Erratum to: A Donsker Delta Functional Approach to Optimal Insider Control and Applications to Finance

1 Erratum to: Commun. Math. Stat. (2015) 3:365–421 DOI 10.1007/s40304-015-0065-y

Unfortunately, in the original publication of the article some equations appeared incorrectly. The corrected equations are given below:

• In (1.1) it should be added that the filtration ${\mathbb {H}}$ is assumed to be right-continuous, i.e., that

  $$\begin{aligned} {\mathcal {H}}_{t}= {\mathcal {H}}_{t^+}=\bigcap _{s>t}{\mathcal {H}}_s. \end{aligned}$$

• The third line of (1.4) should read

  $$\begin{aligned} \gamma (t,x,u,y,\zeta )=\gamma (t,x,u,y,\zeta ,\omega ):[0,T_0]\times \mathbb {R}\times \mathbb {R}\times \mathbb {R}\times \mathbb {R}\times \Omega \mapsto \mathbb {R} \end{aligned}$$

• In (2.5), which reads

  $$\begin{aligned} Y= & {} Y (T_0); \text { where } Y (t)\\= & {} \int _0^t\beta (s)\mathrm{d}B(s)+\int _0^t\int _{{\mathbb {R}}}\psi (s,\zeta ){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta ), \text{ for } t\in [0,T_0], \end{aligned}$$

  we should add the assumption that $\beta \ne 0$.

• In the formulas (2.7)–(2.14) and in (6.107), (6.109), we should replace $\beta$ by $i\beta$, where $i=\sqrt{-1}$.

  When this is done, the correct version of (2.7) becomes

  $$\begin{aligned} \delta _Y(y)= & {} \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\left. +\, \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \mathrm{d}x, \end{aligned}$$

  and similarly, the correct versions of (2.8)–(2.14) become as follows:

  $$\begin{aligned}&{\mathbb {E}}[\delta _Y(y)|{\mathcal {F}}_t]\nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}}{\mathbb {E}} \left[ \exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \right. \nonumber \\&\qquad \left. \left. +\, \int _0^{T_0} \left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] |{\mathcal {F}}_t\right] \mathrm{d}x \nonumber \\&\quad =\frac{1}{2\pi } \int _{{\mathbb {R}}} \exp ^{\diamond }\left[ {\mathbb {E}}\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \right. \nonumber \\&\qquad \left. \left. +\, \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta )) \nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy |{\mathcal {F}}_t\right] \right] \mathrm{d}x \nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}}\exp ^{\diamond }\left[ \int _0^t\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^tix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\qquad +\, \left. \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \mathrm{d}x \nonumber \\&\quad = \frac{1}{2\pi }\int _{{\mathbb {R}}}\left\{ \exp ^{\diamond }\left[ \int _0^t\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )\right] \right\} \diamond \left\{ \exp ^{\diamond }\left[ \int _0^t ix\beta (s)\mathrm{d}B(s)\right] \right\} \nonumber \\&\qquad \diamond \left\{ \exp ^{\diamond }\left[ \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \right\} \mathrm{d}x \nonumber \\&\quad = \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp \left[ \int _0^t\int _{{\mathbb {R}}}ix\psi (s,\zeta ){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta ) +\int _0^t ix\beta (s)\mathrm{d}B(s)\nonumber \right. \\&\qquad \left. +\,\int _t^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )} -1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )\mathrm{d}s-\int _t^{T_0}\frac{1}{2}x^2\beta ^2 (s)\mathrm{d}s-ixy\right] \mathrm{d}x\nonumber \\ \end{aligned}$$

  (2.8)

  Here we have used that (see e.g. [11, Lemma 3.1])

  $$\begin{aligned} \exp ^{\diamond }\left[ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s)\right] =\exp \left[ \int _0^{T_0}ix\beta (s) \mathrm{d}B(s) +\frac{1}{2}\int _0^{T_0}x^2 \beta ^2(s) \mathrm{d}s \right] \end{aligned}$$

  (2.9)

  and

  $$\begin{aligned}&\exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )\right] \nonumber \\&\quad = \exp \left[ \int _0^{T_0}\int _{{\mathbb {R}}}ix\psi (s,\zeta ){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )- \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta ) \right] \end{aligned}$$

  (2.10)

  We proceed to find

  $$\begin{aligned} {\mathbb {E}}[D_{t,z}\delta _Y(y)| {\mathcal {F}}_t], \end{aligned}$$

  where $D_{t,\zeta }$ denotes the Hida–Malliavin derivative at $(t,\zeta ) \in [0,T] \times {\mathbb {R}}$ with respect to the Poisson random measure N:

  First, note that

  $$\begin{aligned} D_{t,z}\delta _Y(y)= & {} \frac{1}{2\pi }\int _{{\mathbb {R}}} D_{t,z} \exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s)\right. \nonumber \\&\left. +\,\int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \mathrm{d}x \nonumber \\= & {} \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\left. +\,\int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \nonumber \\&\diamond \, D_{t,z} \left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\left. +\,\int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \mathrm{d}x \nonumber \\= & {} \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\left. +\, \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \nonumber \\&\times \,(\mathrm{e}^{ix\psi (t,z)}-1)\mathrm{d}x. \end{aligned}$$

  (2.11)

  Here we have used that

  $$\begin{aligned} D_{t,z} \int _0^T \beta (s) \mathrm{d}B(s) = 0, \end{aligned}$$

  which follows from our assumption that B and ${\tilde{N}}$ are independent, so for $D_{t,z}$ the random variable B(s) is like a constant.

  Using Eq. (2.11) and the Wick chain rule we get

  $$\begin{aligned}&{\mathbb {E}}[D_{t,z}\delta _Y(y)|{\mathcal {F}}_t]\nonumber \\&\quad = \frac{1}{2\pi }\int _{{\mathbb {R}}}{\mathbb {E}} \left[ \exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \right. \nonumber \\&\qquad \left. +\, \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )} -1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \nonumber \\&\qquad \left. \,\times \,(\mathrm{e}^{ix\psi (t,z)}-1)\mathrm{d}x|{\mathcal {F}}_t\right] \mathrm{d}x \nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}}{\mathbb {E}} \left[ \exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta ) + \int _0^{T_0}ix\beta (s)\mathrm{d}B(s)\right. \right. \nonumber \\&\qquad \left. \left. +\, \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )} -1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] |{\mathcal {F}}_t\right] \nonumber \\&\qquad \times \, (\mathrm{e}^{ix\psi (t,z)}-1)\mathrm{d}x\nonumber \\&\quad = \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp \left[ \int _0^t\int _{{\mathbb {R}}}ix\psi (s,\zeta ){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta ) +\int _0^t ix\beta (s)\mathrm{d}B(s)\right. \nonumber \\&\qquad \left. +\,\int _t^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )\mathrm{d}s-\int _t^{T_0}\frac{1}{2}x^2\beta ^2(s)\mathrm{d}s-ixy\right] \nonumber \\&\qquad \times (\mathrm{e}^{ix\psi (t,z)}-1)\mathrm{d}x. \end{aligned}$$

  (2.12)

  Next we want to find

  $$\begin{aligned} {\mathbb {E}}[D_{t}\delta _Y(y)| {\mathcal {F}}_t], \end{aligned}$$

  where $D_t$ denotes the Hida–Malliavin derivative at t with respect to Brownian motion B:

  Note that:

  $$\begin{aligned}&D_{t}\delta _Y(y)\nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}} D_{t} \exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s)\right. \nonumber \\&\qquad \left. +\,\int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta ) -\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \mathrm{d}x \nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}}\exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\qquad \left. +\,\int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \nonumber \\&\qquad \diamond \, D_{t} \left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\qquad +\left. \,\int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s-ixy\right] \mathrm{d}x \nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}}\exp ^{\diamond } \left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1) {\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \nonumber \\&\qquad +\left. \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-\!1\!-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )\!-\frac{1}{2}x^2\beta ^2(s)\!\right\} \mathrm{d}s-ixy\!\right] ix\beta (t)\mathrm{d}x.\nonumber \\ \end{aligned}$$

  (2.13)

  Here we have used that

  $$\begin{aligned} D_{t} \int _0^{T_0}\int _{{\mathbb {R}}} (\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta ) =0, \end{aligned}$$

  which follows from the assumption that B and ${\tilde{N}}$ are independent, so for $D_{t}$ the random variable ${\tilde{N}}(s,\zeta )$ is like a constant.

  Using Eq. (2.13) and the Wick chain rule we get

  $$\begin{aligned}&{\mathbb {E}}[D_{t}\delta _Y(y)|{\mathcal {F}}_t] \nonumber \\&\quad = \frac{1}{2\pi }\int _{{\mathbb {R}}}{\mathbb {E}} \left[ \exp ^{\diamond }\left. \left[ \int _0^{T_0}\int _{{\mathbb {R}}} (\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \right. \right. \nonumber \\&\qquad \left. \text { }+ \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta )) \nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s\right. \nonumber \\&\qquad \left. \left. -ixy\right] ix\beta (t)\mathrm{d}x|{\mathcal {F}}_t\right] \mathrm{d}x \nonumber \\&\quad =\frac{1}{2\pi }\int _{{\mathbb {R}}}{\mathbb {E}} \left[ \exp ^{\diamond }\left[ \int _0^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta )+ \int _0^{T_0}ix\beta (s)\mathrm{d}B(s) \right. \right. \nonumber \\&\qquad \left. \left. + \int _0^{T_0}\left\{ \int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )-\frac{1}{2}x^2\beta ^2(s)\right\} \mathrm{d}s\right. \right. \nonumber \\&\qquad \left. \left. -ixy\right] |{\mathcal {F}}_t\right] ix\beta (t)\mathrm{d}x\nonumber \\&\quad = \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp \left[ \int _0^t\int _{{\mathbb {R}}}ix\psi (s,\zeta ){\tilde{N}} (\mathrm{d}s,\mathrm{d}\zeta ) +\int _0^t ix\beta (s)\mathrm{d}B(s)\right. \nonumber \\&\left. \qquad + \int _t^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )} -1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )\mathrm{d}s-\int _t^{T_0} \frac{1}{2}x^2\beta ^2(s)\mathrm{d}s-ixy\right] ix\beta (t)\mathrm{d}x. \end{aligned}$$

  (2.14)

• (2.16) should be corrected to

  $$\begin{aligned} \Vert \beta \Vert ^2_{[t,T]} :=\int _t^T\beta (s)^2\mathrm{d}s>0\quad \text{ for } \text{ all } t\in [0,T]. \end{aligned}$$

• The expression after the last equality sign in (2.19) should be corrected to

  $$\begin{aligned} -\left( 2\pi \Vert \beta \Vert ^2_{[t,T_0]}\right) ^{-\frac{1}{2}}\exp \left[ - \frac{(Y(t)-y)^2}{2\Vert \beta \Vert ^2_{[t,T_0]}}\right] \frac{Y(t)-y}{\Vert \beta \Vert ^2_{[t,T_0]}}\beta (t). \end{aligned}$$

• Section 2.3 should be deleted, because the formulas (2.2)–(2.22) in Sect. 2.3 are not correct as stated. The reason is that with $\beta =0$ in (2.7) these integrals do not converge. However, the integrals do converge for all $\beta \ne 0$, and it can be proved that the limit of the quotient $\psi (t,y):=\frac{(2.12)}{(2.8)}$ with $\psi =1$ exists when $\beta \rightarrow 0$. This limit is the quantity that plays a key role in the corresponding optimal portfolio problem in Sects. 6.3 and 6.4. See (6.62) and (6.82).

• In the last line before (5.3), correct $L2(\lambda \times P)$ to $L^2(\lambda \times P)$.

• In the statement of Theorem 5.1, and also in the proof, we should write $\frac{\mathrm{d}}{\mathrm{d}a}J((u+a\beta )(.,y))|_{a=0}$ in stead of $\frac{\mathrm{d}}{\mathrm{d}a}J(u+a\beta )|_{a=0}$.

• (6.40) should be corrected to

  $$\begin{aligned} \Pi ^*(s)=\frac{b_0(s,Y(T_0))}{\sigma ^2_0(s,Y(T_0))}+\frac{(Y(T_0)-Y(s))\beta (s)}{\sigma _0(s,Y(T_0))\Vert \beta \Vert ^2_{[s,T_0]}}, \quad 0 \le s \le T < T_0. \end{aligned}$$

• From (6.107) to (6.109) we should write:

  Theorem 6.9

  Suppose Y is as in (2.5). Then the processes $\phi (t,y)$ and $\psi (t,y,z)$ in the Eq. (6.106) for the optimal portfolio $\pi (t,y)$ have the following expressions:

  $$\begin{aligned} \phi (t,y)&= \frac{i\beta (t) \int _{{\mathbb {R}}} F(t,x,y) x \mathrm{d}x}{\int _{{\mathbb {R}}} F(t,x,y) \mathrm{d}x} \end{aligned}$$

  (6.107)
  $$\begin{aligned} \psi (t,y,z)&= \frac{\int _{{\mathbb {R}}} F(t,x,y) (\mathrm{e}^{ix\psi (t,z)}-1)\mathrm{d}x}{\int _{{\mathbb {R}}} F(t,x,y) \mathrm{d}x} \end{aligned}$$

  (6.108)

  where

  $$\begin{aligned} F(t,x,y)&=\int _{{\mathbb {R}}}\exp \left[ \int _0^t\int _{{\mathbb {R}}}ix\psi (s,\zeta ){\tilde{N}}(\mathrm{d}s,\mathrm{d}\zeta ) +\int _0^t ix\beta (s)\mathrm{d}B(s)\right. \nonumber \\&\qquad \left. +\int _t^{T_0}\int _{{\mathbb {R}}}(\mathrm{e}^{ix\psi (s,\zeta )}-1-ix\psi (s,\zeta ))\nu (\mathrm{d}\zeta )\mathrm{d}s\right. \nonumber \\&\qquad \left. -\int _t^{T_0}\frac{1}{2}x^2\beta ^2(s)\mathrm{d}s-ixy\right] \mathrm{d}x. \end{aligned}$$

  (6.109)