Optimal life-cycle consumption and investment decisions under age-dependent risk preferences

Abstract

In this article we solve the problem of maximizing the expected utility of future consumption and terminal wealth to determine the optimal pension or life-cycle fund strategy for a cohort of pension fund investors. The setup is strongly related to a DC pension plan where additionally (individual) consumption is taken into account. The consumption rate is subject to a time-varying minimum level and terminal wealth is subject to a terminal floor. Moreover, the preference between consumption and terminal wealth as well as the intertemporal coefficient of risk aversion are time-varying and therefore depend on the age of the considered pension cohort. The optimal consumption and investment policies are calculated in the case of a Black–Scholes financial market framework and hyperbolic absolute risk aversion (HARA) utility functions. We generalize Ye (American control conference, 2008) by adding an age-dependent coefficient of risk aversion and extend Steffensen (J Econ Dyn Control 35(5):659–667, 2011), Hentschel (Planning for individual retirement: optimal consumption, investment and retirement timing under different preferences and habit persistence. Ph.D. thesis, Ulm University, 2016) and Aase (Stochastics 89(1):115–141, 2017) by considering consumption in combination with terminal wealth and allowing for consumption and terminal wealth floors via an application of HARA utility functions. A case study on fitting several models to realistic, time-dependent life-cycle consumption and relative investment profiles shows that only our extended model with time-varying preference parameters provides sufficient flexibility for an adequate fit. This is of particular interest to life-cycle products for (private) pension investments or pension insurance in general.

A Proofs

1.1 A.1 The consumption problem

Proof of Theorem 1

The Lagrangian of the Problem (11) subject to (12) is

$$\begin{aligned} \mathcal {L}(c,\lambda _{1}) = {}&\mathbb {E}\left[ \int _{0}^{T} U_{1}(t,c(t)) dt\right] - \lambda _{1} \left( \mathbb {E}\left[ \int _{0}^{T} \tilde{Z}(t) \left( c(t) - y(t)\right) dt\right] - v_{1}\right) \\ = {}&\mathbb {E}\left[ \int _{0}^{T} U_{1}(t,c(t)) - \lambda _{1} \left( \tilde{Z}(t) \left( c(t) - y(t)\right) - \frac{1}{T} v_{1}\right) dt\right] . \end{aligned}$$

By the structure of the utility function, the optimal $c_{1}$ fulfills $c_{1}(t;v_{1}) > \bar{c}(t)$ and thus the first order conditions involve existence of a Lagrange multiplier $\lambda _{1} = \lambda _{1}(v_{1}) > 0$ such that the optimal $c_{1}$ maximizes $\mathcal {L}(c,\lambda _{1})$ and such that complementary slackness holds true. Hence it can be shown that the Karush–Kuhn–Tucker conditions besides the first derivative condition are satisfied. Following [1], let $\nabla _{h} \mathcal {L}(c,\lambda _{1};h)$ denote the directional derivative of $\mathcal {L}(c,\lambda _{1})$ in the feasible direction h. The directional derivative of a function f in the direction h is generally defined by

$$\begin{aligned} \nabla _{h} f(x) = \lim _{y \rightarrow 0} \frac{f(x + h y) - f(x)}{y}. \end{aligned}$$

If f is differentiable at x this results in

$$\begin{aligned} \nabla _{h} f(x) = f^{\prime }(x) h. \end{aligned}$$

In our case, for the inner function it holds

$$\begin{aligned}&\nabla _{h} \left( U_{1}(t,c(t)) - \lambda _{1} \left( \tilde{Z}(t) \left( c(t) - y(t)\right) - \frac{1}{T} v_{1}\right) \right) \\&\quad = \frac{\partial }{\partial c} \left( U_{1}(t,c(t)) - \lambda _{1} \left( \tilde{Z}(t) \left( c(t) - y(t)\right) - \frac{1}{T} v_{1}\right) \right) h(t)\\&\quad = \left( \frac{\partial }{\partial c} U_{1}(t,c(t)) - \lambda _{1} \tilde{Z}(t)\right) h(t). \end{aligned}$$

By the dominated convergence theorem, which allows interchanging expectation and differentiation, the first order condition gives

$$\begin{aligned} 0 = {}&\mathbb {E}\left[ \int _{0}^{T} \left( \frac{\partial }{\partial c} U_{1}(t,c(t)) - \lambda _{1} \tilde{Z}(t)\right) h(t) dt\right] \\ = {}&\mathbb {E}\left[ \int _{0}^{T} \left( e^{- \beta t} a(t) \left( \frac{1}{1-b(t)} \left( c(t) - \bar{c}(t)\right) \right) ^{b(t)-1} - \lambda _{1} \tilde{Z}(t)\right) h(t) dt\right] \end{aligned}$$

for all feasible h. In order to fulfill this condition for any h, the optimal consumption rate process must be

$$\begin{aligned} c_{1}(t;v_{1}) = (1-b(t)) \left( \lambda _{1} \frac{e^{\beta t}}{a(t)} \tilde{Z}(t)\right) ^{\frac{1}{b(t)-1}} + \bar{c}(t),\ t \in [0,T]. \end{aligned}$$

(A.1)

Since $U_{1}(t,c)$ strictly increases in c, the budget constraint (12) for the optimal solution in (11) turns to equality, i.e.

$$\begin{aligned} \mathbb {E}\left[ \int _{0}^{T} \tilde{Z}(t) \left( c_{1}(t;v_{1}) - y(t)\right) dt\right] = v_{1}. \end{aligned}$$

When plugging in (A.1) and by Fubini, the budget condition after simple calculations turns into

$$\begin{aligned} v_{1} = {}&\mathbb {E}\left[ \int _{0}^{T} \tilde{Z}(t) \left( (1-b(t)) \left( \lambda _{1} \frac{e^{\beta t}}{a(t)} \tilde{Z}(t)\right) ^{\frac{1}{b(t)-1}} + \bar{c}(t) - y(t)\right) dt\right] \\ = {}&\int _{0}^{T} (1-b(t)) \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \lambda _{1}^{\frac{1}{b(t)-1}} dt + F_{1}(0). \end{aligned}$$

Here we used that $\tilde{Z}(t)$ is a log-normal random variable and so is $\tilde{Z}(t)^{\frac{b(t)}{b(t)-1}}$. For any $v_{1} > F_{1}(0) = \int _{0}^{T} e^{- r t} \left( \bar{c}(t) - y(t)\right) dt$, the above equality determines $\lambda _{1} > 0$ uniquely, since the integral in which $\lambda _{1}$ appears strictly decreases in $\lambda _{1}$ and has the limits 0 and $\infty$ as $\lambda _{1}$ approaches $\infty$ and 0. It follows immediately that the condition $v_{1} > \int _{0}^{T} e^{- r t} \left( \bar{c}(t) - y(t)\right) dt$ in (13) is inevitable. The optimal wealth process $V_{1}(t ; v_{1})$ which arises by applying $c_{1}(t ; v_{1})$ is

$$\begin{aligned} V_{1}(t ; v_{1}) = {}&\mathbb {E}\left[ \int _{t}^{T} \frac{\tilde{Z}(s)}{\tilde{Z}(t)} \left( c_{1}(s;v_{1}) - y(s)\right) ds \Big | \mathcal {F}_{t}\right] \\ = {}&\frac{1}{\tilde{Z}(t)} \bigg \{\int _{t}^{T} (1-b(s)) \left( \lambda _{1} \frac{e^{\beta s}}{a(s)}\right) ^{\frac{1}{b(s)-1}} \mathbb {E}\left[ \tilde{Z}(s)^{\frac{b(s)}{b(s)-1}} \Big | \mathcal {F}_{t}\right] ds \\&\quad \quad \quad + \int _{t}^{T} \left( \bar{c}(s) - y(s)\right) \mathbb {E}\left[ \tilde{Z}(s) \Big | \mathcal {F}_{t}\right] ds\bigg \}. \end{aligned}$$

$\tilde{Z}(s)$ can be written as $\frac{\tilde{Z}(s)}{\tilde{Z}(t)} \tilde{Z}(t)$ where $\frac{\tilde{Z}(s)}{\tilde{Z}(t)}$ is independent of $\mathcal {F}_{t}$ and $\tilde{Z}(t)$ is $\mathcal {F}_{t}$-measurable. Therefore it follows

$$\begin{aligned} \mathbb {E}\left[ \tilde{Z}(s)^{\eta } \Big | \mathcal {F}_{t}\right] = \tilde{Z}(t)^{\eta } e^{- \eta \left( r + \frac{1}{2}\Vert \gamma \Vert ^{2}\right) (s-t) + \frac{1}{2} \eta ^{2} \Vert \gamma \Vert ^{2} (s-t)} = \tilde{Z}(t)^{\eta } e^{- \eta \left( r - \frac{1}{2} (\eta - 1) \Vert \gamma \Vert ^{2}\right) (s-t)} \end{aligned}$$

for any $\eta \in \mathbb {R}$, where we used that $\frac{\tilde{Z}(s)}{\tilde{Z}(t)}$ and thus $\left( \frac{\tilde{Z}(s)}{\tilde{Z}(t)}\right) ^{\eta }$ are log-normally distributed. Define the function g by

$$\begin{aligned} g(s,t; v_{1}) = (1-b(s)) \left( \frac{e^{\beta s - b(s) \left( r - \frac{1}{2} \frac{1}{b(s)-1} \Vert \gamma \Vert ^{2}\right) (s-t)}}{a(s)}\right) ^{\frac{1}{b(s)-1}} \lambda _{1}^{\frac{1}{b(s)-1}}, \end{aligned}$$

then the optimal wealth process is given by

$$\begin{aligned} V_{1}(t ; v_{1}) = \int _{t}^{T} g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}} ds + F_{1}(t) \end{aligned}$$

(A.2)

with $F_{1}(t)$ defined in (13). The dynamics can be calculated as

$$\begin{aligned} d V_{1}(t ; v_{1}) = {}&\left( - g(t,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(t)-1}} dt + \int _{t}^{T} d_{t} \left( g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}}\right) ds\right) \\&+ \left( - \left( \bar{c}(t) - y(t)\right) dt + \int _{t}^{T} d_{t} \left( e^{- r (s-t)} \left( \bar{c}(s) - y(s)\right) \right) ds\right) \\ = {}&\left( - g(t,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(t)-1}} - \left( \bar{c}(t) - y(t)\right) + \int _{t}^{T} r e^{- r (s-t)} \left( \bar{c}(s) - y(s)\right) ds\right) dt \\&+ \int _{t}^{T} d_{t} \left( g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}}\right) ds. \end{aligned}$$

Moreover, after differentiation we obtain

$$\begin{aligned} d_{t} g(s,t; v_{1}) = {}&(1-b(s)) \left( \frac{e^{\beta s - b(s) \left( r - \frac{1}{2} \frac{1}{b(s)-1} \Vert \gamma \Vert ^{2}\right) s}}{a(s)}\right) ^{\frac{1}{b(s)-1}} \lambda _{1}^{\frac{1}{b(s)-1}} d_{t} \left( e^{\frac{b(s)}{b(s)-1} \left( r - \frac{1}{2} \frac{1}{b(s)-1} \Vert \gamma \Vert ^{2}\right) t}\right) \\ = {}&\frac{b(s)}{b(s)-1} \left( r - \frac{1}{2} \frac{1}{b(s)-1} \Vert \gamma \Vert ^{2}\right) g(s,t; v_{1}) dt. \end{aligned}$$

By applying Itô’s formula to $d \left( \tilde{Z}(t)^{\frac{1}{b(s)-1}}\right)$, we receive

$$\begin{aligned} d_{t} \left( g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}}\right) = {}&g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}} \left\{ \left( r - \frac{1}{b(s)-1} \Vert \gamma \Vert ^{2}\right) dt - \frac{1}{b(s)-1} \gamma ' dW(t)\right\} . \end{aligned}$$

Define

$$\begin{aligned} Y(t) = \int _{t}^{T} \frac{1}{b(s)-1} g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}} ds. \end{aligned}$$

In summary, the dynamics of the optimal wealth process can be calculated to be

$$\begin{aligned} d V_{1}(t ; v_{1}) = \mu _{V_{1}}(t) dt - Y(t) \gamma ' dW(t) \end{aligned}$$

(A.3)

with drift

$$\begin{aligned} \mu _{V_{1}}(t) = r V_{1}(t ; v_{1}) - g(t,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(t)-1}} - \bar{c}(t) + y(t) - \Vert \gamma \Vert ^{2} Y(t). \end{aligned}$$

By (A.1) it follows

$$\begin{aligned} c_{1}(t;v_{1}) = (1-b(t)) \left( \lambda _{1} \frac{e^{\beta t}}{a(t)} \tilde{Z}(t)\right) ^{\frac{1}{b(t)-1}} + \bar{c}(t) = g(t,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(t)-1}} + \bar{c}(t). \end{aligned}$$

Hence

$$\begin{aligned} \mu _{V_{1}}(t) = r V_{1}(t ; v_{1}) - c_{1}(t;v_{1}) + y(t) - \Vert \gamma \Vert ^{2} Y(t). \end{aligned}$$

In order to determine the optimal investment strategy $\pi _{1}(t ; v_{1})$ to Problem (11) we compare the optimal wealth dynamics in (4) and (A.3):

$$\begin{aligned} d V_{1}(t ; v_{1}) = {}&V_{1}(t ; v_{1}) \left[ \left( r + \hat{\pi }_{1}(t ; v_{1})' \left( \mu - r \mathbf {1}\right) \right) dt + \hat{\pi }_{1}(t ; v_{1})'\sigma dW(t)\right] \\&\quad - c_{1}(t;v_{1}) dt + y(t) dt, \\ d V_{1}(t ; v_{1}) = {}&\left( r V_{1}(t ; v_{1}) - c_{1}(t;v_{1}) + y(t) - \Vert \gamma \Vert ^{2} Y(t)\right) dt - Y(t) \gamma ' dW(t). \end{aligned}$$

Matching the diffusion terms yields the equality

$$\begin{aligned} \hat{\pi }_{1}(t ; v_{1}) = - \frac{Y(t)}{V_{1}(t ; v_{1})} \varSigma ^{-1} (\mu - r \mathbf {1}) \end{aligned}$$

which simultaneously matches the drift terms. Therefore, if we insert the formula for Y(t), we obtain

$$\begin{aligned} \hat{\pi }_{1}(t ; v_{1}) = - \frac{\int _{t}^{T} \frac{1}{b(s)-1} g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}} ds}{V_{1}(t ; v_{1})} \varSigma ^{-1} (\mu - r \mathbf {1}) \end{aligned}$$

(A.4)

which holds true for a general function b(t) that does not have to be continuous. If b(t) is a continuous function, by the first mean value theorem for integrals³ it furthermore follows that there exists $\tilde{t}_{1} \in (t,T)$ such that

$$\begin{aligned} Y(t) = {}&\int _{t}^{T} \frac{1}{b(s)-1} g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}} ds = \frac{1}{b(\tilde{t}_{1})-1} \int _{t}^{T} g(s,t; v_{1}) \tilde{Z}(t)^{\frac{1}{b(s)-1}} ds \\ {\mathop {=}\limits ^{(\hbox {A.2})}} {}&\frac{1}{b(\tilde{t}_{1})-1} \left( V_{1}(t ; v_{1}) - F_{1}(t)\right) . \end{aligned}$$

This determines the optimal investment strategy to be

$$\begin{aligned} \hat{\pi }_{1}(t ; v_{1}) = \frac{1}{1 - b(\tilde{t}_{1})} \varSigma ^{-1} (\mu - r \mathbf {1}) \frac{V_{1}(t ; v_{1}) - F_{1}(t)}{V_{1}(t ; v_{1})}. \end{aligned}$$

(A.5)

$\square$

Proof of Theorem 2

Firstly, after simple calculations the value function of this problem is

$$\begin{aligned} \mathcal {V}_{1}(v_{1}) = {}&\mathbb {E}\left[ \int _{0}^{T} U_{1}(t,c_{1}(t;v_{1})) dt\right] \\ = {}&\int _{0}^{T} \frac{1-b(t)}{b(t)} \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \lambda _{1}^{\frac{b(t)}{b(t)-1}} dt, \end{aligned}$$

where $\lambda _{1}$ is subject to (14). From differentiating both sides of Eq. (14) with respect to $v_{1}$ we derive

$$\begin{aligned} 1 = \int _{0}^{T} (1-b(t)) \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \frac{\partial }{\partial v_{1}} \left( \lambda _{1}^{\frac{1}{b(t)-1}}\right) dt. \end{aligned}$$

(A.6)

This helps to identify $\mathcal {V}_{1}^{\prime }(v_{1})$ to be

$$\begin{aligned} \mathcal {V}_{1}^{\prime }(v_{1}) = {}&\lambda _{1} \int _{0}^{T} (1-b(t)) \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \frac{\partial }{\partial v_{1}} \left( \lambda _{1}^{\frac{1}{b(t)-1}}\right) dt {\mathop {=}\limits ^{(\hbox {A.6})}} \lambda _{1}. \end{aligned}$$

Equation (A.6) further implies concavity of $\mathcal {V}_{1}(v_{1})$ as after differentiation we receive

$$\begin{aligned} 1 = - \lambda _{1}^{\prime } \int _{0}^{T} \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \lambda _{1}^{- \frac{b(t)-2}{b(t)-1}} dt \end{aligned}$$

and thus

$$\begin{aligned} \mathcal {V}_{1}^{\prime \prime }(v_{1}) = \lambda _{1}^{\prime } = - \left( \int _{0}^{T} \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \lambda _{1}^{- \frac{b(t)-2}{b(t)-1}} dt\right) ^{-1} < 0. \end{aligned}$$

$\square$

1.2 A.2 The terminal wealth problem

Proof of Theorem 3

The Lagrangian of the Problem (16) subject to (17) is

$$\begin{aligned} \mathcal {L}(V,\lambda _{2}) = \mathbb {E}\left[ U_{2}(V)\right] - \lambda _{2} \left( \mathbb {E}\left[ \tilde{Z}(T) V\right] - v_{2}\right) = \mathbb {E}\left[ U_{2}(V) - \lambda _{2} \left( \tilde{Z}(T) V - v_{2}\right) \right] . \end{aligned}$$

First of all, it is clear that ${c_{2}(t ; v_{2}) \equiv 0}$. By the structure of the utility function, the optimal $V_{2}$ fulfills $V_{2}(T;v_{2}) > F$ and thus the first order conditions involve existence of a Lagrange multiplier $\lambda _{2} = \lambda _{2}(v_{2}) > 0$ such that the optimal $V_{2}$ maximizes $\mathcal {L}(V,\lambda _{2})$ and such that complementary slackness holds true. Hence it can be shown that the Karush–Kuhn–Tucker conditions besides the first derivative condition are satisfied. By the dominated convergence theorem, the first order condition with respect to the directional derivative gives

$$\begin{aligned} 0 = \mathbb {E}\left[ \left( \frac{\partial }{\partial V} U_{2}(V) - \lambda _{2} \tilde{Z}(T)\right) h\right] = \mathbb {E}\left[ \left( e^{- \beta T} \hat{a} \left( \frac{1}{1-\hat{b}} (V-F)\right) ^{\hat{b}-1} - \lambda _{2} \tilde{Z}(T)\right) h\right] , \end{aligned}$$

which has to be satisfied for all suitable h; hence the optimal terminal wealth has to fulfill

$$\begin{aligned} V_{2}(T ; v_{2}) = (1-\hat{b}) \left( \lambda _{2} \frac{e^{\beta T}}{\hat{a}} \tilde{Z}(T)\right) ^{\frac{1}{\hat{b}-1}} + F. \end{aligned}$$

(A.7)

Since $U_{2}(V)$ strictly increases in V, complementary slackness implies equality for the budget constraint

$$\begin{aligned} \mathbb {E}\left[ \tilde{Z}(T) V_{2}(T ; v_{2})\right] = v_{2}. \end{aligned}$$

Using (A.7) we obtain after straightforward calculations

$$\begin{aligned} v_{2} = {}&\mathbb {E}\left[ \tilde{Z}(T) \left( (1-\hat{b}) \left( \lambda _{2} \frac{e^{\beta T}}{\hat{a}} \tilde{Z}(T)\right) ^{\frac{1}{\hat{b}-1}} + F\right) \right] \\ = {}&(1-\hat{b}) \left( \frac{e^{\left[ \beta - \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T}}{\hat{a}}\right) ^{\frac{1}{\hat{b}-1}} \lambda _{2}^{\frac{1}{\hat{b}-1}} + F_{2}(0). \end{aligned}$$

Solving for $\lambda _{2}$ yields

$$\begin{aligned} \lambda _{2} = e^{-\left[ \beta - \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \left( 1-\hat{b}\right) ^{1-\hat{b}} \hat{a} \left( v_{2} - F_{2}(0)\right) ^{\hat{b}-1} \end{aligned}$$

(A.8)

where ${v_{2} > F_{2}(0) = e^{- r T} F}$ in (18) is required. Plugging this back into (A.7), the optimal terminal wealth becomes

$$\begin{aligned} V_{2}(T ; v_{2}) = \left( v_{2} - F_{2}(0)\right) e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) T} \tilde{Z}(T)^{\frac{1}{\hat{b}-1}} + F. \end{aligned}$$

(A.9)

The optimal wealth process replicates $V_{2}(T ; v_{2})$ and is uniquely given by

$$\begin{aligned} V_{2}(t ; v_{2}) = \mathbb {E}\left[ \frac{\tilde{Z}(T)}{\tilde{Z}(t)} V_{2}(T ; v_{2}) \Big | \mathcal {F}_{t}\right] = \left( v_{2} - F_{2}(0)\right) e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) t} \tilde{Z}(t)^{\frac{1}{\hat{b}-1}} + F_{2}(t) \end{aligned}$$

(A.10)

with $F_{2}(t)$ defined in (18). Recall that

$$\begin{aligned} d \left( \tilde{Z}(t)^{\frac{1}{\hat{b}-1}}\right) = \tilde{Z}(t)^{\frac{1}{\hat{b}-1}} \left\{ \left[ - \frac{1}{\hat{b}-1} r + \frac{1}{2} \frac{1}{\hat{b}-1} \left( \frac{1}{\hat{b}-1} - 1\right) \Vert \gamma \Vert ^{2}\right] dt - \frac{1}{\hat{b}-1} \gamma ' dW(t)\right\} . \end{aligned}$$

It follows by Itô

$$\begin{aligned}&d \left( e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) t} \tilde{Z}(t)^{\frac{1}{\hat{b}-1}}\right) \\&\qquad = e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) t} d \left( \tilde{Z}(t)^{\frac{1}{\hat{b}-1}}\right) + \tilde{Z}(t)^{\frac{1}{\hat{b}-1}} d \left( e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) t}\right) + 0 \\&\qquad = e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) t} \tilde{Z}(t)^{\frac{1}{\hat{b}-1}} \left\{ \left( r - \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) dt - \frac{1}{\hat{b}-1} \gamma ' dW(t)\right\} . \end{aligned}$$

Then, after simple transformations and using (A.10), the optimal wealth dynamics can be calculated as

$$\begin{aligned} d V_{2}(t ; v_{2}) = {}&\left( v_{2} - F_{2}(0)\right) d \left( e^{\frac{\hat{b}}{\hat{b}-1} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) t} \tilde{Z}(t)^{\frac{1}{\hat{b}-1}}\right) + r F_{2}(t) dt \\ = {}&r V_{2}(t ; v_{2}) dt + \left( V_{2}(t ; v_{2}) - F_{2}(t)\right) \left\{ - \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2} dt - \frac{1}{\hat{b}-1} \gamma ' dW(t)\right\} . \end{aligned}$$

Comparing the diffusion term with the one from (4) for $y(t) \equiv 0$ implies

$$\begin{aligned} \hat{\pi }_{2}(t ; v_{2}) = \frac{1}{1-\hat{b}} \varSigma ^{-1} (\mu - r \mathbf {1}) \frac{V_{2}(t ; v_{2}) - F_{2}(t)}{V_{2}(t ; v_{2})} \end{aligned}$$

(A.11)

which automatically matches the drifts iff ${c_{2}(t ; v_{2}) \equiv 0}$. $\square$

Proof of Theorem 4

Using the previous results, the value function of this problem can simply be calculated as

$$\begin{aligned} \mathcal {V}_{2}(v_{2}) = \mathbb {E}\left[ U_{2}(V_{2}(T ; v_{2}))\right] = e^{\left[ - \beta + \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \frac{\left( 1-\hat{b}\right) ^{1-\hat{b}}}{\hat{b}} \hat{a} \left( v_{2} - F_{2}(0)\right) ^{\hat{b}}. \end{aligned}$$

This implies

$$\begin{aligned} \mathcal {V}_{2}^{\prime }(v_{2}) = e^{\left[ - \beta + \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \frac{\left( 1-\hat{b}\right) ^{1-\hat{b}}}{\hat{b}} \hat{a} \hat{b} \left( v_{2} - F_{2}(0)\right) ^{\hat{b}-1} {\mathop {=}\limits ^{(\hbox {A.8})}} \lambda _{2} > 0. \end{aligned}$$

Due to the assumption ${v_{2} - F_{2}(0) > 0}$ in (18), it is straightforward that ${\mathcal {V}_{2}^{\prime \prime }(v_{2}) = \lambda _{2}^{\prime } < 0}$:

$$\begin{aligned} \mathcal {V}_{2}^{\prime \prime }(v_{2}) = - e^{\left[ - \beta + \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \left( 1-\hat{b}\right) ^{2-\hat{b}} \hat{a} \left( v_{2} - F_{2}(0)\right) ^{\hat{b}-2} < 0. \end{aligned}$$

$\square$

1.3 A.3 Optimal merging of the individual solutions

Proof of Theorem 5

The proof can be performed by simply following the line of [44, 49]. $\square$

Proof of Lemma 1

In accordance with Theorem 5 and by expressing $v_{2} = v_{0} - v_{1}$, the candidate for the optimal ${v_{1}^{\star }}$ is the one that satisfies the first order derivative condition on the budget

$$\begin{aligned} 0 = \frac{\partial }{\partial v_{1}} \left( \mathcal {V}_{1}(v_{1}) + \mathcal {V}_{2}(v_{0} - v_{1})\right) = \mathcal {V}_{1}^{\prime }(v_{1}) - \mathcal {V}_{2}^{\prime }(v_{0} - v_{1}) \end{aligned}$$

such that ${v_{1}^{\star } \ge F_{1}(0)}$, ${v_{2}^{\star } = v_{0} - v_{1}^{\star }}$ with ${v_{2}^{\star } \ge F_{2}(0)}$; thus ${F_{1}(0) \le v_{1}^{\star } \le v_{0} - F_{2}(0)}$. Theorems 2 and 4 tell that $\mathcal {V}_{1}(v_{1})$ and $\mathcal {V}_{2}(v_{2})$ are strictly concave functions in $v_{1}$ respectively $v_{2}$. Therefore, it follows

$$\begin{aligned} 0 = \frac{\partial ^{2}}{\partial v_{1}^{2}} \left( \mathcal {V}_{1}(v_{1}) + \mathcal {V}_{2}(v_{0} - v_{1})\right) = \mathcal {V}_{1}^{\prime \prime }(v_{1}) + \mathcal {V}_{2}^{\prime \prime }(v_{0} - v_{1}) < 0. \end{aligned}$$

This implies that the candidates $v_{1}^{\star }$ and ${v_{2}^{\star } = v_{0} - v_{1}^{\star }}$ are the solution when the constraint $F_{1}(0) \le v_{1}^{\star } \le v_{0} - F_{2}(0)$ applies. $\square$

Proof of Lemma 2

In accordance with Theorems 2 and 4 we have

$$\begin{aligned} \mathcal {V}_{1}^{\prime }(v_{1}) = {}&\lambda _{1}, \\ \mathcal {V}_{2}^{\prime }(v_{2}) = {}&\lambda _{2} = e^{-\left[ \beta - \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \left( 1-\hat{b}\right) ^{1-\hat{b}} \hat{a} \left( v_{2} - F_{2}(0)\right) ^{\hat{b}-1}. \end{aligned}$$

By equating $\mathcal {V}_{1}^{\prime }(v_{1})$ and $\mathcal {V}_{2}^{\prime }(v_{0} - v_{1})$ we obtain

$$\begin{aligned} {}(19) \text { in Lemma } 1 \Leftrightarrow&\mathcal {V}_{1}^{\prime }(v_{1}) = \mathcal {V}_{2}^{\prime }(v_{0} - v_{1}) \\ \Leftrightarrow&\lambda _{1} = \lambda _{2} = e^{-\left[ \beta - \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \left( 1-\hat{b}\right) ^{1-\hat{b}} \hat{a} \left( v_{0} - v_{1} - F_{2}(0)\right) ^{\hat{b}-1}. \end{aligned}$$

Inserting $\lambda _{1}$ in Eq. (14), the optimal $v_{1}^{\star }$ is the solution to

$$\begin{aligned} v_{1} - \int _{0}^{T} \chi (t) \left( v_{0} - v_{1} - F_{2}(0)\right) ^{\frac{\hat{b}-1}{b(t)-1}} dt = F_{1}(0), \end{aligned}$$

where the continuous function $\chi (t)$ is defined by

$$\begin{aligned} \chi (t) = (1-b(t)) \left( 1-\hat{b}\right) ^{\frac{1-\hat{b}}{b(t)-1}} \left( \frac{\hat{a}}{a(t)}\right) ^{\frac{1}{b(t)-1}} \left( \frac{e^{\left[ \beta - b(t) \left( r - \frac{1}{2} \frac{1}{b(t)-1} \Vert \gamma \Vert ^{2}\right) \right] t}}{e^{\left[ \beta - \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T}}\right) ^{\frac{1}{b(t)-1}} > 0. \end{aligned}$$

It remains to verify ${F_{1}(0) \le v_{1}^{\star } \le v_{0} - F_{2}(0)}$ and uniqueness of $v_{1}^{\star }$. For this sake, define the function f by

$$\begin{aligned} f: (-\infty ,v_{0} - F_{2}(0)],\ f(x) = x - \int _{0}^{T} \chi (t) \left( v_{0} - x - F_{2}(0)\right) ^{\frac{\hat{b}-1}{b(t)-1}} dt - F_{1}(0). \end{aligned}$$

$v_{1}^{\star }$ is the root of the function f, i.e. ${f(v_{1}^{\star }) = 0}$, if it holds ${v_{1}^{\star } \ge F_{1}(0)}$. f is continuous in x, the exponent ${\frac{\hat{b}-1}{b(t)-1}}$ within the first integral is positive. Furthermore, due to ${v_{0} > F(0)}$ claimed in (9) and ${F(t) = F_{1}(t) + F_{2}(t)}$, we have for the limits

$$\begin{aligned} \lim _{x \searrow F_{1}(0)} f(x) = {}&- \int _{0}^{T} \chi (t) \left( v_{0} - F(0)\right) ^{\frac{\hat{b}-1}{b(t)-1}} dt < 0, \\ \lim _{x \nearrow v_{0} - F_{2}(0)} f(x) = {}&v_{0} - F(0) > 0. \end{aligned}$$

Note, ${F_{1}(0) \le v_{1} = v_{0} - v_{2} \le v_{0} - F_{2}(0)}$ for general $v_{1}$ and $v_{2}$. Additionally, f is strictly monotone increasing in x since

$$\begin{aligned} f^{\prime }(x) = 1 + \int _{0}^{T} \chi (t) \frac{\hat{b}-1}{b(t)-1} \left( v_{0} - x - F_{2}(0)\right) ^{\frac{\hat{b}-b(t)}{b(t)-1}} dt > 0,\ \forall x \le v_{0} - F_{2}(0). \end{aligned}$$

We conclude that there exists a unique root ${x \in [F_{1}(0), v_{0} - F_{2}(0)]}$ such that ${f(x) = 0}$. Therefore, we conclude that the optimal $v_{1}^{*}$ and $v_{2}^{\star } = v_{0} - v_{1}^{*}$ exist and are unique. $v_{1}^{*}$ is the solution to Eq. (20). The optimal Lagrange multiplier $\lambda _{1}^{\star } = \lambda _{1}(v_{1}^{\star })$ is then given by

$$\begin{aligned} \lambda _{1}^{\star } = \left( 1-\hat{b}\right) ^{1-\hat{b}} \hat{a} e^{-\left[ \beta - \hat{b} \left( r - \frac{1}{2} \frac{1}{\hat{b}-1} \Vert \gamma \Vert ^{2}\right) \right] T} \left( v_{0} - v_{1}^{\star } - F_{2}(0)\right) ^{\hat{b}-1}. \end{aligned}$$

$\square$

Proof of Theorem 6

Starting with ${V^{\star }(t; v_{0}) = V_{1}(t;v_{1}^{\star }) + V_{2}(t;v_{2}^{\star })}$ we compare the dynamics of both sides of the equation:

$$\begin{aligned} d V^{\star }(t; v_{0}) = d V_{1}(t;v_{1}^{\star }) + d V_{2}(t;v_{2}^{\star }). \end{aligned}$$

(A.12)

Comparing the diffusion terms in (A.12), by inserting the wealth sde (4) for $V^{\star }(t; v_{0})$, $V_{1}(t;v_{1}^{\star })$ and $V_{2}(t;v_{2}^{\star })$ (with $y(t) \equiv 0$ for $V_{2}(t;v_{2}^{\star })$), gives

$$\begin{aligned} \hat{\pi }^{\star }(t; v_{0}) = \frac{\hat{\pi }_{1}(t;v_{1}^{\star }) V_{1}(t;v_{1}^{\star }) + \hat{\pi }_{2}(t;v_{2}^{\star }) V_{2}(t;v_{2}^{\star })}{V^{\star }(t; v_{0})}. \end{aligned}$$

Inserting this back and comparing the drift terms finally leads to

$$\begin{aligned} c^{\star }(t; v_{0}) = c_{1}(t;v_{1}^{\star }). \end{aligned}$$

Notice that the pair ${(\hat{\pi }^{\star },c^{\star })}$ is admissible, i.e. ${(\hat{\pi }^{\star },c^{\star }) \in \varLambda }$ because ${(\hat{\pi }_{1},c_{1}) \in \varLambda _{1}}$ and ${(\hat{\pi }_{2},0) \in \varLambda _{2}}$ which implies

$$\begin{aligned} V^{\star }(t; v_{0}) = \underbrace{V_{1}(t;v_{1}^{\star })}_{\ge - \int _{t}^{T} e^{- r (s-t)} y(s) ds} + \underbrace{V_{2}(t;v_{2}^{\star })}_{\ge 0} \ge - \int _{t}^{T} e^{- r (s-t)} y(s) ds,\ \mathbb {P}-a.s.,\ \forall t \in [0,T]. \end{aligned}$$

Using Lemma 2 and the solutions in Theorems 1 and 3 we arrive at the stated results in Theorem 6 after straightforward calculations for the utility setup in (8). $\square$

Proof of Remark 4

The formula for the optimal investment strategy is straightforward from Theorem 6 as ${b(\tilde{t}_{1}^{\star }) \equiv \hat{b}}$ and ${V^{\star }(t ; v_{0}) = V_{1}(t ; v_{1}^{\star }) + V_{2}(t ; v_{2}^{\star })}$ for any $t \in [0,T]$. The optimal $v_{1}^{\star }$ and the corresponding $g(s,t; v_{1}^{\star })$ can be determined by Lemma 2, $V_{1}(t ; v_{1}^{\star })$, $V_{2}(t ; v_{2}^{\star })$ and thus $V^{\star }(t ; v_{0})$ then follow from Theorem 6 with simple calculations. Using again Theorem 6 finally leads to the stated optimal consumption rate $c^{\star }(t;v_{0})$. $\square$