In this section, we apply the Lagrange multiplier technique and stochastic control method to obtain a closed-form solution of Problem (6).

### 3.1 Transformation of the original problem

As widely adopted in the literature, we apply the Lagrange multiplier technique to deal with the constraint \(\mathrm{E}[Y(T)]=C\). Define

$$\begin{aligned} J\bigl(Y(t),t;\pi (t),\gamma \bigr) &=\mathrm{E}\bigl[Y(T)-C \bigr]^{2}+2\gamma \mathrm{E}\bigl[Y(T)-C\bigr] \\ &=\mathrm{E}\bigl[Y(T)+\gamma -C\bigr]^{2}-\gamma^{2}, \end{aligned}$$

(7)

where \(\gamma \in \mathrm{R}\) is the Lagrange multiplier. Then by the Lagrange duality theorem (see Luenberger [19]), the original M–V portfolio selection problem (6) is equivalent to the following max–min stochastic control problem:

$$ \textstyle\begin{cases} \max_{\gamma \in \mathrm{R}}\min_{\pi (t)\in \Pi } J(Y(t),t;\pi (t), \gamma )=\mathrm{E}[Y(T)+\gamma -C]^{2}-\gamma^{2}, \\ \mbox{subject to } Y(t) \mbox{ satisfies Eq. }(5). \end{cases} $$

(8)

Clearly, to solve the above max–min stochastic control problem (8), we first need to consider the following quadratic loss minimization problem:

$$ \textstyle\begin{cases} \min_{\pi (t)\in \Pi } J_{0}(Y(t),t;\pi (t),\gamma )=\mbox{E}[Y(T)-u]^{2}, \\ \mbox{subject to } Y(t) \mbox{ satisfies Eq. }(5), \end{cases} $$

(9)

where \(u=C-\gamma \).

Problem (9) can be solved by using the stochastic control method. We now consider a truncated form of Problem (9) beginning at time *t* and define the corresponding value function as

$$ H(t,m,y,l)=\inf_{\pi (t)\in \Pi }\mathrm{E}\bigl\{ \bigl[Y(T)-u \bigr]^{2} \mid m(t)=m,Y(t)=y,L(t)=l \bigr\} $$

(10)

with the boundary condition \(H(T,m,y,l)=(y-u)^{2}\).

If \(H(t,m,y,l)\in \mathrm{C}^{1,2,2,2}([0,T]\times \mathrm{R^{+}} \times \mathrm{R}\times \mathrm{R})\), then by the principle of dynamic programming, \(H(t,m,y,l)\) satisfies the following HJB equation:

$$\begin{aligned}& H_{t}+k(\theta -m)H_{m}+\frac{\sigma^{2}_{m}}{2}mH_{mm}+ \bigl[\alpha (t)+ \lambda_{2}m\bigr]H_{l}+ \frac{\sigma^{2}_{L}}{2}mH_{ll} \\& \quad {}+\rho_{{Sm}}\sigma_{m}\sigma_{L}mH_{ml}+ \bigl[(y+l)r-\alpha (t)-\lambda _{2}m\bigr]H_{y}-m \sigma_{L}[\sigma_{L}H_{yl} \\& \quad {}+\sigma_{m}\rho_{{Sm}}H_{ym}]+\inf _{\pi (t)\in \Pi }\biggl\{ (y+l)\lambda_{1}\pi (t)mH_{y} +(y+l)m\pi (t)[\sigma_{L}H_{yl} \\& \quad {}+\sigma_{m}\rho_{{Sm}}H_{ym}] + \frac{[(y+l)\pi (t)-\sigma_{L}]^{2}m}{2}H_{yy}\biggr\} =0, \end{aligned}$$

(11)

where *H* is short for \(H(t,m,y,l)\), \(H_{t}\), \(H_{m}\), \(H_{y}\), \(H_{l}\), \(H_{mm}\), \(H_{yy}\), \(H_{yl}\), \(H_{lm}\), \(H_{ym}\) and \(H_{ll}\) denote the partial derivatives of first and second orders with respect to *t*, *m*, *y* and *l*, respectively.

We first assume that \(H_{yy}>0\), which will be verified later. The first-order condition for the optimization problem in the HJB Eq. (11) yields the optimal control as

$$ \pi^{*}(t)=\frac{\sigma_{L}H_{yy}-\sigma_{L}H_{yl}-\sigma_{m}\rho_{{Sm}}H_{ym}-\lambda_{1} H_{y}}{(y+l)H_{yy}}. $$

(12)

Substituting (12) into (11), after simplification, we have

$$\begin{aligned}& H_{t}+k(\theta -m)H_{m}+\frac{\sigma^{2}_{m}}{2}mH_{mm}+ \bigl[\alpha (t)+ \lambda_{2}m\bigr]H_{l}+ \frac{\sigma^{2}_{L}}{2}mH_{ll} \\& \quad {}+\rho_{{Sm}}\sigma_{m}\sigma_{L}mH_{ml}+ \bigl[(y+l)r-\alpha (t)+m\lambda _{1}\sigma_{L}- \lambda_{2}m\bigr]H_{y} \\& \quad {}-\frac{m}{2H_{yy}}\bigl[\sigma^{2}_{L}H^{2}_{yl}+ \lambda_{1}^{2}H^{2}_{y}+ \sigma^{2}_{m}\rho^{2}_{{Sm}}H^{2}_{ym}+2 \lambda_{1}\sigma_{L}H_{y}H _{yl} \\& \quad {}+2\sigma_{m}\sigma_{L}\rho_{{Sm}}H_{yl}H_{ym}+2 \lambda_{1}\sigma_{m} \rho_{{Sm}}H_{y}H_{ym} \bigr]=0. \end{aligned}$$

(13)

Therefore, Problem (10) can be transformed into the following non-linear PDE problem:

$$ \textstyle\begin{cases} H_{t}+k(\theta -m)H_{m}+\frac{\sigma^{2}_{m}}{2}mH_{mm}+[\alpha (t)+ \lambda_{2}m]H_{l}+\frac{\sigma^{2}_{L}}{2}mH_{ll} \\ \quad {}+\rho_{{Sm}}\sigma_{m}\sigma_{L}mH_{ml}+[(y+l)r-\alpha (t)+m \lambda_{1}\sigma_{L}-\lambda_{2}m]H_{y} \\ \quad{} -\frac{m}{2H_{yy}}[\sigma^{2}_{L}H^{2}_{yl}+\lambda_{1}^{2}H^{2}_{y}+ \sigma^{2}_{m}\rho^{2}_{{Sm}}H^{2}_{ym}+2\lambda_{1}\sigma_{L}H_{y}H _{yl} \\\quad {} +2\sigma_{m}\sigma_{L}\rho_{{Sm}}H_{yl}H_{ym}+2\lambda_{1}\sigma_{m} \rho_{{Sm}}H_{y}H_{ym}]=0, \\ H(T,m,y,l)=(y-u)^{2}. \end{cases} $$

Here we need to point out that it is very difficult to obtain an explicit solution of the complicated non-linear PDE problem for the market is not self-financing (i.e., the manager has a continuous payment \(dL(t)\) in the interval \((t,t+dt)\)). However, in a particular case, this problem can be solved.

### 3.2 Solution of the auxiliary problem (10)

In this subsection, we shall apply the variable transform techniques and PDE method to a special solution of the auxiliary problem (10). By the terminal condition \(H(T,m,y,l)=(y-u)^{2}\), we may look for a candidate solution in the form

$$ H(t,m,y,l)=f(t,m)\bigl[y+l-g(t,m,l)\bigr]^{2}, $$

(14)

where \(f(t,m)\) and \(g(t,m,l)\) are two undetermined functions with the boundary conditions \(f(T,m)=1\) and \(g(T,m,l)=l+u\).

After substituting (14) into (13) and by some tedious calculations, we have

$$\begin{aligned}& [y+l-g]^{2}\biggl\{ f_{t}+\bigl[k(\theta -m)-2 \lambda_{1}\sigma_{m} \rho_{{Sm}}m \bigr]f_{m}+\frac{\sigma^{2}_{m}}{2}mf_{mm} \\& \quad {}-m\sigma^{2}_{m}\rho^{2}_{{Sm}} \frac{f_{m}^{2}}{f} +\bigl[2r-m\lambda_{1} ^{2}\bigr]f\biggr\} -2f[y+l-g]\biggl\{ g_{t}+\biggl[k(\theta -m) \\& \quad {}-m\sigma^{2}_{m}\bigl(\rho^{2}_{{Sm}}-1 \bigr)\frac{f_{m}}{f} -m\lambda_{1} \sigma_{m} \rho_{{Sm}}\biggr]g_{m}+\frac{\sigma^{2}_{m}}{2}mg_{mm} + \frac{ \sigma^{2}_{L}}{2}mg_{ll}+\bigl[\alpha (t) \\& \quad {}+\lambda_{2}m-m\sigma_{L}\lambda_{1} \bigr]g_{l} +m\sigma_{L}\sigma_{m} \rho_{{Sm}}g_{lm}-rg\biggr\} +mfg^{2}_{m} \sigma^{2}_{m}\bigl(1-\rho ^{2}_{{Sm}} \bigr)=0, \end{aligned}$$

(15)

where *f* and *g* are short for \(f(t,m)\) and \(g(t,m,l)\), respectively.

Note that Eq. (15) can be taken as a polynomial of variable \(y+l-g\). Thus, by the boundary conditions \(f(T,m)=1\) and \(g(T,m,l)=l+u\), we have

$$\begin{aligned}& \textstyle\begin{cases} f_{t}+[k(\theta -m)-2\lambda_{1}\sigma_{m}\rho_{{Sm}}m]f_{m}+\frac{ \sigma^{2}_{m}}{2}mf_{mm} \\ \quad {}-m\sigma^{2}_{m}\rho^{2}_{{Sm}}\frac{f_{m}^{2}}{f}+[2r-m\lambda_{1} ^{2}]f=0, \\ f(T,m)=1, \end{cases}\displaystyle \end{aligned}$$

(16)

$$\begin{aligned}& \textstyle\begin{cases} g_{t}+[k(\theta -m)-m\sigma^{2}_{m}(\rho^{2}_{{Sm}}-1) \frac{f_{m}}{f} -m\lambda_{1}\sigma_{m}\rho_{{Sm}}]g_{m} +\frac{ \sigma^{2}_{m}}{2}mg_{mm} \\ \quad {}+\frac{\sigma^{2}_{L}}{2}mg_{ll}+[\alpha (t)-m\sigma_{L}\lambda _{1}+m\lambda_{2}]g_{l} +m\sigma_{L}\sigma_{m}\rho_{{Sm}}g_{lm}-rg=0, \\ g(T,m,l)=l+u, \end{cases}\displaystyle \end{aligned}$$

(17)

$$\begin{aligned}& mfg^{2}_{m}\sigma^{2}_{m}\bigl(1- \rho^{2}_{{Sm}}\bigr)=0. \end{aligned}$$

(18)

In what follows, we aim to solve the two terminal value problems (16) and (17) based on Eq. (18). We first solve the problem (16) and postulate that \(f(t,m)\) has the following exponential affine form:

$$ f(t,m)=\mathrm{e}^{A_{1}(t)m+A_{2}(t)}, $$

(19)

where \(A_{1}(t)\) and \(A_{2}(t)\) are two undetermined functions with the boundary conditions \(A_{1}(T)=0\) and \(A_{2}(T)=0\).

Note that \(\sigma_{m}>0\), \(m>0\) and \(f>0\). Then Eq. (18) is equivalent to \(g_{m}=0\) (false) or \(\rho^{2}_{{Sm}}=1\). In such a case, substituting (19) into (16) and by some simplifications, Problem (16) can be decomposed into the following ordinary differential equation (ODE) problems:

$$\begin{aligned}& \textstyle\begin{cases} \frac{dA_{1}(t)}{dt}=\frac{\sigma^{2}_{m}}{2}A^{2}_{1}(t)+(k+2\lambda _{1}\sigma_{m}\rho_{{Sm}})A_{1}(t)+\lambda_{1}^{2}, \\ A_{1}(T)=0, \end{cases}\displaystyle \end{aligned}$$

(20)

$$\begin{aligned}& \textstyle\begin{cases} \frac{dA_{2}(t)}{dt}+k\theta A_{1}(t)+2r=0, \\ A_{2}(T)=0. \end{cases}\displaystyle \end{aligned}$$

(21)

The following proposition presents an explicit solution for Problem (20).

### Proposition 1

*The solution to Problem* (20) *is given by*

$$ A_{1}(t)= \textstyle\begin{cases} \frac{n_{1}n_{2}(1-\mathrm{e}^{\sqrt{\Delta }(T-t)})}{n_{1}-n_{2} \mathrm{e}^{\sqrt{\Delta }(T-t)}},&\rho^{2}_{{Sm}}=1 \textit{ and } \Delta >0, \\ \frac{\sigma^{2}_{m}(T-t)n^{2}}{\sigma^{2}_{m}(T-t)n-2},&\rho^{2} _{{Sm}}=1 \textit{ and } \Delta =0, \\ \sqrt{\frac{-\Delta }{\sigma^{4}_{m}}}\tan (\arctan (\frac{k+2\lambda_{1}\sigma_{m}\rho_{{Sm}}}{\sqrt{-\Delta }})-\frac{\sqrt{- \Delta }(T-t)}{2})+n, &\rho^{2}_{{Sm}}=1 \textit{ and } \Delta < 0, \end{cases} $$

(22)

*where*

$$\begin{aligned}& \Delta =(k+2\lambda_{1}\sigma_{m}\rho_{{Sm}})^{2}-2 \lambda_{1}^{2} \sigma^{2}_{m}, \\& n_{1}=\frac{-(k+2\lambda_{1}\sigma_{m}\rho_{{Sm}})+\sqrt{ \Delta }}{\sigma^{2}_{m}}, \\& n_{2}=\frac{-(k+2\lambda_{1}\sigma_{m}\rho_{{Sm}})-\sqrt{\Delta }}{ \sigma^{2}_{m}}, \\& n=\frac{-(k+2\lambda_{1}\sigma_{m}\rho_{{Sm}})}{ \sigma^{2}_{m}}. \end{aligned}$$

### Proof

See the Appendix. □

According to the result of Proposition 1, the solution to Problem (21) can be expressed in terms of \(A_{1}(t)\), that is,

$$ A_{2}(t)= \int_{t}^{T} k\theta A_{1}(s)\,ds+2r(T-t). $$

(23)

Furthermore, we have the following proposition.

### Proposition 2

*The solution to Problem* (16) *is given by*

$$ f(t,m)=\mathrm{e}^{A_{1}(t)m+A_{2}(t)}, $$

*where*
\(A_{1}(t)\)
*and*
\(A_{2}(t)\)
*are given by Eqs*. (22) *and *(23), *respectively*.

We now solve Problem (17). Let

$$ g(t,m,l)=A_{3}(t)l+A_{4}(t)m+A_{5}(t), $$

(24)

where \(A_{3}(t)\), \(A_{4}(t)\) and \(A_{5}(t)\) are three undetermined functions with the boundary conditions \(A_{3}(T)=1\), \(A_{4}(T)=0\) and \(A_{5}(T)=u\).

Substituting (24) into (17) and by some simplifications, Problem (17) can be decomposed into the following ODE problems:

$$\begin{aligned}& \textstyle\begin{cases} \frac{dA_{3}(t)}{dt}-rA_{3}(t)=0, \\ A_{3}(T)=1, \end{cases}\displaystyle \end{aligned}$$

(25)

$$\begin{aligned}& \textstyle\begin{cases} \frac{dA_{4}(t)}{dt}-(k+\lambda_{1}\sigma_{m}\rho_{{Sm}}+r)A_{4}(t)-( \lambda_{1} \sigma_{L}-\lambda_{2})A_{3}(t)=0, \\ A_{4}(T)=0, \end{cases}\displaystyle \end{aligned}$$

(26)

$$\begin{aligned}& \textstyle\begin{cases} \frac{dA_{5}(t)}{dt}-rA_{5}(t)+k\theta A_{4}(t)+\alpha (t) A_{3}(t)=0, \\ A_{5}(T)=u. \end{cases}\displaystyle \end{aligned}$$

(27)

By a simple calculation, we have

$$ A_{3}(t)=\mathrm{e}^{r(t-T)}. $$

(28)

Then the solutions of Problems (26)–(27) are given by

$$ A_{4}(t)= \textstyle\begin{cases} -(\lambda_{1} \sigma_{L}-\lambda_{2})\mathrm{e}^{r(t-T)}(T-t),&k+\lambda_{1}\sigma_{m}\rho_{{Sm}}=0, \\ \frac{(\lambda_{1} \sigma_{L}-\lambda_{2})\mathrm{e}^{r(t-T)}[ \mathrm{e}^{(k+\lambda_{1}\sigma_{m}\rho_{{Sm}})(t-T)}-1]}{k+\lambda _{1}\sigma_{m}\rho_{{Sm}}}, &k+\lambda_{1}\sigma_{m}\rho_{{Sm}} \neq 0, \end{cases} $$

(29)

and

$$ A_{5}(t)=\biggl\{ u+ \int_{t}^{T}\bigl[k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr] \mathrm{e}^{r(T-s)}\,ds\biggr\} \mathrm{e}^{r(t-T)}, $$

(30)

respectively. Furthermore, we have the following conclusion.

### Proposition 3

*The solution to Problem* (17) *is given by*

$$ g(t,m,l)=A_{3}(t)l+A_{4}(t)m+A_{5}(t), $$

*where*
\(A_{3}(t)\), \(A_{4}(t)\)
*and*
\(A_{5}(t)\)
*are given by Eqs*. (28), (29) *and* (30), *respectively*.

Based on the results of Propositions 2–3 and the expression for \(H(t,m,y,l)\) in Eq. (14), we have

$$ H(t,m,y,l)=\mathrm{e}^{A_{1}(t)m+A_{2}(t)}\bigl[y-A_{4}(t)m+ \bigl(1-A_{3}(t)\bigr)l-A _{5}(t)\bigr]^{2}. $$

(31)

Obviously, \(H_{yy}=2\mathrm{e}^{A_{1}(t)m+A_{2}(t)}>0\), which means that Problem (10) does have the optimal solution. Substituting (31) into (12) further reveals the optimal investment strategy of the optimal control problem (10) as

$$ \pi^{*}(t)=\frac{\sigma_{L} A_{3}(t)+\sigma_{m} \rho_{{Sm}}A_{4}(t)- \lambda_{1}[X(t)-A_{3}(t)l-A_{4}(t)m-A_{5}(t)]}{X(t)}. $$

(32)

To sum up, we have the following theorem.

### Theorem 1

*For any*
\(t\in [0,T]\), *the value function of the optimization problem* (10) *under the condition of*
\(\rho^{2}_{{Sm}}=1\)
*is given by Eq*. (31), *while the corresponding optimal solution is given by Eq*. (32).

### 3.3 Efficient investment strategy and efficient frontier

In this subsection, we shall apply the Lagrange duality theorem (see Luenberger [19]) to derive the efficient investment strategy of Problem (6). Since the optimal control problem (9) is the same as the optimal control problem (10) when \(t=0\), the value function of Problem (9) is

$$ H(0,m_{0},y_{0},l_{0})=\mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)} \bigl[y_{0}-A _{4}(0)m_{0}+\bigl(1-A_{3}(0) \bigr)l_{0}-A_{5}(0)\bigr]^{2}, $$

(33)

where \(y_{0}=Y(0)\) and \(l_{0}=L(0)\). By the analysis of Sect. 3.1, to solve the original M–V portfolio selection problem, we only need to maximize the following function:

$$\begin{aligned} J\bigl(Y(0),0;\pi (t),\gamma \bigr)&=\mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)} \bigl[y_{0}-A _{4}(0)m_{0} \\ &\quad {}+\bigl(1-A_{3}(0)\bigr)l_{0}-A_{5}(0) \bigr]^{2}-\gamma^{2} \end{aligned}$$

(34)

over *γ*. By the expression of \(A_{5}(t)\) and \(u=C-\gamma \), Eq. (34) can be reduced to

$$\begin{aligned} J\bigl(Y(0),0;\pi (t),\gamma \bigr) &=\gamma^{2}\bigl[ \mathrm{e}^{A_{1}(0)m_{0}+A _{2}(0)-2rT}-1\bigr]-2\gamma \mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)-2rT}\biggl\{ C \\ &\quad {}-\mathrm{e}^{rT}\biggl[y_{0}-A_{4}(0)m_{0}+ \bigl(1-A_{3}(0)\bigr)l_{0} \\ &\quad {}- \int_{0}^{T}\bigl(k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr)\mathrm{e}^{-rs}\,ds\biggr] \biggr\} \\ &\quad {}+\mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)-2rT}C^{2} \\ &\quad {}+\mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)}\biggl[y_{0}-A_{4}(0)m_{0}+ \bigl(1-A_{3}(0)\bigr)l _{0} \\ &\quad {}- \int_{0}^{T}\bigl(k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr)\mathrm{e}^{-rs}\,ds\biggr]^{2} \\ &\quad {}-2C\mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)-rT}\biggl[y_{0}-A_{4}(0)m_{0}+ \bigl(1-A _{3}(0)\bigr)l_{0} \\ &\quad {}- \int_{0}^{T}\bigl(k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr)\mathrm{e}^{-rs}\,ds\biggr]. \end{aligned}$$

(35)

Note that Eq. (35) is a quadratic function with respect to *γ*, which implies that for (35) there may exist a finite maximum value while the existence of finite maximum value depends on the coefficient of \(\gamma^{2}\). For this purpose, in this paper, we need to add one more assumption as follows.

### Assumption 2

There exists a constant \(m_{0}\) such that \(\mathrm{e}^{A_{1}(0)m_{0}+A_{2}(0)-2rT}-1<0\).

Under Assumption 2, the optimal value of \(J(Y(0),0;\pi (t),\gamma )\) can be achieved when *γ* is given by

$$ \gamma^{*}=\frac{C-[y_{0}-A_{4}(0)m_{0}+(1-A_{3}(0))l_{0}-\int_{0} ^{T}(k\theta A_{4}(s)+\alpha (s) A_{3}(s))\mathrm{e}^{-rs}\,ds] \mathrm{e}^{rT}}{1-\mathrm{e}^{-[A_{1}(0)m_{0}+A_{2}(0)-2rT]}}. $$

(36)

Substituting (36) into (35), the optimal investment strategy and the minimum variance of the M–V ALM problem (9) are given by

$$\begin{aligned} \pi^{*}(t) &=\frac{1}{X(t)}\biggl\{ \sigma_{L} A_{3}(t)+\sigma _{m} \rho_{{Sm}}A_{4}(t)- \lambda_{1}\biggl[X(t)-A_{3}(t)l-A_{4}(t)m \\ &\quad {}-\biggl\{ C-\gamma^{*}+ \int_{t}^{T}\bigl(k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr)\mathrm{e}^{r(T-s)}\,ds\biggr\} \mathrm{e}^{r(t-T)}\biggr]\biggr\} \end{aligned}$$

(37)

and

$$\begin{aligned} \operatorname{Var}^{*}\bigl[Y(T)\bigr] &=\frac{1}{\mathrm{e}^{-[A_{1}(0)m_{0}+A_{2}(0)]}-\mathrm{e}^{-2rT}}\biggl\{ C \mathrm{e}^{-rT}-[y_{0}-A_{4}(0)m _{0} \\ &\quad {}+\bigl(1-A_{3}(0)\bigr)l_{0}- \int_{0}^{T}\bigl(k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr) \mathrm{e}^{-rs}\,ds \biggr\} ^{2}, \end{aligned}$$

(38)

respectively. Furthermore, setting

$$ C^{*}=\mathrm{e}^{rT}\biggl[y_{0}-A_{4}(0)m_{0}+ \bigl(1-A_{3}(0)\bigr)l_{0}- \int_{0} ^{T}\bigl(k\theta A_{4}(s)+\alpha (s) A_{3}(s)\bigr)\mathrm{e}^{-rs}\,ds\biggr], $$

then we can get the global minimum variance \(\mathrm{Var^{*}_{{\min}}}[Y(T)]=0\). Moreover, according to the definition of efficient investment strategy, the rational investors should not select the expected terminal wealth less than \(C^{*}\).

In conclusion, we summarize the above results in the following theorem.

### Theorem 2

*Under Assumptions*
1*–*2, *the efficient investment strategy and the efficient frontier of this M–V ALM problem under the condition of*
\(\rho^{2}_{{Sm}}=1\)
*for*
\(C\geq C^{*}\)
*are given by Eqs*. (37) *and *(38), *respectively*.

### Remark 2

Based on the result of Theorem 2, we can derive the efficient investment strategy and efficient frontier for the corresponding asset allocation problem (i.e., the case of no liability).