In this section, one aims at illustrating Theorems 3.1 and 3.2 by three examples. For convenience, one still adopts the state equation, the observation equation, and the corresponding assumptions introduced in Sects. 2 and 3 unless noted otherwise.
Example 4.1
Find an admissible control to minimize
$$ \begin{aligned}[b] \mathcal{J}[v]={}&\frac{1}{2} \mathbb{E}\biggl\{ \int ^{T}_{0} \bigl[A_{t} \bigl(x ^{v}_{t} \bigr)^{2}+\bar{A}_{t} \bigl( \mathbb{E}x^{v}_{t} \bigr)^{2}+B_{t}v^{2}_{t}+ \bar{B}_{t}(\mathbb{E}v_{t})^{2} \bigr]\,dt \\ & {} +D \bigl(x^{v} _{T} \bigr)^{2}+\bar{D} \bigl( \mathbb{E}x^{v}_{T} \bigr)^{2} \biggr\} \end{aligned} $$
(17)
over \(\mathcal{U}^{c}_{\mathrm {ad}}\) with \(U=\mathbb{R}\) and the terminal constraint
$$ \mathbb{E}x^{v}_{T}=\gamma , \quad \gamma \in \mathbb{R}, $$
(18)
subject to
$$ \textstyle\begin{cases} dx^{v}_{t} = (a_{t}x^{v}_{t}+\bar{a}_{t}\mathbb{E}x^{v}_{t}+b_{t}v _{t}+\bar{b}_{t}\mathbb{E}v_{t} )\,dt+c_{t}\,d\omega _{t}+\tilde{c} _{t}\,d\tilde{\omega }_{t},\\ x^{v}_{0}=\xi , \end{cases} $$
(19)
and
$$ \textstyle\begin{cases} dy^{v}_{t} = (f_{t}x^{v}_{t}+\bar{f}_{t}\mathbb{E}x^{v}_{t}+g_{t}v _{t}+\bar{g}_{t}\mathbb{E}v_{t} )\,dt+h_{t}\,d\tilde{\omega }_{t},\\ y _{0}=0, \end{cases} $$
(20)
where \(A,\bar{A}, B, \bar{B}\in L^{\infty }(0,T;\mathbb{R})\), \(A_{t}>0\), \(A_{t}+\bar{A}_{t}\geq 0\), \(B_{t}>0\), \(B_{t}+\bar{B}_{t}>0\), \(D\geq 0\), \(D+\bar{D}\geq 0\), \(b\neq 0\) and \(b+\bar{b}\neq 0\).
Define an auxiliary cost functional without terminal constraint
$$ \mathcal{J}_{\kappa }[v]=J_{\kappa }[v]-\kappa \gamma $$
with
$$ \begin{aligned}[b] J_{\kappa }[v]={}& \frac{1}{2}\mathbb{E}\biggl\{ \int ^{T}_{0} \bigl[A_{t} \bigl(x ^{v}_{t} \bigr)^{2}+\bar{A}_{t} \bigl( \mathbb{E}x^{v}_{t} \bigr)^{2}+B_{t}v^{2}_{t}+ \bar{B}_{t}(\mathbb{E}v_{t})^{2} \bigr]\,dt \\ & {} +D \bigl(x^{v} _{T} \bigr)^{2}+\bar{D} \bigl( \mathbb{E}x^{v}_{T} \bigr)^{2}+2\kappa x^{v}_{T} \biggr\} , \quad \kappa \in \Delta . \end{aligned} $$
(21)
Since both κ and γ are constants, it is enough to minimize (21) over \(\mathcal{U}_{\mathrm {ad}}\) subject to (19) and (20). One will use three steps to explicitly solve the example.
Step 1 Candidate optimal control of the auxiliary problem without terminal constraint.
With the data, the Hamiltonian function is
$$ \begin{aligned} H(t,x,\bar{x},v;p_{\kappa },q_{\kappa }, \tilde{q}_{\kappa })={}& (a _{t}x+\bar{a}_{t} \bar{x}+b_{t}v+\bar{b}_{t}\bar{v} )p_{\kappa }+c _{t}q_{\kappa }+\tilde{c}_{t}\tilde{q}_{\kappa } \\ & {} +\frac{1}{2} \bigl[A_{t}x^{2}+ \bar{A}_{t}( \bar{x})^{2}+B_{t}v^{2}+ \bar{B}_{t}( \bar{v})^{2} \bigr], \end{aligned} $$
where \((p_{\kappa }, q_{\kappa },\tilde{q}_{\kappa })\) is determined by the Hamiltonian system
$$ \textstyle\begin{cases} dx^{u_{\kappa }}_{t} = (a_{t}x^{u_{\kappa }}_{t}+\bar{a}_{t}\mathbb{E}x ^{u_{\kappa }}_{t}+b_{t}u_{\kappa ,t}+\bar{b}_{t}\mathbb{E}u_{\kappa ,t} )\,dt+c _{t}\,d\omega _{t}+\tilde{c}_{t}\,d\tilde{\omega }_{t},\\ -dp_{\kappa ,t}= [a_{t}p_{\kappa ,t}+A_{t}x^{u}_{t}+\mathbb{E}(\bar{a}_{t}p _{\kappa ,t}+\bar{A}_{t}x^{u}_{t} ) ]-q_{\kappa ,t}\,d\omega _{t}-\tilde{q}_{\kappa ,t}\,d\tilde{\omega }_{t},\\ x^{u_{\kappa }}_{0}= \xi , \quad p_{\kappa ,T}=Dx^{u_{\kappa }}_{T}+\bar{D}\mathbb{E}x^{u_{ \kappa }}_{T}+\kappa . \end{cases} $$
(22)
If \(u_{\kappa }\) is an optimal control of the auxiliary problem, then it follows from Theorem 3.1 that
$$ B_{t}u_{\kappa ,t}+b_{t}\hat{p}_{\kappa ,t}+ \bar{B}_{t}\mathbb{E}u_{ \kappa ,t}+\bar{b}_{t}\mathbb{E}p_{\kappa ,t}=0. $$
Solving it, we get
$$ u_{\kappa ,t}=-B^{-1}_{t} \bigl\{ b_{t}\hat{p}_{\kappa ,t}+ \bigl[\bar{b} _{t}- \bar{B}_{t}(B_{t}+\bar{B}_{t})^{-1}(b_{t}+ \bar{b}_{t}) \bigr]\mathbb{E}p _{\kappa ,t} \bigr\} . $$
(23)
Step 2 Feedback form of (23).
Inserting (23) into (22) and taking expectations, one gets an ordinary differential equation
$$ \textstyle\begin{cases} \frac{d}{dt}\mathbb{E}x^{u_{\kappa }}_{t} =(a_{t}+\bar{a}_{t})\mathbb{E}x ^{u_{\kappa }}_{t}-(B_{t}+\bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2}\mathbb{E}p _{\kappa ,t},\\ \frac{d}{dt}\mathbb{E}p_{\kappa ,t}=-(A_{t}+\bar{A} _{t})\mathbb{E}x^{u_{\kappa }}_{t}-(a_{t}+\bar{a}_{t})\mathbb{E}p_{\kappa ,t},\\ \mathbb{E}x^{u_{\kappa }}_{0}=\mu _{0}, \quad \mathbb{E}p_{\kappa ,T}=(D+ \bar{D})\mathbb{E}x^{u_{\kappa }}_{T}+\kappa . \end{cases} $$
(24)
Note that the first equation and the second equation in (24) are coupled. Since
$$ -(B_{t}+\bar{B}_{t})^{-1}(b_{t}+ \bar{b}_{t})^{2}< 0, $$
the assumption condition of Theorem 2.6 in Peng and Wu [17] is satisfied, and hence (24) has a unique solution \((\mathbb{E}x^{u_{\kappa }}, \mathbb{E}p_{\kappa })\). Noticing the terminal condition of (24), one sets
$$ \mathbb{E}p_{\kappa ,t}=\alpha _{t} \mathbb{E}x^{u_{\kappa }}_{t}+ \beta _{\kappa ,t}, $$
(25)
where α and \(\beta _{\kappa }\) are deterministic and differential functions such that \(\alpha _{T}=D+ \bar{D}\) and \(\beta _{\kappa ,T}=\kappa \). Using the chain rule for computing the derivative of (25), one has
$$ \begin{aligned} \frac{d}{dt}\mathbb{E}p_{\kappa ,t}={}&\dot{ \alpha }_{t}\mathbb{E}x^{u _{\kappa }}_{t}+\alpha _{t}\frac{d}{dt}\mathbb{E}x^{u_{\kappa }}_{t}+ \dot{ \beta }_{\kappa ,t} \\ ={}& \bigl[\dot{\alpha } _{t}+(a_{t}+\bar{a}_{t}) \alpha _{t}-(B_{t}+\bar{B}_{t})^{-1}(b_{t}+ \bar{b}_{t})^{2}\alpha ^{2}_{t} \bigr] \mathbb{E}x^{u_{\kappa }}_{t} \\ & {} + \dot{\beta }_{\kappa ,t}-\alpha _{t}(B_{t}+ \bar{B} _{t})^{-1}(b_{t}+\bar{b}_{t})^{2} \beta _{\kappa ,t}. \end{aligned} $$
Comparing the equality with the second equation in (24), one deduces
$$ \textstyle\begin{cases} \dot{\alpha }_{t}+2(a_{t}+\bar{a}_{t})\alpha _{t}-(B_{t}+\bar{B}_{t})^{-1}(b _{t}+\bar{b}_{t})^{2}\alpha ^{2}_{t}+A_{t}+\bar{A}_{t}=0,\\ \alpha _{T}=D+\bar{D}, \end{cases} $$
(26)
and
$$ \textstyle\begin{cases} \dot{\beta }_{\kappa ,t}+ [a_{t}+\bar{a}_{t}-(B_{t}+\bar{B}_{t})^{-1}(b _{t}+\bar{b}_{t})^{2}\alpha _{t} ]\beta _{\kappa ,t}=0,\\ \beta _{\kappa ,T}=\kappa . \end{cases} $$
(27)
It is easy to see (26) and (27) admit unique solutions, respectively. Inserting (25) into the first equation of (24), one derives
$$ \textstyle\begin{cases} \frac{d}{dt}\mathbb{E}x^{u_{\kappa }}_{t}= [a_{t}+\bar{a}_{t}-(B _{t}+\bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2}\alpha _{t} ]\mathbb{E}x ^{u_{\kappa }}_{t}\\ \hphantom{\frac{d}{dt}\mathbb{E}x^{u_{\kappa }}_{t}=}{}-(B_{t}+\bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2} \beta _{\kappa ,t},\\ \mathbb{E}x^{u}_{0}=\mu _{0}. \end{cases} $$
(28)
Using Theorem 3.2 to (22) with (23), one gets a forward–backward stochastic differential filtering equation of mean-field type,
$$\begin{aligned} \textstyle\begin{cases} d\hat{x}^{u_{\kappa }}_{t}= [a_{t}\hat{x}^{u_{\kappa }}_{t}-b ^{2}_{t}B^{-1}_{t}\hat{p}_{\kappa ,t}+\bar{a}_{t}\mathbb{E}x^{u_{\kappa }}_{t}-(B_{t}+\bar{B}_{t})^{-1} (\bar{b}^{2}_{t}+2b_{t}\bar{b} _{t}-B^{-1}_{t}\bar{B}_{t}b^{2}_{t} )\mathbb{E}p_{\kappa ,t} ]\,dt \\ \hphantom{d\hat{x}^{u_{\kappa }}_{t}=}{}+ (\tilde{c}_{t}+\varSigma _{t}f_{t}h^{-1}_{t} )\,d \hat{\omega }_{t},\\ -d\hat{p}_{\kappa ,t}= [A_{t}\hat{x}^{u _{\kappa }}_{t}+a_{t}\hat{p}_{\kappa ,t}+\mathbb{E}(\bar{a}_{t}p _{\kappa ,t}+\bar{A}_{t}x^{u_{\kappa }}_{t} ) ]\,dt-Q_{t}\,d \hat{\omega }_{t},\\ \hat{x}^{u_{\kappa }}_{0}= \mu _{0}, \quad \hat{p} _{\kappa ,T}=D \hat{x}^{u_{\kappa }}_{T}+\bar{D}\mathbb{E}x^{u_{\kappa }}_{T}+\kappa , \end{cases}\displaystyle \end{aligned}$$
(29)
where Σ and ω̂ satisfy (15) with (16), and \(\mathbb{E}x^{u_{\kappa }}\) and \(\mathbb{E}p_{\kappa }\) solve (28) and (25), respectively. Since \(b\neq 0\), (29) admits a unique solution \((\hat{x}^{u_{\kappa }},\hat{p} _{\kappa },Q)\in L^{2}_{\mathcal{F}^{y^{u_{\kappa }}}}(0,T;\mathbb{R}^{3})\) by Theorem 2.6 in [17] again. Similarly, let
$$ \hat{p}_{\kappa ,t}=\varGamma _{t} \hat{x}^{u_{\kappa }}_{t}+\bar{\varGamma } _{t}\mathbb{E}\hat{x}^{u_{\kappa }}_{t}+\varLambda _{\kappa ,t}, $$
(30)
where Γ, Γ̄ and \(\varLambda _{\kappa }\) are three deterministic and differential functions satisfying \(\varGamma _{T}=D\), \(\bar{\varGamma }_{T}=\bar{D}\) and \(\varLambda _{\kappa ,T}=\kappa \). It follows from Itô’s formula that
$$ \begin{aligned} d\hat{p}_{\kappa ,t}={}& \dot{\varGamma }_{t}\hat{x}^{u_{\kappa }}_{t}\,dt+ \varGamma _{t}\,d \hat{x}^{u_{\kappa }}_{t}+\dot{\bar{\varGamma }}_{t}\mathbb{E}\hat{x}^{u_{\kappa }}_{t}\,dt+\bar{\varGamma }_{t}\,d\mathbb{E}\hat{x}^{u_{ \kappa }}_{t}+\dot{\varLambda }_{\kappa ,t}\,dt \\ ={}& \bigl(\dot{\varGamma } _{t}+a_{t}\varGamma _{t}-B^{-1}_{t}b^{2}_{t} \varGamma ^{2}_{t} \bigr)\hat{x} ^{u_{\kappa }}_{t} \,dt \\ & {} + \bigl\{ \dot{\bar{\varGamma }}_{t}+ \bigl[a_{t}+ \bar{a}_{t}-2(B_{t}+\bar{B}_{t})^{-1}(b_{t}+ \bar{b}_{t})^{2}\varGamma _{t} \bigr]\bar{ \varGamma }_{t}-(B_{t}+\bar{B}_{t})^{-1}(b_{t}+ \bar{b} _{t})^{2}\bar{\varGamma }^{2}_{t} \\ & {} -(B_{t}+\bar{B}_{t})^{-1} \bigl( \bar{b}^{2}_{t}+2b_{t}\bar{b}_{t}-B^{-1}_{t} \bar{B}_{t}b^{2} _{t} \bigr)\varGamma ^{2}_{t}+\bar{a}_{t}\varGamma _{t} \bigr\} \mathbb{E}\hat{x}^{u_{\kappa }}_{t}\,dt \\ & {} + \bigl\{ \dot{\varLambda }_{\kappa ,t}-( \varGamma _{t}+\bar{ \varGamma }_{t}) (B_{t}+\bar{B}_{t})^{-1}(b_{t}+ \bar{b} _{t})^{2}\varLambda _{\kappa ,t} \bigr\} \,dt \\ & {} +\varGamma _{t} \bigl(\tilde{c} _{t}+\varSigma _{t}f_{t}h^{-1}_{t} \bigr)\,d\hat{ \omega }_{t}. \end{aligned} $$
Comparing it with the second equation in (29), one deduces
$$\begin{aligned}& \textstyle\begin{cases} \dot{\varGamma }_{t}+2a_{t}\varGamma _{t}-B^{-1}_{t}b^{2}_{t}\varGamma ^{2}_{t}+A _{t}=0,\\ \varGamma _{T}=D, \end{cases}\displaystyle \end{aligned}$$
(31)
$$\begin{aligned}& \textstyle\begin{cases} \dot{\bar{\varGamma }}_{t}+2 [a_{t}+\bar{a}_{t}-(B_{t}+\bar{B}_{t})^{-1}(b _{t}+\bar{b}_{t})^{2}\varGamma _{t} ]\bar{\varGamma }_{t}-(B_{t}+ \bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2}\bar{\varGamma }^{2}_{t}\\ \quad {}-(B _{t}+\bar{B}_{t})^{-1}(\bar{b}^{2}_{t}+2b_{t}\bar{b}_{t}-B^{-1}_{t} \bar{B}_{t}b^{2}_{t})\varGamma ^{2}_{t}+2\bar{a}_{t}\varGamma _{t}+\bar{A} _{t}=0,\\ \bar{\varGamma }_{T}=\bar{D}, \end{cases}\displaystyle \end{aligned}$$
(32)
and
$$ \textstyle\begin{cases} \dot{\varLambda }_{\kappa ,t}+ [a_{t}+\bar{a}_{t}-(\varGamma _{t}+\bar{ \varGamma }_{t})(B_{t}+\bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2} ] \varLambda _{\kappa ,t}=0,\\ \varLambda _{\kappa ,T}=\kappa , \end{cases} $$
(33)
which admit a unique solution, respectively. Plugging (30) into (23), one gets
$$ \begin{aligned}[b] u_{\kappa ,t}={}&-B^{-1}_{t} \bigl\{ b_{t}\varGamma _{t}\hat{x}^{u_{\kappa }} _{t}+ \bigl[b_{t}\bar{\varGamma }_{t}+(B_{t}+ \bar{B}_{t})^{-1}(B_{t}\bar{b} _{t}- \bar{B}_{t}b_{t}) (\varGamma _{t}+\bar{\varGamma }_{t}) \bigr]\mathbb{E}\hat{x} ^{u_{\kappa }}_{t} \\ & {} +B_{t}(B_{t}+\bar{B}_{t})^{-1}(b _{t}+\bar{b}_{t})\varLambda _{\kappa ,t} \bigr\} , \end{aligned} $$
(34)
where Γ, Γ̄, \(\varLambda _{\kappa }\) and \(\hat{x} ^{u_{\kappa }}\) solve (31), (32), (33) and
$$ \textstyle\begin{cases} d\hat{x}^{u_{\kappa }}_{t}= \{ (a_{t}-b^{2}_{t}B^{-1}_{t} \varGamma _{t} )\hat{x}^{u_{\kappa }}_{t}+ [\bar{a}_{t}-(B_{t}+ \bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2}\bar{\varGamma }_{t} \\ \hphantom{d\hat{x}^{u_{\kappa }}_{t}=}{}-(B_{t}+\bar{B}_{t})^{-1}(\bar{b}^{2}_{t}+2b_{t}\bar{b} _{t}-B^{-1}_{t}\bar{B}_{t}b^{2}_{t})\varGamma _{t} ]\mathbb{E}\hat{x} ^{u_{\kappa }}_{t}\\ \hphantom{d\hat{x}^{u_{\kappa }}_{t}=}{}-(B_{t}+\bar{B}_{t})^{-1}(b_{t}+\bar{b} _{t})^{2}\varLambda _{\kappa ,t} \}\,dt\\ \hphantom{d\hat{x}^{u_{\kappa }}_{t}=}{}+ (\tilde{c}_{t}+ \varSigma _{t}f_{t}h^{-1}_{t} )\,d\hat{\omega }_{t},\\ \hat{x}^{u_{ \kappa }}_{0}=\mu _{0}, \end{cases} $$
respectively.
Step 3 Optimal control of Example 4.1.
Solving (27) and (28), one gets
$$\begin{aligned}& \beta _{\kappa ,t}=\kappa e^{\int ^{T}_{t}\rho _{s}\,ds}, \\& \mathbb{E}\hat{x}^{u_{\kappa }}_{t}=\mu _{0}e^{\int _{0}^{t}\rho _{s}\,ds}- \kappa \int _{0}^{t}(B_{s}+\bar{B}_{s})^{-1}(b_{s}+ \bar{b}_{s})^{2}e ^{\int _{s}^{T}\rho _{r}\,dr+\int _{s}^{t}\rho _{r}\,dr}\,ds, \end{aligned}$$
with
$$ \rho _{t}=a_{t}+\bar{a}_{t}-(B_{t}+ \bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2} \alpha _{t}. $$
Recalling the terminal constraint (18), it yields
$$ \kappa _{0}=\frac{\mu _{0}e^{\int _{0}^{T}\rho _{t}\,dt}-\gamma }{\int _{0} ^{T}(B_{t}+\bar{B}_{t})^{-1}(b_{t}+\bar{b}_{t})^{2}e^{2\int _{t}^{T} \rho _{s}\,ds}\,dt}. $$
(35)
Then Proposition 2.2 implies the desired conclusion. The above deduction is summarized as follows.
Proposition 4.1
The optimal feedback control of Example 4.1
is given by (34) with
κ
being replaced by (35).
Example 4.2
In particular, if one lets the coefficients of (17), (19) and (20) \(\bar{A}=\bar{B}=\bar{D}= \bar{a}=\bar{b}=c=f=\bar{f}=g=\bar{g}=0\) on \([0, T]\), then Example 4.1 is reduced to an LQ optimal control with terminal constraint and complete information. Further, let \(v\in L^{2}(0, T; \mathbb{R})\). Since v is deterministic, Proposition 4.1 implies that the optimal feedback control is
$$ u_{\kappa _{0},t}=-B^{-1}_{t}b_{t} \bigl(\alpha _{t}\mathbb{E}x^{u_{\kappa _{0}}}_{t}+\beta _{\kappa _{0},t} \bigr), $$
where α, \(\beta _{\kappa _{0}}\) and \(x^{u_{\kappa _{0}}}\) satisfy
$$\begin{aligned}& \textstyle\begin{cases} \dot{\alpha }_{t}+2a_{t}\alpha _{t}-B^{-1}_{t}b^{2}_{t}\alpha ^{2} _{t}+A_{t}=0,\\ \alpha _{T}=D, \end{cases}\displaystyle \\& \beta _{\kappa _{0},t}=\kappa _{0}e^{\int _{t}^{T} (a_{s}-B^{-1}_{s}b ^{2}_{s}\alpha _{s} )\,ds}, \end{aligned}$$
and
$$\begin{aligned} \textstyle\begin{cases} dx^{u_{\kappa _{0}}}_{t} = (a_{t}x^{u_{\kappa _{0}}}_{t}-B^{-1} _{t}b^{2}_{t}\alpha _{t}\mathbb{E}x^{u_{\kappa _{0}}}_{t}-B^{-1}_{t}b^{2} _{t}\beta _{\kappa _{0}, t} )\,dt+\tilde{c}_{t}\,d\tilde{\omega }_{t}, \\ x^{u_{\kappa _{0}}}_{0}=\xi , \end{cases}\displaystyle \end{aligned}$$
(36)
with
$$ \kappa _{0}=\frac{\mu _{0}e^{\int _{0}^{T} (a_{t}-B^{-1}_{t}b^{2} _{t}\alpha _{t} )\,dt}}{\int _{0}^{T}B^{-1}_{t}b^{2}_{t}e^{2\int _{t}^{T} (a_{s}-B^{-1}_{s}b^{2}_{s}\alpha _{s} )\,ds}\,dt}, $$
respectively.
Note that (36) is an SDE of mean-field type. It shows that one begins with a classical control system without mean-field term, but one ends up with a control system of mean-field type. This is a very interesting phenomenon indeed.
Example 4.3
One denotes by v the rate of capital withdrawal or injection of a firm, and by \(x^{v}\) the cash-balance process on \([0, T]\). Assume that the liability of the firm is governed by
$$ -d\bar{L}^{v}_{t}=b_{t}v_{t} \,dt+c_{t}\,d \omega _{t}+\tilde{c}_{t}\,d \tilde{\omega }_{t}, $$
where \(c_{t}\,d\omega _{t}\) and \(\tilde{c}_{t}\,d\tilde{\omega }_{t}\) describe the liability risk. Assume that the firm owns an initial investment ξ, and only invests in a money account with compounded interest rate a. Then the cash-balance is denoted by
$$ x^{v}_{t}=e^{\int ^{t}_{0}a_{s}\,ds} \biggl(\xi - \int ^{t}_{0}e^{-\int ^{s} _{0}a_{r}\,dr}\,d \bar{L}^{v}_{s} \biggr), $$
whose differential form is the same as (6) with \(\bar{a}= \bar{b}=0\). Due to the discreteness of the account information, the firm partially observes the cash-balance by the corresponding stock price
$$ \textstyle\begin{cases} dS^{v}_{t} =S^{v}_{t} [ (f_{t}x^{v}_{t}+g_{t}+\frac{1}{2}h ^{2}_{t} )\,dt+h_{t}\,d\tilde{\omega }_{t} ],\\ S^{v}_{0}=1. \end{cases} $$
Set
$$ y^{v}_{t}=\log S^{v}_{t}. $$
It follows from Itô’s formula that \(y^{v}\) is governed by (7) with \(\bar{f}_{t}=\bar{g}_{t}=0\). The firm hopes to find a suitable v such that
$$ \mathcal{J}[v]=\frac{1}{2}\mathbb{E}\biggl[ \int _{0} ^{T}v^{2}_{t}\,dt+ \bigl(x^{v}_{T}-\mathbb{E}x^{v}_{T} \bigr)^{2} \biggr] $$
is minimized over \(\mathcal{U}^{c}_{\mathrm {ad}}\) with terminal constraint (18). The model implies that the firm wants to minimize the risk of \(x^{v}_{T}\) and v under a fixed terminal cash-balance level. Since
$$ \mathbb{E}\bigl(x^{v}_{T}-\mathbb{E}x^{v}_{T} \bigr)^{2}=\mathbb{E}\bigl(x^{v}_{T} \bigr)^{2}- \bigl(\mathbb{E}x ^{v}_{T} \bigr)^{2}, $$
Example 4.3 is also a special case of Example 4.1. The following result is an immediate one of Proposition 4.1.
Corollary 4.1
The optimal rate of capital withdrawal or injection of the firm is
$$ u_{\kappa _{0},t}=-b_{t} \bigl(\varGamma _{t} \hat{x}^{u_{\kappa _{0}}}_{t}+\bar{ \varGamma }_{t}\mathbb{E}\hat{x}^{u_{\kappa _{0}}}_{t}+\varLambda _{\kappa _{0},t} \bigr), $$
where
Γ, Γ̄, Λ
and
\(\hat{x}^{u_{\kappa _{0}}}\)
are the solutions to
$$\begin{aligned}& \textstyle\begin{cases} \dot{\varGamma }_{t}+2a_{t}\varGamma _{t}-b^{2}_{t}\varGamma ^{2}_{t}=0,\\ \varGamma _{T}=1, \end{cases}\displaystyle \\& \textstyle\begin{cases} \dot{\bar{\varGamma }}_{t}+2 (a_{t}-b^{2}_{t}\varGamma _{t} )\bar{ \varGamma }_{t}-b^{2}_{t}\bar{\varGamma }^{2}_{t}=0,\\ \bar{\varGamma }_{T}=-1, \end{cases}\displaystyle \\& \textstyle\begin{cases} \dot{\varLambda }_{\kappa _{0},t}+ [a_{t}-(\varGamma _{t}+\bar{\varGamma }_{t})b^{2}_{t} ]\varLambda _{\kappa _{0},t}=0,\\ \varLambda _{\kappa _{0}, T}=\kappa _{0}, \end{cases}\displaystyle \end{aligned}$$
and
$$ \textstyle\begin{cases} d\hat{x}^{u_{\kappa _{0}}}_{t}= [ (a_{t}-b^{2}_{t}\varGamma _{t} )\hat{x}^{u_{\kappa _{0}}}_{t}-b^{2}_{t}\bar{\varGamma }_{t}\mathbb{E}\hat{x}^{u_{\kappa _{0}}}_{t}-b_{t}^{2}\varLambda _{\kappa _{0},t} ]\,dt+ (\tilde{c}_{t}+\varSigma _{t}f_{t}h^{-1}_{t} )\,d\hat{\omega } _{t},\\ \hat{x}^{u_{\kappa _{0}}}_{0}= \mu _{0}, \end{cases} $$
with
$$ \kappa _{0}=\frac{\mu _{0}e^{\int _{0}^{T}a_{t}\,dt}-\gamma }{\int _{0}^{T}b ^{2}_{t}e^{2\int _{t}^{T}a_{s}\,ds}\,dt}. $$
One remarks that the model in Example 4.3 is inspired by Huang et al. [18], where the variance of v does not enter the performance functional.