Three Essays on Exponential Hedging with Variable Exit Times

Abstract

This paper addresses three main problems that are intimately related to exponential hedging with variable exit times. The first problem consists of explicitly parameterizing the exponential forward performances and describing the optimal solution for the corresponding utility maximization problem. The second problem deals with the horizon-unbiased exponential hedging. Precisely, we are interested in describing the dynamic payoffs for which there exists an admissible strategy that minimizes the risk—in the exponential utility framework—whenever the investor exits the market at stopping times. Furthermore, we explicitly describe this optimal strategy when it exists. Our last contribution is concerned with the optimal sale problem, where the investor is looking simultaneously for the optimal portfolio and the optimal time to liquidate her assets.

Appendices

Appendix 1: Some Auxiliary Lemmas

This section is contains six lemmas, which were used in previous sections. We note that some of these lemmas are interesting on their own right.

Lemma 5

Let Q be a σ-martingale measure for S, and θ∈L(S) be such that

(40)

Then the process θ⋅S is a Q-local martingale and the process exp[θ⋅S] is a positive Q-submartingale.

Proof

Since Q is a σ-martingale measure for S, there exists a positive, bounded and predictable process ϕ such that ϕ⋅S is a Q-local martingale. As a result, θ⋅S is σ-martingale under Q. On the other hand, it is clear that

$$X_{t}:=\exp\left(\frac{1}{2}(\theta\cdot S)_{t}\right) $$

is a positive special semimartingale under Q with the Doob–Meyer decomposition

$$ X=X_{0}+{\overline{N}}+{\overline{B}} $$

where ${\overline{N}}$ is a Q-local martingale and ${\overline{B}}$ is a predictable process with finite variation such that ${\overline{N}}_{0}={\overline{B}}_{0}=0$. Let (T _n ) _n≥1 be a sequence of stopping times that increases stationarily to T and

$$ E^Q\left([{\overline{N}},{\overline{N}}]^{1/2}_{T_{n}}+{\rm Var}_{T_{n}}{\overline{B}}\right)<\infty. $$

For any predictable process φ such that |φ|≤1, we have:

$$ E^Q|\varphi\cdot X_{T_{n}}|\leq cE^Q\left([{\overline{N}},{\overline{N}}]^{1/2}_{T_{n}}+{\rm Var}_{T_{n}}({\overline{B}})\right), $$

(41)

where c is a constant that does not depend on φ.

Using Ito’s formula, we obtain

$$X=1+\frac{1}{2}X_{-}\cdot(\theta\cdot S)+X_{-}\cdot V(\theta), $$

where V(θ) is a non-decreasing process given by

$$ V(\theta):=\frac{1}{8}\theta^{T}c\theta\cdot A+\left(\exp\left(\frac{1}{2}\theta^{T}x\right)-1-\frac{1}{2}\theta ^{T}x\right)\star\mu. $$

Since θ⋅S is a σ-martingale under Q, then there exists a predictable process ϕ with values in the interval (0,1] such that ϕθ⋅S is a Q-local martingale. Consider a sequence of stopping times, (σ _n ) _n≥1 increasing stationarily to T such that $(\phi\theta\cdot S)^{\sigma_{n}}$ is a true Q-martingale. Then, for any ε>0, the process

$$\left(\frac{\phi}{\phi+\varepsilon+\varepsilon X_{-}}X_{-}\theta\cdot S\right)^{\sigma_{n}} $$

is also a true Q-martingale. As a result,

$$\begin{aligned} E^Q\int_{0}^{\sigma_{n}\wedge T_{n}}X_{s-}\mathrm{d}V_{s}(\theta) =&\lim_{\varepsilon\downarrow0} E^Q\int_{0}^{\sigma_{n}\wedge T_{n}}\frac{\phi_{s}}{\phi_{s}+\varepsilon+\varepsilon X_{s-}}X_{s-}\mathrm{d}V_{s}(\theta) \\ =&\lim_{\varepsilon\downarrow 0}E^Q\left(\frac{\phi}{\phi+\varepsilon+\varepsilon X_{-}}\cdot X_{\sigma_{n}\wedge T_{n}}\right)<\infty. \end{aligned}$$

The first equality follows from the monotone convergence theorem, while the finiteness of the last quantity is due to (41).

Hence, V(θ) is Q-locally integrable and thus (θ⋅S) is Q-locally integrable. This proves that (θ⋅S) is really a Q-local martingale. Furthermore, $\exp\bigl(\frac{1}{2}\theta \cdot S\bigr)$ is a positive Q-local submartingale. Then, the condition (40) and de la Vallée Poussin’s argument imply that $\exp\bigl(\frac{1}{2}\theta\cdot S\bigr)$ is a positive Q-submartingale which is square integrable. Now the lemma follows from Jensen’s inequality. □

For a random variable H, we denote by ${\widetilde{Q}}^{(H)}$ the minimal entropy martingale measure for S with respect to P ^(H):=e^H(E(e^H))⁻¹⋅P. Also, Θ ₁ denotes the set of strategies considered in [6]:

Lemma 6

Suppose that S is locally bounded and . Let H be a random variable bounded from below with

$$ E\left[\mathrm{e}^{pH}\right]<\infty, $$

for some p∈(1,∞), and let $\widetilde{\theta}\in\varTheta_{1}$. Then the assertions (i) and (ii) are equivalent:

(i)

$$1-u_0:=\inf_{\theta\in\varTheta_1}E\exp\bigl(H-(\theta\cdot S)_T\bigr)=E\exp\bigl(H-(\widetilde{\theta}\cdot S)_T\bigr). $$

(ii) For any stopping time σ≤T, we have

Proof

Using the results in [6], we change the probability and work under Q instead of P, where

$$Q:=\frac{\exp(H)}{E[\exp(H)]} P. $$

Suppose that the assertion (i) holds. Putting

(42)

where

we obtain the existence of $\underline{\xi}$ that belongs to the set

$$\varXi:=\Big\{ \xi>0:\ E(\xi)=1,\ \ E(\xi\eta)=0,\ \ \mbox{for any}\ \ \eta:=(\theta\cdot S)_T,\ \theta\in\varTheta_1\Big\} $$

and satisfies

$$J_0=\min_{\xi\in\varXi}E^{Q}(\xi\log\xi)=E^{Q}(\underline{\xi}\log\underline{\xi}). $$

Thus, Theorem 3.5 of [17] implies that

$$ \underline{\xi}=\exp\bigl(-\log\bigl(E^Q\mathrm{e}^{-(\widetilde{\theta}\cdot S)_T}\bigr)-\widetilde{\theta}\cdot S_{T}\bigr)\quad \mbox{and} \quad u_0=1-\mathrm{e}^{-J_0}. $$

(43)

It is clear that the set is stable under concatenation (for more detail about this see [17]), and due to Proposition 4.1 in [17] we conclude that the optimizer of J _t is given by . Denoting $P^{*}:=Z^{*}_{T} Q$ and using the first equation in (43), we derive that

Equivalently, we have

$$ {{Z^{*}_{T}}\over{Z^{*}_{\sigma}}}=\exp\left[-\int_{\sigma}^T \widetilde{\theta}_u \mathrm{d}S_u+J_{\sigma}\right]. $$

(44)

Due to Young’s inequality (xy≤e^x+ylogy−y), we obtain that

$$\left(-\int_{\sigma}^{T}\theta_{u}dS_{u}\right)\left(\frac {Z^{*}_{T}}{Z^{*}_{\sigma}} \mathrm{e}^{-J_{\sigma}}\right) \leq \mathrm{e}^{-\int _{\sigma}^{T}\theta_{u}\mathrm{d}S_{u}}+\frac{Z^{*}_{T}}{Z^{*}_{\sigma}} \mathrm{e}^{-J_{\sigma}}\log\left(\frac{Z^{*}_{T}}{Z^{*}_{\sigma}} \mathrm{e}^{-J_{\sigma }}\right) -\frac{Z^{*}_{T}}{Z^{*}_{\sigma}} \mathrm{e}^{-J_{\sigma}}. $$

Therefore, by taking conditional expectation on both sides, and using the equalities

and (44), we derive that

Since there is equality for $\theta=\widetilde{\theta}$, due to

$$\frac{Z^{*}_{T}}{Z^{*}_{\sigma}} \mathrm{e}^{-J_{\sigma}}=\exp\left(-\int_{\sigma }^{T}\widetilde{\theta}_{u}\mathrm{d}S_{u}\right), $$

assertion (ii) follows. The converse is immediate by putting σ=0. This ends the proof of the lemma. □

Lemma 7

Let Z be a given positive local martingale such that ZlogZ is locally integrable. There is a RCLL semimartingale X such that ZX is a local martingale and

$$\log Z=X+h^{E}(Z,P). $$

Proof

Since Z is a positive local martingale, there exists a local martingale N such that N ₀=0 and . Due to Ito’s formula, we deduce that

$$ \begin{array}{l}\log Z=\displaystyle{\left(N-\frac{1}{1+\Delta N}\cdot[N,N]\right) +\frac{1}{2}\langle N^{c}\rangle+\sum\frac{(1+\Delta N)\log (1+\Delta N)-\Delta N}{1+\Delta N}}. \end{array} $$

Next, we note that h ^E(Z,P) is the compensator of the process (1+ΔN)⋅V, where $Y:=N-\frac{1}{1+\Delta N}\cdot[N,N]$ and V:=log(Z)−Y. Again Ito’s formula implies that

and

where ${\overline{V}}:=V- h^{E}(Z,P)$. Thus, the conclusion follows immediately. □

Lemma 8

Consider a positive σ-martingale density with

$$ {N}={\lambda}\cdot S^{c}+{W}\star(\mu-\nu),\,\,{W}_{t}(x)= \bigl(\mathrm{e}^{{\lambda}_t^{T}x}-1\bigr)\Bigl(1-a_{t}+\int \mathrm{e}^{{\lambda}_t^{T}x}\nu(\{t\},\mathrm{d}x)\Bigr)^{-1}. $$

(45)

Then

$$\begin{aligned} h^{E}({Z},P)&=\frac{1}{2}{\lambda}^{T}c{\lambda}\cdot A+\phi({\gamma }^{-1}\mathrm{e}^{\lambda^T x}-1)\star\nu+\sum(1-a)\phi(\gamma^{-1}-1) \end{aligned}$$

(46)

$$\begin{aligned} &=\frac{1}{2}{\lambda}^{T}c{\lambda}\cdot A+{1\over{\gamma}}\phi\left(\mathrm{e}^{{\lambda}^{T}x}-1\right)\star\nu- \sum{1\over{\gamma}}\left[{\gamma}\log\left({\gamma}\right)-{\gamma }+1\right], \end{aligned}$$

(47)

$$\begin{aligned} \Delta h^{E}({Z},P)&=-\log\gamma, \end{aligned}$$

(48)

where ${\gamma}_{t}:=1-a_{t}+\int \mathrm{e}^{{\lambda}_{t}^{T}x}\nu(\{t\},\mathrm{d}x)$ and ϕ(z):=(1+z)log(1+z)−z.

Proof

Notice that h ^E(Z,P) is the compensator of V ^E(N), where

$$ V^{E}\left({N}\right)=\frac{1}{2}\langle {N}^{c}\rangle+\sum\left[\left(1+\Delta{N} \right)\log\left(1+\Delta{N}\right)-\Delta{N}\right]. $$

(49)

From (45) we derive that

$$1+\Delta{N}_{t}=\frac{\mathrm{e}^{{\lambda}_t^{T}\Delta S_{t}}}{{\gamma}_{t}}I_{\{\Delta S_{t}\neq 0\}}+\frac{1}{{\gamma}_{t}}I_{\{\Delta S_{t}=0\}}. $$

After simplification, this leads to the identity

$$\begin{aligned} & \sum\left[\left(1+\Delta{N} \right)\log\left(1+\Delta{N}\right)-\Delta N\right]\\ &\quad =\phi\left ({\gamma}^{-1}\mathrm{e}^{{\lambda}^{T}x}-1\right)\star\mu+ \sum\phi\left({\gamma}^{-1}-1\right)I_{\{\Delta S=0\}}. \end{aligned}$$

By plugging this representation into (49) and compensating, we obtain (46).

Inserting the expression

$$\phi\left({\gamma}^{-1}\mathrm{e}^{{\lambda}^{T}x}-1\right)\star\nu=\gamma ^{-1}\phi\left(\mathrm{e}^{{\lambda}^{T}x}-1\right)\star\nu+\sum{{(1-a-\gamma )\log(\gamma)+(\gamma-1)a}\over{\gamma}} $$

into (46) and simplifying, we get (47).

Calculating the jumps in both sides of (47), we have

$$\begin{aligned} \Delta h^{E}\left({Z},P\right)&=\frac{1}{ {\gamma}}\left(a-\int \mathrm{e}^{{\lambda}^{T}x}F(\mathrm{d}x)\Delta A\right)-\frac{{\gamma}\log\left ({\gamma}\right)-{\gamma}+1}{{\gamma}}\\ &=\frac{-{\gamma}+1}{{\gamma}}-\frac{{\gamma}\log\left ({\gamma}\right)-{\gamma}+1}{{\gamma}} =-\log\left({\gamma}\right). \end{aligned}$$

Note that the first equality follows because

$$\displaystyle\int x\mathrm{e}^{{\lambda}^{T}x}F(\mathrm{d}x)\Delta A=\displaystyle \int x\mathrm{e}^{{\lambda}^{T}x}\nu(\{.\},\mathrm{d}x)=0 $$

in virtue of the fact that Z is a σ-martingale density for S. This ends the proof. □

Lemma 9

Suppose that (3) holds. Then the function

$${\overline{K}}(\lambda):=b^{T}\lambda+\frac{1}{2}\lambda^{T}c\lambda+\int \left(\mathrm{e}^{\lambda^{T}x}-1-\lambda^{T}h(x)\right)F(\mathrm{d}x),\quad\lambda\in \mathbb{R}^d, $$

is convex, proper, closed, and continuously differentiable with

$$\nabla{\overline{K}}(\lambda)=b+c\lambda+\int\left(x\mathrm{e}^{\lambda ^{T}x}-h(x)\right)F(\mathrm{d}x),\quad\lambda\in\mathbb{R}^d. $$

Proof

The proof of this lemma is obvious. For the definitions of proper and closed convex functions, we refer the reader to [25]. □

Lemma 10

The following assertions are equivalent:

(i) For any $\lambda\in\mathbb{R}^{d}$,

$$ \int_{\{\vert x\vert>1\}}\mathrm{e}^{\lambda^T x} F(\mathrm{d}x)<\infty. $$

(ii) For any $\lambda\in\mathbb{R}^{d}$,

$$ \int_{\{\vert x\vert>1\}}\vert x\vert \mathrm{e}^{\lambda^T x} F(\mathrm{d}x)<\infty. $$

As a result, if (i) holds, then for any $\lambda\in\mathbb{R}^{d}$ and q∈(0,∞),

$$ \int_{\{\vert x\vert>1\}}\vert x\vert^q \mathrm{e}^{\lambda^T x} F(\mathrm{d}x)<\infty. $$

Proof

The implication (ii)⟹(i) is obvious. We focus on proving the reverse. Let e _i be the element of $\mathbb{R}^{d}$ that has the i ^th component equal to one and the other components null. Due to the equivalence of norms in $\mathbb{R}^{d}$, we may work with the norm $\vert x\vert=\sum_{i=1}^{d} \vert x_{i}\vert$. We get that

$$\begin{aligned} \int_{\{\vert x\vert>1\}}\vert x\vert \mathrm{e}^{\lambda^T x} F(\mathrm{d}x)&=\sum_{i=1}^d \int_{\{\vert x\vert>1\}} \left((e_i^T x)^+ +(-e_i^T x)^+\right) \mathrm{e}^{\lambda^T x} F(\mathrm{d}x)\\ & \leq \sum_{i=1}^d \int_{\{\vert x\vert >1\}}\mathrm{e}^{(e_i+\lambda)^T x} F(\mathrm{d}x)+\sum_{i=1}^d\int_{\{\vert x\vert >1\}}\mathrm{e}^{(-e_i+\lambda)^T x} F(\mathrm{d}x). \end{aligned}$$

Due to (i) the last term in the rhs of the above string is finite for any $\lambda\in\mathbb{R}^{d}$. The proof of the remaining part of the lemma follows by the same arguments. □

Appendix 2: MEH σ-Martingale Density Under Change of Probability

In this section, we focus on describing the MEH σ-martingale density when we change probability. This case can be derived easily from a more general case where one works with respect to a positive local martingale density, Z, that may not be uniformly integrable. First, we generalize the characterization of the MEH σ-martingale density for the case when S may not be bounded nor quasi-left continuous. For the case of bounded and quasi-left continuous S, a more elaborate result is given in [3].

Theorem 9

Suppose that and (3) holds. If is the MEH σ-martingale density then there exists $\widetilde{H}\in L(S)$ such that

$$ \log(\widetilde{Z})=\widetilde{H}\cdot S+h^E(\widetilde{Z},P). $$

(50)

Furthermore, $\widetilde{H}$ can be described as root of the equation

$$ 0=\left\{ \begin{array}{l@{\quad}l} b+c\theta+\displaystyle \int\bigl[\mathrm{e}^{\theta^T x}x-h(x)\bigr] F(\mathrm{d}x),& \mathit{on}\ \{\Delta A=0\}\\ \displaystyle\int \mathrm{e}^{\theta^T x}x F(\mathrm{d}x),& \mathit{on}\ \{\Delta A\not=0\}. \end{array} \right. $$

(51)

Proof

Notice that the assumptions of Theorem 3.3 in [4] are fulfilled. Hence, a direct application of this theorem implies that $\widetilde{Z}$ is given by

where $\widetilde{\beta}$ is a root of (51). Therefore, in the remaining part of this proof we will focus on showing (50). Thus,

$$\begin{aligned} \log(\widetilde{Z})&=\widetilde{N}-\frac{1}{2}\langle \widetilde{N}\rangle+\sum[\log(1+\Delta \widetilde{N})-\Delta\widetilde{N}]\\ & =\widetilde{\beta}\cdot S^{c}+\widetilde{W}\star(\mu-\nu)-\frac{1}{2}\widetilde{\beta }^{T}c\widetilde{\beta}\cdot A+\sum\bigg[\log\frac{\mathrm{e}^{\widetilde{\beta}^{T}x}}{\widetilde{\gamma }}-\frac{\mathrm{e}^{\widetilde{\beta}^{T}x}}{\widetilde{\gamma}}+1\bigg]I_{\{ \Delta S\neq0\}}\\ &\quad {}+\sum\bigg[\log\frac{1}{\widetilde{\gamma}}-\frac {1}{\widetilde{\gamma}}+1\bigg]I_{\{\Delta S=0\}}\\ &=\widetilde{\beta}\cdot S^{c}+\widetilde{W}\star(\mu-\nu)-\frac{1}{2}\widetilde{\beta }^{T}c\widetilde{\beta}\cdot A+\sum\bigg[\frac{-\widetilde{\gamma}\log\widetilde{\gamma}+\widetilde {\gamma}-1}{\widetilde{\gamma}}\bigg]\\ &\quad {}+ \frac{\widetilde{\gamma}\widetilde{\beta}^{T}x-\mathrm{e}^{\widetilde{\beta }^{T}x}+1}{\widetilde{\gamma}}\star\mu. \end{aligned}$$

Note that

$$\begin{aligned} & {1\over{\widetilde{\gamma}}}\bigl(\widetilde{\gamma}\widetilde {\beta}^{T}x-\mathrm{e}^{\widetilde{\beta}^{T}x}+1\bigr)\star\mu\\ &\quad = \widetilde{\beta}^{T}\bigl(x-h(x)\bigr)\star\mu+{1\over{\widetilde{\gamma }}}\bigl(\widetilde{\gamma}\widetilde{\beta}^{T}h(x)-\mathrm{e}^{\widetilde{\beta }^{T}x}+1\bigr)\star(\mu-\nu)\\ &\qquad {}+ {\widetilde{\gamma}}^{-1}\bigl(\widetilde{\gamma}\widetilde{\beta}^{T}h(x)-\mathrm{e}^{\widetilde{\beta}^{T}x}+1\bigr)\star\nu, \end{aligned}$$

since the function ${\widetilde{\gamma}}^{-1}(\widetilde{\gamma}\widetilde{\beta }^{T}h(x)-\mathrm{e}^{\widetilde{\beta}^{T}x}+1)$ is (μ−ν)-integrable which is due to the (μ−ν)−integrability of ${\widetilde{\gamma}}^{-1}(\mathrm{e}^{\widetilde{\beta}^{T}x}-1)=W(x)$ and the boundedness of h(x). Therefore, we get that

$$\begin{aligned} \log(\widetilde{Z})&=\widetilde{\beta}\cdot S^{c}+\widetilde{\beta}^{T}h(x)\star(\mu-\nu)+\widetilde{\beta }^{T}(x-h(x))\star\mu\\ &\quad {}+{\widetilde{\gamma}}^{-1}\bigl(\widetilde{\gamma}\widetilde{\beta }^{T}h(x)-\mathrm{e}^{\widetilde{\beta}^{T}x}+1\bigr)\star\nu-\frac{1}{2}\widetilde {\beta}^{T}c\widetilde{\beta}\cdot A\\ &\quad {}+\sum{\widetilde{\gamma}}^{-1}(-\widetilde{\gamma}\log(\widetilde {\gamma})+\widetilde{\gamma}-1). \end{aligned}$$

Equivalently, we deduce that

$$ \log\widetilde{Z}=\widetilde{\beta}\cdot S+\frac{1}{2}\widetilde{\beta}^{T}c\widetilde{\beta}\cdot A+ \frac{\widetilde{\beta}^{T}x\mathrm{e}^{\widetilde{\beta}^{T}x}-\mathrm{e}^{\widetilde {\beta}^{T}x}+1}{\widetilde{\gamma}}\star\nu+ \sum{\widetilde{\gamma}}^{-1}(-\widetilde{\gamma}\log(\widetilde{\gamma })+\widetilde{\gamma}-1), $$

since

$$\widetilde{\beta}\cdot S=\widetilde{\beta}\cdot S^{c}+\widetilde{\beta}^{T}b\cdot A+\widetilde{\beta}^{T}h(x)\star(\mu-\nu)+\widetilde{\beta }^{T}(x-h(x))\star\mu. $$

Therefore, by a direct application of Lemma 8 for $\lambda=\widetilde{\beta}$, (50) follows immediately from putting $\widetilde{H}=\widetilde{\beta}$. This ends the proof of the theorem. □

In what follows, we denote by a positive local martingale with

$$ N:=\beta\cdot S^{c}+W\star(\mu-\nu)+g\star\mu+\overline{N},\qquad W_{t}(x):=f_{t}(x)+\frac{\widehat{f_{t}}}{1-a_{t}}I_{\{a_{t}<1\}}, $$

(52)

where $\left(\beta,f,g,\overline{N}\right)$ are Jacod’s components of N. Here, we define:

where is given by (2).

Theorem 10

Consider Z defined in (52) and suppose that

Then the minimization problem

(53)

admits a solution given by

$$\widetilde{N}=\widetilde{\beta}\cdot S^{c,Z}+\widetilde{W}\star(\mu-\nu ^{Z}),\qquad \widetilde{W}_{t}(x)=\frac{\mathrm{e}^{\widetilde{\beta }_t^{T}x}-1}{1-a_{t}^{Z}+\int \mathrm{e}^{\widetilde{\beta}_t^{T}y}\nu^{Z}(\{t\},\mathrm{d}y)}, $$

where $\widetilde{\beta}$ is the root of the equation

$$ 0=b^{Z}+c\lambda+\int(\mathrm{e}^{\lambda^{T}x}x-h(x))F^{Z}(\mathrm{d}x). $$

(54)

Here S ^c,Z, b ^Z, a ^Z, ν ^Z and F ^Z are given by

$$\begin{aligned} S^{c,Z}&:=S^{c}-c\beta\cdot A,\qquad b^{Z}:=b+c\beta-\int f(x)h(x)F(\mathrm{d}x),\\ a^{Z}_{t}&:=\nu^{Z}(\{t\},\mathbb{R}^{d}\setminus\{0\}) \end{aligned}$$

and

$$\nu^{Z}(\mathrm{d}t,\mathrm{d}x):=F^{Z}_t(\mathrm{d}x)\mathrm{d}A_{t},\qquad F^{Z}_{t}(\mathrm{d}x):=(1+f_{t}(x))F_{t}(\mathrm{d}x). $$

Proof

Consider a sequence of stopping times, (T _n ) _n≥1, stationarily increasing to T and such that $Z^{T_{n}}$ is a true martingale. For a fixed but arbitrary n we denote $Q:=Z_{T_{n}}\cdot P$. Note that all the equations in the theorem are robust under stopping. Due to Lemma 1, it is sufficient to prove that the theorem holds on 〚0,T _n 〛. Then, we obtain that $\nu^{Q}(\mathrm{d}t,\mathrm{d}x)=(1+f_{t}(x))I_{\{t\leq T_{n}\}}\nu(\mathrm{d}t,\mathrm{d}x)$, , and $\int_{\{\vert x\vert>1\}}\mathrm{e}^{\lambda^{T}x}F^{Q}(\mathrm{d}x)<\infty$ for $\lambda\in\mathbb{R}^{d}$.

Therefore, the assumptions of Theorem 3.3 in [4] are fulfilled. By direct application of this theorem for $S^{T_{n}}$ and under the measure $Q=Z_{T_{n}}\cdot P$, we deduce that the problem defined in (53) admits a solution , where $\widetilde{N}^{Q}$ is given, on 〚0,T _n 〛, by

$$\widetilde{N}^{Q}=\widetilde{\beta}\cdot S^{c,Q}+\widetilde{W}\star(\mu -\nu^{Q}),\qquad \widetilde{W}_{t}(x)=\frac{\mathrm{e}^{\widetilde{\beta }_t^{T}x}-1}{1-a_{t}^{Q}+\int \mathrm{e}^{\widetilde{\beta}_t^{T}y}\nu^{Q}(\{t\},\mathrm{d}y)}. $$

Herein S ^c,Q is the continuous local martingale part of S under Q and ν ^Q is the Q-compensator measure of μ, and $a^{Q}_{t}=\nu^{Q}(\{t\},\mathbb{R}^{d}\setminus\{ 0\})$. Moreover, $\widetilde{\beta}$ is given by the equation

(55)

It is then clear that $\widetilde{N}^{Q}$ coincides with $\widetilde{N}^{T_{n}}$ of the theorem and that the equation (55) is exactly the equation (54) on 〚0,T _n 〛. This ends the proof of theorem. □

Theorem 11

Let Z be a positive local martingale and let . If the assumptions of Theorem 10 are fulfilled and $\widetilde{Z}$ is the MEH local martingale density with respect to Z, then

$$ \log\widetilde{Z}=\widetilde{\beta}\cdot S+h^E(\widetilde{Z}, Z) $$

and $\widetilde{\beta}$ is a root of (54).

Proof

The proof of this theorem follows from the same arguments as in the proofs of Theorems 9 and 10. □