sim \mathbf{P}^{\alpha}_T$. Proof. We apply Proposition 1 and then we use the proof of the same proposition with exchanging of $\mathbb{Q}_T$, $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)$, $\mathbf{Q}^{\alpha}_T$ and $\mathbb{P}_T$, $\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$, $\mathbf{P}^{\alpha}_T$. Corollary 1 is proved. In what follows we assume that the Lévy processes $X^{(j)}$, $j=1,\dots, N$, are integrable, i.e., for $t\in[0,T]$, $\mathbf{E}_{\mathbb{P}}(|X^{(j)}_t|)< +\infty$. Let $P_T^{(j)}$ and $Q_T^{(j)}$ be the laws of the process $X^{(j)}$ under $\mathbb{P}_T$ and $\mathbb{Q}_T$, respectively, and let $(\beta^{(j)},Y^{(j)})$ be the Girsanov parameters of the change of the measure $P_T^{(j)}$ into $Q_T^{(j)}$ (for the details about the Girsanov parameters see [23], [24]). We denote also by $\mathcal M^{(j)}$ the set of equivalent to $P_T^{(j)}$ martingale measures of the process $X^{(j)}$, i.e., the measures under which the process $X^{(j)}$ is a $\mathbb{G}$-martingale. Let $\mathcal M$ be the set of all equivalent to $\mathbb{P}_T$ martingale measures $\mathbb{Q}_T$ of the process $(X,\alpha)$ w.r.t. $\widehat{\mathbb{F}}$. We denote by $(\beta, Y)$ the Girsanov parameters of the change of the conditional measure $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)$ into $\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$. We suppose that they satisfy the usual conditions: $\mathbf{P}^{\alpha}_T$-a.s.
$$
\begin{equation*}
\int_0^T\int_{\mathbf R^d}|Y_s(x)-1|\, \nu_X(dx)\, ds<+\infty\quad\text{and} \quad \int_0^T\|\beta_s\|^2\, ds<+\infty,
\end{equation*}
\notag
$$
where $\nu_X$ is the compensator of the measure of jumps of the process $X$ w.r.t. $(\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha), \mathbb{G})$. Further, we introduce also the process $m=(m_t)_{0\leqslant t\leqslant T}$ such that
$$
\begin{equation}
m_t=\int_0^t {}^{\top}\!\beta_s\, dX_s^c+\int_0^t\int_{\mathbf R^d} (Y_s(x)-1)\bigl(\mu_X(ds,dx)-\nu_X(ds,dx)\bigr),
\end{equation}
\tag{3}
$$
where $\mu_X$ is the jump measure of the process $X$. Proposition 2. The set $\mathcal M \neq \varnothing $ if and only if for all $j=1,\dots,N$, $\mathcal M^{(j)} \neq \varnothing$. If $\mathbb{Q}_T\in\mathcal{M}$, then Moreover, the Girsanov parameters $(\beta, Y)$ of change of the measure $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)$ into $\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$ coincide on the set $\{\alpha_{t-} =j\}$ for all $t\in [0,T]$ and $j=1,\dots, N$ with the corresponding Girsanov parameters $(\beta^{(j)},Y^{(j)})$ of the change of the measure $P_T^{(j)}$ into $Q_T^{(j)}\in\mathcal M^{(j)}$ and Proof. Let $Z=(Z_t)_{0\leqslant t\leqslant T}$ be the Radon–Nikodym process of an equivalent martingale measure $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)$ w.r.t. $\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)$. This process can be always written as $Z_t=\mathcal E(m)_t$, where $m=(m_t)_{t\geqslant 0}$ is a $(\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha),\mathbb{G})$ local martingale of the form (3). So, the equality (4) follows from Proposition 1.
Let us denote by $(\mathbb{Z}_t)_{0\leqslant t\leqslant T}$ the Radon–Nikodym process of an equivalent martingale measure $\mathbb{Q}_T$ w.r.t. $\mathbb{P}_T$. Then, for $t\in[0,T]$,
$$
\begin{equation}
\mathbb{Z}_t=\mathbf{E}_{\mathbb{P}}\biggl(\frac{d\mathbb{Q}_T}{d\mathbb{P}_T} \biggm| \widehat{\mathcal{F}}_t\biggr)= \xi^{(\alpha)}_T \mathbf{E}_{\mathbf{P}^X}\biggl( \frac{d\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)}{d\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha)} \biggm| \mathcal{G}_t \biggr)= \xi^{(\alpha)}_T Z_t.
\end{equation}
\tag{6}
$$
Using the formulas for the Girsanov parameters of the change of the measure $\mathbb{Q}_T$ into $\mathbb{P}_T$ (cf. [24; Chapter 3, Part 3, Theorem 3.24, p. 159]) we see that they coincide ($\mathbb{P}_T$-a.s.) with the Girsanov parameters $(\beta, Y)$ of the change of the measure $\mathbf{Q}_T^X(\,{\cdot}\,|\,\alpha)$ into $\mathbf{P}_T^X(\,{\cdot}\,|\,\alpha)$ when we replace a fixed trajectory of $\alpha$ by the corresponding stochastic process. So, the drift of the process $X$ under $\mathbb{Q}_T$ can be obtained from the drift of the process $X$ under $\mathbf{Q}_T^X(\,{\cdot}\,|\,\alpha)$ with the same replacement. The same is true for the compensator of jump measure of the process $X$ w.r.t. $(\mathbb{P}, \widehat{\mathbb{F}})$.
According to the Girsanov theorem (see [24; Chapter 3, Part 3, Theorem 3.24, p. 159]), the drift $A^{\mathbf{Q}^X}$ of the process $X$ under $\mathbf{Q}_T^X(\,{\cdot}\,|\,\alpha)$ is equal to
$$
\begin{equation*}
A_t^{\mathbf{Q}^X}=A_t+\int_0^tc_s\beta_s\, ds +\int_0^t\int_{\mathbf R^d}x\cdot (Y_s(x)-1)\, \nu_X(dx, ds),
\end{equation*}
\notag
$$
where $c=(c_s)_{s\geqslant0}$ is the density of the quadratic variation of the continuous martingale part of $X$ w.r.t. Lebesgue measure and $\nu_X$ is the compensator of the jump measure of $X$ w.r.t. $(\mathbf{P}^X_T(\,{\cdot}\,|\,\alpha), \mathbb{G})$. In order the measure $\mathbf{Q}^X_T(\,{\cdot}\,|\,\alpha)$ to be a martingale measure of $X$, $A_t^{\mathbf{Q}^X}$ should be equal to zero for all $t\in[0,T]$ $( \mathbf{P}^{\alpha}$-a.s.).
Let us express $ c_t$ and $\nu_X$ to identify $\beta$ and $Y$ from the identity $A^{\mathbf{Q}^X}_t=0 $ with $t\in[0,T]$. Since the continuous martingale part $X^c$ of $X$ is equal to $\bigl(\sum_{j=1}^N\int_0^t\mathbf{I}_{\{\alpha_{s-}=j\}} \, dX_s^{(j),c}\bigr)_{0\leqslant t\leqslant T}$, the quadratic variation $C=(C_t)_{0\leqslant t\leqslant T}$ of $X^c$ and its density $c$ satisfy relations
$$
\begin{equation}
C_t=\sum_{j=1}^Nc^{(j)}\int_0^t\mathbf{I}_{\{\alpha_{s-}=j\}}\, ds,\qquad c_t=\sum_{j=1}^Nc^{(j)}\mathbf{I}_{\{\alpha_{t-}=j\}}.
\end{equation}
\tag{7}
$$
At the same time,
$$
\begin{equation*}
\Delta X_t=\Delta \biggl(\sum_{j=1}^N\int_0^t\mathbf{I}_{\{\alpha_{s-}=j\}}\, dX_s^{(j)}\biggr) =\sum_{j=1}^N\mathbf{I}_{\{\alpha_{t-}=j\}}\Delta X_t^{(j)},
\end{equation*}
\notag
$$
which implies that
$$
\begin{equation}
\nu_X(dt,dx)=\sum_{j=1}^N\mathbf{I}_{\{\alpha_{t-}=j\}}\, \nu^{(j)}(dx)\, dt,
\end{equation}
\tag{8}
$$
where $\nu^{(j)}$ is the Lévy measure of the process $X^{(j)}$.
Taking into account the previous calculus, we finally get that, for all $t\in[0,T]$ ($\mathbf{P}^{\alpha}$-a.s.),
$$
\begin{equation*}
A_t^{\mathbf{Q}^X}=\sum_{j=1}^ N \int_0^ t \biggl( a^{(j)}+c^{(j)}\beta_s + \int_{\mathbf R^d}x\cdot (Y_s(x)-1)\, \nu^{(j)}(dx)\biggr) \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds.
\end{equation*}
\notag
$$
Since $X=(X_t)_{0\leqslant t\leqslant T}$ is a martingale w.r.t. $(\mathbf{Q}_T^X(\,{\cdot}\,|\,\alpha), \mathbb{G})$, $A_t^{\mathbf{Q}^X}=0$ for all $t\in[0,T]$. We take a right-hand derivative in $t$ in the previous expression and since $\mathbf{I}_{\{\alpha_{t-}=i\}}\times \mathbf{I}_{\{\alpha_{t-}=j\}}=0$ for $i\neq j$, we get that on the set $\{\alpha_{t-}=j\}$
$$
\begin{equation*}
a^{(j)}+c^{(j)}\beta_t + \int_{\mathbf R^d}x\cdot (Y_t(x)-1)\, \nu^{(j)}(dx)=0.
\end{equation*}
\notag
$$
But this equality holds if and only if the Girsanov parameters $(\beta,Y)$ of the change of the measure $\mathbf{P}_T^X(\,{\cdot}\,|\,\alpha)$ into $\mathbf{Q}_T^X(\,{\cdot}\,|\,\alpha)$ coincide with the Girsanov parameters $(\beta^{(j)},Y^{(j)})$ of the change of the measure $P_T^{(j)}$ into $Q_T^{(j)}$. The last result implies (5) and this finishes the proof of Proposition 2. Corollary 2. For $t\in[0,T]$, we have with Proof. The first relation follows from (5). The second one holds because the processes $X^{(1)}, \dots, X^{(d)}$are independent and since $\mathbf{I}_{\{\alpha_{t-}=i\}} \times \mathbf{I}_{\{\alpha_{t-}=j\}}=0$ for all $i \neq j$ and $0\leqslant t\leqslant T$. Corollary 2 is proved.
3. $F$-divergence minimal equivalent martingale measures of exponential Lévy switching processesAt the beginning of this section we will present some useful information about dual method. The idea of this method consists in replacing the problem of utility maximization by the problem of minimization of the corresponding $f$-divergence over the set $\mathcal{M}$ of all, equivalent to the law $\mathbb{P}_T$ martingale measures $\mathbb{Q}_T$. The function $f$ is nothing else as the dual function of $u$, which can be obtained by Fenchel–Legendre transform Simple calculations show that
$$
\begin{equation*}
\begin{aligned} \, f(x) &=-\ln x-1,\quad x>0, \quad \text{if }u\text{ is logarithmic}, \\ f(x) &=-\frac{p-1}{p}x^{p/(p-1)},\quad x>0,\quad p\in (-\infty,0)\cup (0,1),\quad \text{if }u\text{ is power}, \\ f(x) &=1-x+x\ln x,\quad x>0, \quad \text{if }u\text{ is exponential}. \end{aligned}
\end{equation*}
\notag
$$
For two equivalent measures $\mathbb{Q}_T$ and $\mathbb{P}_T$ the $f$-divergence of $\mathbb{Q}_T$ w.r.t. $\mathbb{P}_T$ is defined as
$$
\begin{equation*}
f(\mathbb{Q}_T\,|\,\mathbb{P}_T)= \mathbf E_{\mathbb{P}} \biggl(f\biggl(\frac{d\mathbb{Q}_T}{d\mathbb{P}_T}\biggr)\biggr).
\end{equation*}
\notag
$$
Definition 1. An equivalent to $\mathbb{P}_T$ measure $\mathbb{Q}^*_T$ is called $f$-divergence minimal if $\mathbf E_{\mathbb{P}}|f(d\mathbb{Q}^*_T/d\mathbb{P}_T)|<+\infty$ and if In the proof of Theorem 1 we will use the following properties of $f$-divergence minimal equivalent martingale measures. Definition 2. An $f$-divergence minimal measure $\mathbb{Q}_T^*$ is called scale invariant if, for all $x \in \mathbf R^+$, Definition 3. An $f$-divergence minimal measure $\mathbb{Q}_T^*$ is time horizon invariant if, for all $t \in]0,T]$, where $\mathbb{Q}_t$ and $\mathbb{P}_t$ are the restrictions of $\mathbb{Q}_T$ and $\mathbb{P}_T$ on the $\sigma $-algebra $\widehat{\mathcal{F}}_t$. Definition 4. An $f$-divergence minimal measure $\mathbb{Q}_T^*$ is said to preserve the Lévy property of some Lévy process if this process remains a Lévy process under $\mathbb{Q}_T^*$. Proposition 3 (cf. [5]). For any Lévy process and HARA utilities for which the $f$-divergence minimal equivalent martingale measure exists, this measure is scale invariant, time horizon invariant and it preserves Lévy property. To formulate Theorem 1 we need some additional notations. Let $1\leqslant j\leqslant N$ be fixed and let $Q_T^{(j),*}$ be an $f$-divergence minimal measure of the Lévy process $X^{(j)}$. We denote by $(\beta^{(j),*}, Y^{(j),*})$ the Girsanov parameters corresponding to the change of the measure $P^{(j)}_T$ into $Q^{(j),*}_T$ and we put $\xi^{(j),*}_T= dQ^{(j),*}_T/dP^{(j)}_T$. Let $Z_T^{(j),*}=\mathcal{E}(m^{(j),*})_T$, where $m^{(j),*}$ is defined by (10) with the replacement of $(\beta^{(j)}, Y^{(j)})$ by $(\beta^{(j),*}, Y^{(j),*})$. We also put
$$
\begin{equation}
T_j^{(\alpha)}= \int_0^T \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds.
\end{equation}
\tag{11}
$$
We need also some propositions related to by a specific choice of HARA utility. Let the utility be power, i.e., $u(x)=x^p/p$, $x>0$, with $p\in(-\infty,0)\cup(0,1)$. Then the corresponding $f$-divergence is $f(x)=-x^{\gamma}/\gamma$ with $\gamma=p/(p-1)$ and this $f$-divergence is related to the Hellinger processes. Proposition 4 (cf. [24], [14], [30]). The Hellinger process $H^{(j),*}{(\gamma)}=(H_t^{(j),*}{(\gamma)})_{0\leqslant t\leqslant T}$ of order $\gamma$ for the change of the measure $P^{(j)}_T$ into $Q^{(j),*}_T$ is given by the following expression: $H_t^{(j),*}{(\gamma)}= t\cdot h^{(j),*}{(\gamma)}$ with Let the utility be exponential, i.e., $u(x)=1-\exp(-x)$, $x\in\mathbf{R}$. Then the corresponding $f$-divergence is $f(x) = 1-x+x\ln x$ and this $f$-divergence is related to the Kulback–Leibler processes. Proposition 5 (cf. [26], [14], [30]). The Kulback–Leibler process $K^{(j),*}=(K_t^{(j),*})_{0\leqslant t\leqslant T}$ for the change of the measure $P^{(j)}_T$ into $Q^{(j),*}_T$ is given by the expression $K_t^{(j),*}= t\cdot\kappa^{(j),*}$ with and it satisfies $\mathbf{E}_P(\xi^{(j),*}_T\ln(\xi^{(j),*}_T))= K_T^{(j),*}$. Moreover, we have Now we are ready to formulate Theorem 1. Theorem 1. Suppose $\mathcal M^{(j)} \neq \varnothing$ for all $j=1,\dots,N$. Then for HARA utilities, the minimal martingale measure $\mathbb{Q}^*_T$ exists and its Radon–Nikodym derivative w.r.t. $\mathbb{P}_T$ is given by Proof. 1) First of all we prove that for any density $\xi^{(\alpha)}_T$ from Proposition 2
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl(f(\mathbb{Z}_T)\bigr)= \mathbf{E}_{\mathbb{P}}\biggl( f\biggl(\xi^{(\alpha)}_T\prod_{j=1}^NZ_t^{(j)}\biggr)\biggr) \geqslant \mathbf{E}_{\mathbb{P}} \biggl( f\biggl(\xi^{(\alpha)}_T\prod_{j=1}^NZ_t^{(j),*}\biggr)\biggr).
\end{equation*}
\notag
$$
According to Corollary 2 the density process $Z=(Z_t)_{0\leqslant t\leqslant T}$ of the conditional martingale measure $\mathbf{Q}_T^X(\,{\cdot}\,|\,\alpha)$ w.r.t. $\mathbf{P}_T^X(\,{\cdot}\,|\,\alpha)$ satisfies $Z_t= \mathcal{E}(m)_t$, where $m$ is defined by (3). Since $\sum_{k=0}^{+\infty}\mathbf{I}_{\{\tau_k\leqslant t<\tau_{k+1}\}}=1$ for each $t>0$ and $\tau_0=0$, we get
$$
\begin{equation*}
\begin{aligned} \, m_T &= \sum_{k=0}^{+\infty}\int_{\tau_k\wedge T}^{\tau_{k+1}\wedge T} \biggl({}^{\top}\!\beta_t^{(j_k)}\, dX_t^{(j_k),c} \\ &\qquad+\int_{\mathbf{R}^d}\bigl(Y_t^{(j_k)}(x) -1\bigr) \, \bigl(\mu^{(j_k)}(dt,dx)-\nu^{(j_k)}(dx)\, dt\bigr)\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $j_k$ is the state of the process $\alpha$ in the interval $[\tau_k, \tau_{k+1}[$ and $(\beta^{(j_k)},Y^{(j_k)})$ are the Girsanov parameters of the change of the measure $P_T^{(j_k)}$ into $Q_T^{(j_k)}$. The last formula means that on the set $\{\tau_k\leqslant T<\tau_{k+1}\}$ with fixed $k$
$$
\begin{equation*}
\begin{aligned} \, &m_T = \sum_{m=0}^{k-1} \int_{\tau_m}^{\tau_{m+1}} \biggl({}^{\top}\!\beta_t^{(j_m)}\, dX_t^{(j_m),c} \\ &\quad\qquad+\int_{\mathbf{R}^d}\bigl(Y_t^{(j_m)}(x) -1\bigr)\, \bigl(\mu^{(j_m)}(dt,dx) -\nu^{(j_m)}(dx) \, dt\bigr)\biggr) \\ &\quad+\int_{\tau_k}^T \biggl({}^{\top}\!\beta_t^{(j_k)}\, dX_t^{(j_k),c} +\int_{\mathbf{R}^d}\bigl(Y_t^{(j_k)}(x) -1\bigr)\, \bigl(\mu^{(j_k)}(dt,dx)-\nu^{(j_k)}(dx)\, dt\bigr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
As a conclusion, on the set $\{\tau_k\leqslant T<\tau_{k+1}\}$ with fixed $k$ the Radon–Nikodym density $\mathbb{Z}_T$ of the measure $\mathbb{Q}_T$ w.r.t. the measure $\mathbb{P}_T$ satisfies the relation where $\xi^{(j_m)}_{t}= dQ_{t}^{(j_k)}/dP_{t}^{(j_k)}$ for $0\leqslant m\leqslant k$ and all $t\in[0,T]$.
We denote the value of $\mathbb{Z}_T$ on the set $\{\tau_k\leqslant T<\tau_{k+1}\}$ by $Z_T(k)$. Then for any $f$-divergence corresponding to the HARA utility
$$
\begin{equation*}
\begin{aligned} \, \mathbf{E}_{\mathbb{P}}\bigl(f(\mathbb{Z}_T)\bigr) &=\sum_{k=0}^{\infty}\mathbf{E}_{\mathbb{P}} \bigl(f(\mathbb{Z}_T)\,\mathbf{I}_{\{\tau_k\leqslant T<\tau_{k+1}\}}\bigr) \\ &=\sum_{k=0}^{\infty}\mathbf{E}_{\mathbb{P}}\bigl( \mathbf{I}_{\{\tau_k\leqslant T<\tau_{k+1}\}}\,\mathbf{E}_{\mathbb{P}}\bigl(f(Z_T(k))\bigm| \alpha\bigr)\bigr). \end{aligned}
\end{equation*}
\notag
$$
For any fixed $k$ we take the conditional expectation w.r.t. the processes $X^{(j_1)},\dots, X^{(j_{k-1})}$:
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl(f(Z_T(k))\bigm| \alpha\bigr) =\mathbf{E}_{\mathbb{P}} \bigl( \mathbf{E}_{\mathbb{P}} \bigl(f(Z_T(k))\bigm| \alpha,X^{(j_1)},\dots, X^{(j_{k-1})} \bigr) \bigm| \alpha\bigr).
\end{equation*}
\notag
$$
Using scale invariance and time horizon invariance of the $f$-divergence minimal martingale measure for Lévy processes proved in [4], [5] together with the fact that conditionally to $\alpha$, $X^{(j_1)},\dots, X^{(j_{k-1})}$, the identity in law $\xi^{(j_k)}_T/\xi^{(j_k)}_{\tau_k}\stackrel{d}{=}\xi^{(j_k)}_{T-\tau_k}$ holds, we can replace $\xi^{(j_k)}$ by the density $\xi^{(j_k),*}$ of the $f$-divergence minimal martingale measure and this procedure will decrease $\mathbf{E}_{\mathbb{P}}(f(Z_T(k))\,|\,\alpha)$.
The same can be done for any $j_m$ with $1\leqslant m\leqslant k-1$. For this aim one can take the conditional expectation w.r.t. the process $\alpha$ and then w.r.t. the processes $X^{(j_1)},\dots, X^{(j_{m-1})},X^{(j_{m+1})},\dots, X^{(j_k)}$ and use scale invariance and time horizon invariance of the $f$-divergence minimal martingale measure as well as the identity in law $\xi^{(j_m)}_{\tau_m}/\xi^{(j_m)}_{\tau_{m-1}}\stackrel{d}{=}\xi^{(j_m)}_{\tau_m-\tau_{m-1}}$. This permits to replace $\xi^{(j_m)}$ by $\xi^{(j_m,*)}$ and it will again decrease $\mathbf{E}_{\mathbb{P}}(f(Z_T(k))\,|\,\alpha)$. Finally, performing the above procedure for each $k$ we get that
$$
\begin{equation*}
\begin{aligned} \, \mathbf{E}_{\mathbb{P}}\bigl(f(\mathbb{Z}_T)\bigr) &\geqslant \sum_{k=0}^{+\infty} \mathbf{E}_{\mathbb{P}} \biggl(\mathbf{I}_{\{\tau_k\leqslant t<\tau_{k+1}\}} f\biggl(\xi^{(\alpha)}_T\biggl(\prod_{m=1}^{k-1} \frac{\xi^{(j_m),*}_{\tau_m}}{\xi^{(j_m),*}_{\tau_{m-1}}}\biggr) \frac{\xi^{(j_k),*}_T}{\xi^{(j_k),*}_{\tau_k}}\biggr)\biggr) \\ &=\mathbf{E}_{\mathbb{P}} \biggl(f\biggl(\xi^{(\alpha)}_T \prod_{j=1}^NZ_T^{(j),*}\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
Now we will obtain the expression for $\xi^{(\alpha),*}$ minimizing each $f$-divergence.
2) Logarithmic utility. For the logarithmic utility $f(x)= -\ln x-1$ and we get that
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl(f(\mathbb{Z}_T^*)\bigr)= -\mathbf{E}_{\mathbb{P}} \bigl(\ln \xi^{(\alpha)}_T\bigr) -\sum_{j=1}^N \mathbf{E}_{\mathbb{P}}\bigl(\ln Z^{(j),*}_T\bigr)-1.
\end{equation*}
\notag
$$
By the Jensen inequality we deduce that
$$
\begin{equation*}
-\mathbf{E}_{\mathbb{P}}\bigl(\ln \xi^{(\alpha)}_T\bigr)\geqslant -\ln \bigl(\mathbf{E}_{\mathbb{P}}\xi^{(\alpha)}_T\bigr)=0,
\end{equation*}
\notag
$$
so that the minimum of the $f$-divergence is achieved on $\xi^{(\alpha),*}_T=1$.
3) Power utility. For the power utility $u(x) = x^p/p$ we have $f(x) = -(1/\gamma)x^{\gamma}$ with $\gamma=p/(p-1)$. Since conditionally to the process $\alpha$, the processes $Z^{(j),*}$, $j=1,\dots, N$, are independent, we have
$$
\begin{equation*}
\begin{aligned} \, \mathbf{E}_{\mathbb{P}}\bigl(f(\mathbb{Z}_T)\bigr) &\geqslant -\frac{1}{\gamma}\, \mathbf{E}_{\mathbb{P}}\biggl( \bigl(\xi^{(\alpha)}_T\bigr)^{\gamma} \prod_{j=1}^N \bigl(Z_t^{(j),*}\bigr)^{\gamma}\biggr) \\ &= -\frac{1}{\gamma}\, \mathbf{E}_{\mathbb{P}}\biggl(\bigl(\xi^{(\alpha)}_T\bigr)^{\gamma} \prod_{j=1}^N \mathbf{E}_{\mathbb{P}}\bigl(\bigl(Z_t^{(j),*}\bigr)^{\gamma}\bigm| \alpha\bigr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
From Proposition 4 we deduce that, for $1\leqslant j\leqslant N$,
$$
\begin{equation}
\mathbf{E}_{\mathbb{P}}\bigl(\bigl(Z_t^{(j),*}\bigr)^{\gamma}\bigm| \alpha\bigr)= \exp \bigl(-T_j^{(\alpha)} h^{(j),*}(\gamma)\bigr)
\end{equation}
\tag{14}
$$
and that
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl( f(\mathbb{Z}_T^*)\bigr) =-\frac{1}{\gamma}\mathbf{E}_{\mathbb{P}}\biggl(\bigl(\xi^{(\alpha)}_T\bigr)^{\gamma} \sum_{1\leqslant j\leqslant N}\exp \bigl(- T_j^{(\alpha)} h^{(j),*}(\gamma)\bigr)\biggr).
\end{equation*}
\notag
$$
Finally, the Lagrange method of minimization subject to the constraint $\mathbf{E}_{\mathbb{P}}(\xi^{(\alpha)}_T)=1$ gives the mentioned expression for $\xi^{(\alpha),*}_T $.
4) Exponential utility. For exponential utility $u(x)= 1-\exp(-x)$, as known, $f(x) = 1-x+x\ln x$. Since
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\biggl(\xi^{(\alpha)}_T \prod_{j=1}^N Z_t^{(j),*}\biggr)=1,\qquad \mathbf{E}_{\mathbb{P}}\bigl( Z_T^{(j),*}\bigm| \alpha \bigr)=1,
\end{equation*}
\notag
$$
for all $j$ and since the processes $Z^{(1),*}, \dots, Z^{(N),*}$ are independent conditionally to the process $\alpha$, we have
$$
\begin{equation*}
\begin{aligned} \, \mathbf{E}_{\mathbb{P}} \biggl( f\biggl(\xi_T^{(\alpha)} \prod_{j=1}^NZ_T^{(j),*}\biggr) \biggr) &=\mathbf{E}_{\mathbb{P}}\bigl(\xi_T^{(\alpha)}\ln \xi_T^{(\alpha)}\bigr) \\ &\qquad +\mathbf{E}_{\mathbb{P}}\biggl( \xi_T^{(\alpha)}\sum_{j=1}^N \mathbf{E}_{\mathbb{P}} \bigl( Z_T^{(j),*}\ln Z_T^{(j),*}\bigm| \alpha\bigr) \biggr). \end{aligned}
\end{equation*}
\notag
$$
We apply now Proposition 5 to deduce that, for $j=1,\dots, N$,
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl( Z_T^{(j),*}\ln Z_T^{(j),*}\bigm| \alpha\bigr)= T_j^{(\alpha)} \kappa_T^{(j),*}.
\end{equation*}
\notag
$$
So, finally we have to minimize the expression
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl( \xi_T^{(\alpha)}\ln \xi_T^{(\alpha)}\bigr) + \mathbf{E}_{\mathbb{P}}\biggl( \xi_T^{(\alpha)}\sum_{j=1}^NT_j^{(\alpha)} \kappa^{(j),*}\biggr)
\end{equation*}
\notag
$$
subject to the constraint $\mathbf{E}_{\mathbb{P}}(\xi_T^{(\alpha)})=1$. Again, using the Lagrange method of the minimization we get the expression for $\xi^{(\alpha),*}$. This ends the proof of Theorem 1. 4. Utility maximizing strategies for Lévy switching modelsIn this section we will find optimal strategies for the utility maximization. These strategies will be automatically adapted to the progressive filtration, but for technical reasons we will search them in the initially enlarged filtration as it was mentioned in the introduction. To find optimal strategies, we will use Theorem 1 and the following proposition. Proposition 6 (cf. [17], [4]). Let $\mathbb{Z}^*_T$ be Radon–Nikodym derivative of $f$-divergence minimal equivalent martingale measure $\mathbb{Q}^*_T$ w.r.t. $\mathbb{P_T}$. Let $x_0\,{>}\,0$ be an initial capital. We suppose that for $\lambda _0>0$ such that Since for $1\leqslant k\leqslant d$ the processes $\bigl(\int_0^{\cdot} \widehat{\phi}^{\,(k)}_s \, dS^{(k)}_s\bigr)$ are $\mathbb{Q}^*_T$-martingales w.r.t. the initially enlarged filtration $\widehat{\mathbb{F}}$, from Proposition 6 it follows that
$$
\begin{equation}
-\mathbf{E}_{\mathbb{Q}^*}\bigl(f'(\lambda _0\mathbb{Z}^*_T)\bigm| \widehat{\mathcal{F}}_t\bigr) =x_0+\sum_{k=1}^d\int_0^t \widehat{\phi}^{\,(k)}_s\, dS^{(k)}_s.
\end{equation}
\tag{16}
$$
Using the Itô formula we obtain another expression for $-\mathbf{E}_{\mathbb{Q}^*}[f'(\lambda _0\mathbb{Z}^*_T)\,|\,\widehat{\mathcal{F}}_t)]$ to identify $\widehat{\phi}^{\,(k)}$ for $1\leqslant k\leqslant d$. To do it in an efficient way we introduce the shifted process $\alpha^{t}$ as $\alpha^t(s) = \alpha(s+t)$ and we put
$$
\begin{equation}
\mathbb{Z}_t^*= \mathbf{E}_{\mathbb{P}}\bigl(\mathbb{Z}^*_T\bigm| \widehat{\mathcal{F}}_t\bigr) \quad \text{and also}\quad Z_t(\alpha)=\frac{\mathbb{Z}_t^*}{\mathbb{Z}_0^*}.
\end{equation}
\tag{17}
$$
Then, since the processes $X^{(1)},\dots, X^{(N)}$ and $\alpha$ are independent, and since the processes $X^{(1)},\dots, X^{(N)}$ are homogeneous in time, we have that
$$
\begin{equation}
\frac{\mathbb{Z}_T^*}{\mathbb{Z}_t^*}\stackrel{d}{=} Z_{T-t}(\alpha^t).
\end{equation}
\tag{18}
$$
Proof. We change the measure in the conditional expectation:
$$
\begin{equation*}
\mathbf{E}_{\mathbb{Q}^*} \bigl( f'(\lambda\mathbb{Z}_T^*) \bigm| \widehat{\mathcal{F}}_t\bigr) =\mathbf{E}_{\mathbb{P}} \biggl( \frac{\mathbb{Z}_T^*}{\mathbb{Z}_t^*}f'(\lambda\mathbb{Z}_T^*) \biggm| \widehat{\mathcal{F}}_t\biggr).
\end{equation*}
\notag
$$
Since the process $X$, conditionally to the process $\alpha$, has independent increments, $\mathbb{Z}_T^*/\mathbb{Z}_t^*$ and $\mathbb{Z}_t^*$ are also conditionally to the process $\alpha$ independent for each $t\in[0,T]$. As a conclusion,
$$
\begin{equation*}
\mathbf{E}_{\mathbb{Q}^*} \bigl( f'(\lambda\mathbb{Z}_T^*) \bigm| \widehat{\mathcal{F}}_t \bigr) =\mathbf{E}_{\mathbb{P}} \biggl( \frac{\mathbb{Z}_T^*}{\mathbb{Z}_t^*}f'\biggl(\lambda x \frac{\mathbb{Z}_T^*}{\mathbb{Z}_t^*}\biggr) \biggm| \widehat{\mathcal{F}}_t\biggr) \bigg|_{x=\mathbb{Z}_t^*}
\end{equation*}
\notag
$$
and (18) gives the result of lemma. The lemma is proved. We recall that according to Theorem 1, the Girsanov parameters of $f$-divergence minimal martingale measure of the process $X$ are defined in the following way: for $t\in[0,T]$ and $x \in \mathbf R^d$,
$$
\begin{equation}
\beta_t^*=\sum_{j=1}^N\beta^{(j),*}\mathbf{I}_{\{\alpha_{t-}=j\}},\quad Y_t^*(x)=\sum_{j=1}^N Y^{(j),*}(x)\, \mathbf{I}_{\{\alpha_{t-}=j\}},
\end{equation}
\tag{19}
$$
where $(\beta^{(j),*},Y^{(j),*})$ are the Girsanov parameters of $f$-divergence minimal equivalent martingale measure $Q^{(j),*}$ of the Lévy process $X^{(j)}$ and they do not depend on the time $t$. 4.1. When the utility is logarithmicLet us suppose that the utility $u(x)=\ln x$. Then the corresponding $f$-divergence is also logarithm up to a constant, i.e., $f(x)=-\ln x -1$ with $f'(x)=-1/x$. Proposition 7. Suppose that, for $1 \leqslant j \leqslant N$, the matrices $c^{(j)}$ are invertible and $\mathbf{E}_{\mathbb{P}}|{\ln\mathbb{Z}^*_T}|<+\infty$. Then for an initial capital $x_0 >0$, there exists an optimal strategy such that, for $1 \leqslant k \leqslant d $ and $0\leqslant t\leqslant T$, Proof. Since $f'(x)=-1/x$ we have
$$
\begin{equation*}
-\mathbf{E}_{\mathbb{P}} \bigl( Z_{T-t}(\alpha^t)f'\bigl(\lambda_0 x Z_{T-t}(\alpha^t)\bigr) \bigm| \alpha\bigr) =\frac{1}{\lambda_0x}.
\end{equation*}
\notag
$$
Then by Lemma 1
$$
\begin{equation*}
-\mathbf{E}_{\mathbb Q^*}\bigl(f'(\lambda_0 \mathbb Z^*_T)\bigm| \widehat{\mathcal F}_t \bigr) =\frac{1}{\lambda_0 Z^*_t}=\frac{1}{\lambda_0 Z_t(\alpha)},
\end{equation*}
\notag
$$
since $\mathbb{Z}^{*}_0=\xi^{(\alpha),*}_T=1$. From condition (15) we get that $\lambda_0=1/x_0$ and hence,
We notice that
$$
\begin{equation*}
\begin{aligned} \, Z_t(\alpha)&=\mathcal E(m^*)_t \\ &=\exp\biggl(m^*_t-\frac{1}{2}\langle m^{*,c}\rangle_t+\int_0^t\int_{\mathbf R^d} \bigl(\ln (Y_s^*(x))-Y_s^*(x)\,{+}\,1\bigr) \, \mu_X(ds,dx)\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $m^*$ is given by (9) with the replacement of $(\beta^{(j)}, Y^{(j)})$ by $(\beta^{(j),*}, Y^{(j),*})$ for all $1\leqslant j\leqslant N$. From the mentioned formula for $m^*$ we get using the compensating formula and doing the simplifications for discontinuous martingale part that where $M=(M_t)_{t\geqslant 0}$ is $\mathbb{Q}^*$-martingale with
$$
\begin{equation*}
M_t=\int_0^t {}^\top\!\beta_s^*\, dX^{c,\mathbb{Q}^*}_{s} +\int_0^t\int_{\mathbf R^d} \ln(Y_s^*(x))\, (\mu_X-\nu_X^{\mathbb{Q}^*})(dt,dx)
\end{equation*}
\notag
$$
and $B=(B_t)_{t \geqslant 0}$ is a predictable process with
$$
\begin{equation*}
\begin{aligned} \, B_t &= - \int_0^t\frac{1}{2}\langle c_s\beta_s^*,\beta_s^*\rangle\, ds +\int_0^t\int_{\mathbf R^d} (Y_s^*(x)-1)^2\, \nu_X^{\mathbb{P}}(ds,dx) \\ &\qquad +\int_0^t\int_{\mathbf R^d} \bigl(\ln Y_s^*(x)-Y_s^*(x)+1\bigr) \nu^{\mathbb{Q}^*}_X(ds,dx). \end{aligned}
\end{equation*}
\notag
$$
Let $\tau_n=\inf\{t\geqslant 0 \,|\, Z_t(\alpha)\leqslant 1/n\}$ with $n\geqslant 1$ and $\inf\{\varnothing \}=+\infty $. By the Itô formula applied to the function $g(x)= \exp(-x)$ we get
$$
\begin{equation*}
\begin{aligned} \, \frac{1}{Z_{t\wedge\tau_n}(\alpha)} &=1-\int_0^{t\wedge\tau_n}\frac{1}{Z_{s-}(\alpha)}\, (dM_s+dB_s) +\frac{1}{2}\int_0^{t\wedge\tau_n} \frac{1}{Z_{s-}(\alpha)}\, d\langle M^c\rangle_s \\ &\qquad+\int_0^{t\wedge\tau_n} \int_{\mathbf R^d} \frac{1}{Z_{s-}(\alpha)}\bigl((Y_s^*(x))^{-1}-1+\ln(Y_s^*(x))\bigr)\, \mu_X(ds,dx). \end{aligned}
\end{equation*}
\notag
$$
Since $\tau_n\to+\infty$, we deduce after limit passage the similar expression for $1/Z_{t}(\alpha)$ with $t\wedge\tau_n$ replaced by $t$. Using again the compensating formula and taking into account that $(Z_t(\alpha))^{-1}_{t \geqslant 0}$ is a $\mathbb{Q}^*$-martingale, we get that its drift part
$$
\begin{equation*}
\begin{aligned} \, &-\int_0^t\frac{1}{Z_{s-}(\alpha)}\, dB_s +\frac{1}{2}\int_0^t \frac{1}{Z_{s-}(\alpha)}\, d\langle M^c\rangle_s \\ &\qquad+\int_0^t \int_{\mathbf R^d} \frac{1}{Z_{s-}(\alpha)} \bigl((Y_s^*(x))^{-1}-1+\ln(Y_s^*(x))\bigr)\, \nu^{\mathbb{Q}^*}_X(ds,dx)=0. \end{aligned}
\end{equation*}
\notag
$$
Further, from (20) and (16) we find that, for $0 \leqslant t \leqslant T$,
$$
\begin{equation*}
\begin{aligned} \, \frac{x_0}{Z_t(\alpha)} &= x_0-\int_0^t \frac{x_0}{Z_{s-}(\alpha)}{}^\top\!\beta_s^* \, dX^{c,\mathbb{Q}^*}_{s} \\ &\qquad+\int_0^t \int_{\mathbf R^d} \frac{x_0}{Z_{s-}(\alpha)}\bigl((Y^*_s(x))^{-1}-1\bigr) \, (\mu_X-\nu_X^{\mathbb{Q}^*})(ds,dx) \\ &\qquad=x_0 + \sum_{k=1}^d\int_0^t \widehat \varphi_s^{\,(k)}\, dS_s^{(k)}. \end{aligned}
\end{equation*}
\notag
$$
But $dS_s^{(k)}=S^{(k)}_{s-}\, d\chi^{(k)}_s$ and the last equality implies that, for $0 \leqslant t \leqslant T$,
$$
\begin{equation*}
\begin{aligned} \, &\sum_{k=1}^d\int_0^t\biggl( \frac{x_0}{Z_{s-}(\alpha)}\beta_s^{*,(k)}+\widehat \varphi_s^{\,(k)}S_{s-}^{(k)}\biggr)\, d\chi_s^{c,\mathbb{Q}^*,(k)} \\ &\quad=\int_0^t \int_{\mathbf R^d} \biggl(\frac{x_0}{Z_{s-}(\alpha)}\bigl((Y_s^*)^{-1}(x)-1\bigr) -\sum_{k=1}^d\widehat \varphi_s^{\,(k)}S_{s-}^{(k)}\biggr)\, (\mu_X-\nu_X^{\mathbb{Q}^*})(ds,dx). \end{aligned}
\end{equation*}
\notag
$$
The left-hand side of the above equality represents a continuous martingale and the right-hand side defines a purely discontinuous martingale, which is orthogonal to any continuous martingale. Hence, the quadratic variation of the continuous martingale in the left-hand side of the previous equality is equal to zero, i.e., for $0 \leqslant t \leqslant T$,
$$
\begin{equation}
\biggl\langle \sum_{k=1}^d\int_0^{\cdot}\biggl( \frac{x_0}{Z_{s-}(\alpha)}\beta_s^{*,(k)}+\widehat \varphi_s^{\,(k)}S_{s-}^{(k)}\biggr)\, d\chi_s^{c,\mathbb{Q}^*,(k)} \biggr\rangle_t=0.
\end{equation}
\tag{22}
$$
We calculate then the quadratic variation of the left-hand side of (22). Since the density $c_t=(c^{(k,l)}_{t})_{1\leqslant k\leqslant d,\, 1\leqslant l\leqslant d}$ of quadratic variation of a continuous martingale $X^{c,\mathbb{Q}^*}$ is an invertible matrix, we deduce from (22) that, for $1\leqslant k\leqslant d$,
$$
\begin{equation*}
\frac{x_0}{Z_{t-}(\alpha)}\beta_t^{*,(k)}+\widehat \varphi_t^{\,(k)}S_{t-}^{(k)}=0.
\end{equation*}
\notag
$$
Then our result for the optimal strategy follows.
Using again Proposition 6, we get the formula for the optimal expected utility in the logarithmic case. The proposition is proved. 4.2. When the utility is powerWhen the utility is power $u(x)=x^p/p$, the dual function $f$ is also power, i.e., $f(x)=(-1/\gamma)x^{\gamma}$ and $f'(x)=-x^{\gamma-1}$ with $\gamma=p/(p-1)$. Proposition 8. Suppose that $\mathbf E_{\mathbb P}((\mathbb{Z}^*_T)^{\gamma}) \,{<}\, {+}\infty$ and that the matrices $c^{(j)}$, $1\leqslant j\leqslant N$, are invertible. Then for an initial capital $x_0 >0$, there exists an optimal strategy and, for $1 \leqslant k \leqslant d$ and $0 \leqslant t \leqslant T$, it is given by Proof. When the utility is power,
$$
\begin{equation*}
-\mathbf{E}_{\mathbb P}\bigl(Z_{T-t}(\alpha^t)f'\bigl(\lambda_0 x Z_{T-t}(\alpha^t)\bigr)\bigm| \alpha \bigr) =\lambda_0^{\gamma-1}x^{\gamma-1}\mathbf{E}_{\mathbb P}\bigl( Z^{\gamma}_{T-t}(\alpha^t) \bigm| \alpha\bigr).
\end{equation*}
\notag
$$
According to the information of Section 3 on the Hellinger processes, the last conditional expectation is given by
$$
\begin{equation*}
\begin{aligned} \, \mathbf{E}_{\mathbb P}\bigl(Z^{\gamma}_{T-t}(\alpha^t)\bigm|\alpha\bigr) &=\exp\biggl(-\sum_{j=1}^N h^{(j),*}(\gamma)\int_0^{T-t}\mathbf{I}_{\{\alpha_{(s+t)-}=j\}}\, ds\biggr) \\ &=\exp\biggl(-\sum_{j=1}^N h^{(j),*}(\gamma)\int_t^T\mathbf{I}_{\{\alpha_{s-}=j\}}\, ds\biggr). \end{aligned}
\end{equation*}
\notag
$$
Then, using Lemma 1 we get
We denote by $N=(N_t)_{t \geqslant 0}$ the process with $N_t=(Z_t(\alpha))^{\gamma-1} g_t$, where
$$
\begin{equation*}
g_t=\exp\biggl(-\sum_{j=1}^N h^{(j),*}(\gamma)\int_t^T\mathbf{I}_{\{\alpha_{s-}=j\}}\, ds\biggr).
\end{equation*}
\notag
$$
Then (23) and (17) imply that
$$
\begin{equation}
-\mathbf{E}_{\mathbb{Q}^*}\bigl(f'(\lambda_0 \mathbb Z^*_T)\bigm|\widehat{\mathcal F}_t \bigr) =\lambda_0^{\gamma-1}(\mathbb{Z}_0^*)^{\gamma-1} N_t.
\end{equation}
\tag{24}
$$
By integration by part formula we have
$$
\begin{equation}
N_t=g_0+\int_0^t Z^{\gamma-1}_s(\alpha)\, dg_s+\int_0^t g_s\, dZ_s^{\gamma-1}(\alpha),
\end{equation}
\tag{25}
$$
since $Z_0(\alpha)=1$. Using the representation (21) for $Z_s(\alpha)$ from the previous proposition, we get that
$$
\begin{equation*}
Z_s^{\gamma-1}(\alpha)=\exp\bigl((\gamma-1)(M_s+B_s)\bigr).
\end{equation*}
\notag
$$
Again by the Itô formula with $f(x) = \exp((\gamma-1)x)$ via a localization procedure with the stopping times
$$
\begin{equation*}
\tau_n=\inf\biggl\{t\geqslant 0 \biggm| Z_t(\alpha)\leqslant \frac{1}{n} \text{ or } Z_t(\alpha)\geqslant n\biggr\},
\end{equation*}
\notag
$$
where $n\geqslant 1$ and $\inf\{\varnothing\}=+\infty $, and further limit passage $n\to+\infty$, we find that
$$
\begin{equation*}
\begin{aligned} \, &Z_t^{\gamma-1}(\alpha)= 1+(\gamma-1)\int_0^tZ_{s-}^{\gamma-1}(\alpha)\, (dM_s+dB_s) \\ &\qquad+\frac{1}{2}(\gamma-1)^2\int_0^tZ_{s-}^{\gamma-1}(\alpha)\, d\langle M^c\rangle_s \\ &\qquad+\int_0^t\int_{\mathbf R^d}Z_{s-}^{\gamma-1}(\alpha ) \bigl((Y_s^*)^{\gamma-1}(x)-1-(\gamma-1)\ln Y_s^*(x)\bigr)\, \mu_X(ds,dx). \end{aligned}
\end{equation*}
\notag
$$
We deduce from the above formula and (25) that
$$
\begin{equation*}
\begin{aligned} \, N_t &=g_0+ \int_0^t Z^{\gamma-1}_s(\alpha)\, dg_s+ (\gamma-1)\int_0^t g_s Z_{s-}^{\gamma-1}(\alpha)\, dM_s \\ &\qquad+(\gamma-1)\int_0^t g_s Z_{s-}^{\gamma-1}(\alpha)\, dB_s +\frac{1}{2}(\gamma-1)^2\int_0^t g_sZ_{s-}^{\gamma-1}(\alpha)\, d\langle M^c\rangle_s \\ &\qquad+\int_0^t\int_{\mathbf R^d}g_sZ_{s-}^{\gamma-1}\bigl( (Y_s^*)^{\gamma-1}(x)-1-(\gamma-1) \ln Y_s^*(x)\bigr)\, \mu_X(ds,dx). \end{aligned}
\end{equation*}
\notag
$$
From the previous formula, using a compensation and taking into account that $N$ is $\mathbb{Q}^*$-martingale, we find that the drift part of $N$ is equal to zero, and
$$
\begin{equation*}
\begin{aligned} \, N_t &= g_0+(\gamma-1)\int_0^t g_s Z_{s-}^{\gamma-1}(\alpha) {}^\top\!\beta_s^*\, dX^{c,\mathbb{Q}^*}_{s} \\ &\qquad +\int_0^t\int_{\mathbf R^d}g_sZ_{s-}^{\gamma-1}\bigl( (Y_s^*)^{\gamma-1}(x)-1\bigr) \, (\mu_X-\nu_X^{\mathbb{Q}^*})(ds,dx). \end{aligned}
\end{equation*}
\notag
$$
In addition, according to (24) and (16)
$$
\begin{equation}
(\lambda_0\mathbb Z_0^*)^{\gamma-1}N_t=x_0+\sum_{k=1}^N\int_0^t \widehat \varphi_s^{\,(k)}S_{s-}^{(k)}\, d\chi_s^{(k)}.
\end{equation}
\tag{26}
$$
Now, to find $\lambda_0$ we write
$$
\begin{equation*}
\mathbf{E}_{\mathbb{P}}\bigl(\lambda_0^{\gamma-1}( \mathbb{Z}^*_T)^{\gamma}\bigr) = \mathbf{E}_{\mathbb{P}}\bigl(\lambda_0^{\gamma-1}(\mathbb{Z}^*_0 )^{\gamma} \mathbb{Z}_T^{\gamma}(\alpha)\bigr).
\end{equation*}
\notag
$$
We take conditional expectation w.r.t. $\alpha$ in the last mathematical expectation and by Proposition 6, (24), (26) we get that $\mathbf{E}_{\mathbb P} (\lambda_0^{\gamma-1}(\mathbb Z_0^*)^{\gamma }g_0)=x_0$. Then (12) gives
$$
\begin{equation*}
\lambda_0^{\gamma -1}= x_0\biggl( \mathbf{E}_{\mathbb{P}}\biggl( \exp\biggl(\frac{1}{\gamma-1}\sum_{i=1}^N T^{(\alpha)}_j h^{(j),*}(\gamma)\biggr) \biggr) \biggr)^{\gamma-1}.
\end{equation*}
\notag
$$
Further, we calculate $(\lambda_0\mathbb Z_0^*)^{\gamma-1}$ to obtain that it is equal to $x_0/g_0$. Then we separate continuous and purely discontinuous martingale parts in (26) to find that for $0\leqslant t\leqslant T$
$$
\begin{equation*}
\sum_{k=1}^d \int_0^t\biggl( \frac{x_0(\gamma-1)g_s}{g_0} Z_{s-}^{\gamma-1}(\alpha)\beta_s^{*,(k)} -\varphi_s^{(k)}S_{s-}^{(k)}\biggr)\, d\chi^{c,\mathbb{Q}^*,(k)}_s= V_t,
\end{equation*}
\notag
$$
where $V=(V_t)_{0\leqslant t\leqslant T}$ is purely discontinuous $\mathbb{Q}^*$-martingale which is orthogonal to any continuous martingale. This fact implies that the quadratic variation of the mentioned continuous martingale is equal to zero for $0 \leqslant t \leqslant T$:
$$
\begin{equation}
\biggl\langle \sum_{k=1}^d\int_0^t\biggl( \frac{x_0(\gamma-1)g_s}{g_0} Z_{s-}^{\gamma-1}(\alpha) \beta_s^{*,(k)}-\widehat \varphi_s^{\,(k)}S_{s-}^{(k)}\biggr) \, d\chi_s^{c,\mathbb{Q}^*,(k)} \biggr\rangle_t=0.
\end{equation}
\tag{27}
$$
We write the quadratic variation of the left-hand side of (27). Since the density $c_t=(c^{(k,l)}_{t})_{1\leqslant k\leqslant d,\, 1\leqslant l\leqslant d}$ of quadratic variation of a continuous martingale $X^{c,\mathbb{Q}^*}$ is an invertible matrix, we deduce from (27) that, for $1\leqslant k\leqslant d$,
$$
\begin{equation*}
\widehat \varphi_t^{\,(k)} =\frac{x_0(\gamma-1) g_t Z_{t-}^{\gamma-1}(\alpha) \beta_{t}^{*,(k)}}{g_0 S_{t-}^{(k)}}
\end{equation*}
\notag
$$
and this gives the formula for the optimal strategy for power utility.
Using again Proposition 6 and doing simple calculus, we get the formula for the optimal expected utility. The proposition is proved. 4.3. When the utility is exponentialWhen the utility is exponential, $f(x)=x\ln x-x+1$ and $f'(x)=\ln x$. Proposition 9. Suppose that $\mathbf E_{\mathbb P}(|\mathbb Z^*_T \ln \mathbb Z^*_T|) < +\infty$ and that the matrices $c^{(j)}$ are invertible for $1\leqslant j\leqslant N$. Then for an initial capital $x_0 >0$, there exists an asymptotically optimal strategy such that, for $1 \leqslant k \leqslant d$ and $0 \leqslant t \leqslant T$, Proof. In the exponential case, by Proposition 5 and $\mathbf{E}_{\mathbb{P}}(Z_{T-t}(\alpha^t)\,|\,\alpha)\,{=}\,1$, we have
$$
\begin{equation*}
\begin{aligned} \, &\mathbf{E}_{\mathbb P}\bigl( Z_{T-t}(\alpha^t)f'(\lambda_0 x Z_{T-t}(\alpha^t)) \bigm|\alpha\bigr) \\ &\qquad=\ln(\lambda_0x)\, \mathbf{E}_{\mathbb P} \bigl(Z_{T-t}(\alpha^t)\bigm|\alpha\bigr) +\mathbf{E}_{\mathbb P}\bigl(Z_{T-t}(\alpha^t) \ln Z_{T-t}(\alpha^t)\bigm|\alpha\bigr) \\ &\qquad= \ln (\lambda_0 x)+\sum_{j=1}^N \kappa^{(j),*}\int_t^T \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds. \end{aligned}
\end{equation*}
\notag
$$
Hence, by Lemma 1,
$$
\begin{equation}
\begin{aligned} \, &-\mathbf{E}_{\mathbb{Q}^*}\bigl(f'(\lambda_0 \mathbb Z_T^*)\bigm| \widehat{\mathcal F}_t\bigr) =-\ln \lambda_0 -\ln \mathbb Z^*_t -\sum_{j=1}^N \kappa^{(j),*}\int_t^T \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds \nonumber \\ &\qquad= -\ln \lambda_0 -\ln \mathbb Z^*_0 -\ln \mathbb Z_t(\alpha) -\sum_{j=1}^N \kappa^{(j),*}\int_t^T \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds. \end{aligned}
\end{equation}
\tag{28}
$$
As it was mentioned $ Z_t(\alpha)=\exp(M_t+B_t)$ w.r.t. the measure $\mathbb{Q}^*$ and this fact implies Moreover, the right-hand side of (28) should be a $\mathbb{Q}^*$-martingale. Since the processes $B=(B_t)_{t \geqslant 0}$ and $\bigl(\sum_{j=1}^N \int_0^t \mathbf{I}_{\{\alpha_{s-}=j\}}\kappa_s^{(j)}\,ds\bigr)_{t \geqslant 0}$ are predictable processes, we obtain from (16) that
$$
\begin{equation}
\begin{aligned} \, &-\int_0^t {}^{T}\!\beta_s^*dX_s^{c,\mathbb{Q}^*}-\int_0^t\int_{\mathbf R^d} \ln(Y_s^*(x))\, (\mu_X-\nu_X^{\mathbb{Q}^*})(ds,dx) \nonumber \\ &=\sum_{k=1}^d \int_0^t \widehat \varphi_s^{\,(k)}S_{s-}^{(k)}\, d\chi_s^{(k)} \end{aligned}
\end{equation}
\tag{30}
$$
and that
$$
\begin{equation*}
\ln \lambda_0 +\mathbf{E}_{\mathbb{P}}(\ln \mathbb Z_0^*) +\mathbf{E}_{\mathbb{P}}\biggl(\sum_{j=1}^N \kappa^{(j),*}\int_0^T \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds\biggr)=-x_0
\end{equation*}
\notag
$$
with $\mathbb Z_0^*$ defined by (13). Performing the separation of continuous and purely discontinuous martingale part in (30), and using the orthogonality of continuous and purely discontinuous martingales, we deduce that the quadratic variation of continuous martingale part is zero, and, hence, for $0\leqslant t\leqslant T$, Since $c_t=(c^{(k,l)}_{t})_{1\leqslant k\leqslant d,\, 1\leqslant l\leqslant d}$ is an invertible matrix, it proves the formula for the optimal strategy for exponential utility.
Using again Proposition 6 and doing simple calculus, we get the formula for the optimal expected utility in the exponential case. 5. ExampleIn this section, we shall consider one important example when each Lévy process is a Brownian motion with drift. More precisely, where $(W^{(j)})_{1\leqslant j\leqslant N}$ states for independent $d$-dimensional Brownian motions, $(b^{(j)})_{1\leqslant j\leqslant N}$ are $d$-dimensional vectors, and $(\sigma^{(j)})_{ 1\leqslant j\leqslant N}$ are $d\times d$ real-valued matrix. We assume that $\sigma^{(j)}$ is invertible for each $j$. In this case, the set $\mathcal M^{(j)}$ of the equivalent martingale measures consists of only one element with the Girsanov parameter
$$
\begin{equation*}
\beta^{(j),*}= -\bigl({}^{\top}\!\sigma^{(j)}\sigma^{(j)}\bigr)^{-1}b^{(j)}.
\end{equation*}
\notag
$$
Then we see that the derivatives of the Hellinger process and of the Kulback–Leibler process are
$$
\begin{equation*}
h^{(j),*}(\gamma)=\frac{\gamma(1-\gamma)}{2} \bigl\langle \bigl({}^{\top}\!\sigma^{(j)}\sigma^{(j)}\bigr)^{-1}b^{(j)}, b^{(j)}\bigr\rangle
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\kappa^{(j),*}=\frac{1}{2}\big\langle \bigl({}^{\top}\!\sigma^{(j)}\sigma^{(j)}\bigr)^{-1}b^{(j)}, b^{(j)}\big\rangle.
\end{equation*}
\notag
$$
For the power utility from Proposition 8 we deduce that the optimal investment strategy $\widehat \varphi$ is given by the formulae, for $1\leqslant k\leqslant d$,
$$
\begin{equation*}
\widehat \varphi_t^{\,(k)} =\frac{x_0(\gamma-1) Z_{t-}^{\gamma-1}(\alpha) \beta_{t}^{*,(k)} \exp\bigl(-\sum_{j=1}^N h^{(j),*}(\gamma)\int_0^t \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds \bigr)}{S_{t-}^{(k)}},
\end{equation*}
\notag
$$
where $\beta^*$ is defined by formula (19). Moreover, we have
$$
\begin{equation}
(\lambda _0\mathbb{Z}^*_T)^{\gamma-1}=x_0+\sum_{k=1}^d\int_0^T \widehat{\phi}^{\,(k)}_s \, dS^{(k)}_s.
\end{equation}
\tag{31}
$$
Denote by $V^{\widehat \varphi}=(V_t^{\widehat \varphi})_{0 \leqslant t \leqslant T}$ the corresponding value process of the optimal portfolio. Then
$$
\begin{equation*}
V_T^{\widehat \varphi}=(\lambda _0\mathbb{Z}^*_T)^{\gamma-1}.
\end{equation*}
\notag
$$
Since the stochastic integral in (31) is a $\mathbb{Q}^*$-martingale, we have
$$
\begin{equation*}
V_t^{\widehat \varphi}=\mathbf E_{\mathbb{Q}^*}\bigl((\lambda_0\mathbb Z_T^*)^{\gamma-1}\bigm| \widehat{\mathcal F}_t\bigr) =\lambda_0^{\gamma-1}\mathbb Z_t^{*,\gamma-1}\mathbf E_{\mathbb P} \bigl(Z^{\gamma}_{T-t}(\alpha^t)\bigm|\alpha\bigr).
\end{equation*}
\notag
$$
In the proof of Proposition 8, we have already showed that
$$
\begin{equation*}
V_t^{\widehat \varphi}=x_0Z_{t}^{\gamma-1}(\alpha)\exp\biggl(-\sum_{j=1}^N h^{(j),*}(\gamma)\int_0^t \mathbf{I}_{\{\alpha_{s-}=j\}}\, ds\biggr).
\end{equation*}
\notag
$$
Thus, the optimal strategy can also be written as
$$
\begin{equation*}
\widehat \varphi^{\,(k)}_t =\frac{(\gamma-1)\beta_t^{*,(k)}V_{t-}^{\widehat \varphi}}{S_{t-}^{(k)}}.
\end{equation*}
\notag
$$
We can also rewrite the optimal value process for the logarithmic utility in a similar form as
$$
\begin{equation*}
\widehat \varphi^{\,(k)}_t =\frac{\beta_t^{*,(k)}V_{t-}^{\widehat \varphi}}{S_{t-}^{(k)}}.
\end{equation*}
\notag
$$
In fact, from the discussion above, we see that the same is true for the general case where the Lévy processes admits jumps but have a nondegenerated diffusion.
Цитирование в формате AMSBIB
\RBibitem{DonVos24} |
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