1 Correction to: Appl Math Optim https://doi.org/10.1007/s00245-017-9425-1

The original version of this article unfortunately contained a few mistakes in Theorems and notation. The corrected information is given below.

  • In Definition 1.1 the space \(\mathcal {M}\) is not a Hilbert space, but a pre-Hilbert space. Throughout the paper \(\mathcal {M}\) denotes the completion of this space.

  • The proofs of Theorems 4.5 and 4.6 are not correct as stated. However, the proofs can easily be corrected—and even simplified—as follows: For a given Fréchet differentiable operator \(G: \mathcal {M}_0 \mapsto \mathbb {L}^1(\mathbb {P})\) let \(\nabla _{m}G\) denote its Fréchet derivative at the measure \(m \in \mathcal {M}_0\), and for \(X \in \mathbb {L}^2(\mathbb {P})\) let \(M=\mathcal {L}(X)\) denote the law of X. Define the dual operator \(\nabla _m^{*}G(\cdot ) \in \mathbb {L}^2(\mathbb {P})\) by the property

    $$\begin{aligned} \mathbb {E}[\langle \nabla _m G, M\rangle ] = \mathbb {E}[ \nabla _m^{*}G X];\quad X \in \mathbb {L}^2(\mathbb {P}), \end{aligned}$$
    (0.1)

    and let the dual operator \(\nabla _{\bar{m}}H^{t}\) be defined in the similar way as \(\nabla _{\bar{x}}H^{t}\) in (4.6)–(4.7) in the paper. The existence and the uniqueness of \(\nabla _m^{*}G\) and \(\nabla _{\bar{m}}H^{t}\) follow by the Riesz representation theorem, as explained in the paper. With the introduction of these dual operators the proofs of Theorems 4.5 and 4.6 can be corrected and simplified. Specifically, there is no need for the adjoint variables \(p^1,q^1\) and \(r^1\), if we modify the backward stochastic differential equation (BSDE) for the adjoint variables \(p^0,q^0,r^0\) accordingly. More precisely, everything concerning the processes \(p^1,q^1,r^1\) can be deleted throughout the paper if we change the BSDE (4.13) to

    $$\begin{aligned} dp^0(t) =&- \Bigg \{ \frac{\partial H}{\partial x}(t) + \mathbb {E}[\nabla _{\bar{x}}H^{t}|\mathcal {F}_t] + \nabla _m^{*}H(t) + \mathbb {E}[\nabla _{\bar{m}}H^{t} | \mathcal {F}_t] \Bigg \} dt\nonumber \\&+ q^0(t) dB(t) + \int _{\mathbb {R}_0} r^0(t,\zeta )\tilde{N}(dt,d\zeta ); \quad t \in [0,T] \nonumber \\ p^0(t)=&\,\frac{\partial h}{\partial x}(X(T),M(T))+\nabla _{m}^{*}h(X(T),M(T)); \quad t \ge T \nonumber \\ q^0(t)=&\,0;\quad t> T\nonumber \\ r^0(t,\cdot )=&0; \quad t > T, \end{aligned}$$
    (0.2)
  • The BSDE (4.14) is not needed and should be deleted.

  • The expression (4.23) should be corrected to

    $$\begin{aligned}&= \mathbb {E} \Bigg [ \int _0^T \hat{p}^{0}(t) \tilde{b}(t) \nonumber \\&\quad - \int _0^T \Bigg \{ \frac{\partial \hat{H}}{\partial x}(t) +\mathbb {E}[\nabla _{\bar{x}}\hat{H}^{t}|\mathcal {F}_t] + \nabla _m^{*}\hat{H}(t) + \mathbb {E}[\nabla _{\bar{m}}\hat{H}^{t} |\mathcal {F}_t] \Bigg \} \tilde{X}(t)dt\nonumber \\&\quad + \int _0^T \hat{q}^{0}(t) \tilde{\sigma }(t) dt + \int _0^T \int _{\mathbb {R}_0} \bar{r}^{0}(t,\zeta ) \tilde{\gamma }(t,\zeta ) \nu (d\zeta ) dt \Bigg ] \end{aligned}$$
    (0.3)
  • Equation (4.24) can be deleted.

  • The first part of the proof of Theorem 4.6 (up to and including (4.29)) can be deleted. All terms in the proof involving \(p^1\) should be deleted. The equation for \(\mathbb {E}[p^0(T)Z(T)]\) should be corrected to

    $$\begin{aligned}&\mathbb {E}[p^{0}(T)Z(T)] \\&\quad =\mathbb {E}\left[ \int _{0}^{T}p^{0}(t)dZ(t)+\int _{0}^{T}Z(t)dp^{0}(t)+[p^{0},Z]_{T}\right] \\&\quad =\mathbb {E}\left[ \int _{0}^{T}p^{0}(t)(\nabla b(t))^{T}\text { }(Z(t),Z_{t},DM(t),DM_{t},\pi (t),\pi _{t})dt\right. \\&\qquad -\int _{0}^{T}\left\{ \tfrac{\partial H}{\partial x}(t)+\mathbb {E}[\nabla _{\overline{x}}H^{t}|\mathcal {F}_t] +\nabla _m^{*}H(t) + \mathbb {E}[\nabla _{\bar{m}}H^{t}|\mathcal {F}_t] \right\} Z(t)dt\\&\qquad +\int _{0}^{T} q^{0}(t)(\nabla \sigma \left( t\right) )^{T}\text { }(Z(t),Z_{t},DM(t),DM_{t},\pi \left( t\right) ,\pi _{t})dt\\&\qquad \left. +\int _{0}^{T}\int _{\mathbb {R}_{0}}r^{0}(t,\zeta )(\nabla \gamma \left( t,\zeta \right) )^{T}\text { }(Z(t),Z_{t},DM(t),DM_{t},\pi (t),\pi _{t})\nu (d\zeta )dt\right] . \end{aligned}$$