Adaptive basket liquidation

Abstract

We consider the infinite-horizon optimal basket portfolio liquidation problem for a von Neumann–Morgenstern investor in a multi-asset extension of the liquidity model of Almgren (Appl. Math. Finance 10:1–18, 2003) with cross-asset impact. Using a stochastic control approach, we establish a “separation theorem”: the sequence of portfolios held during an optimal liquidation depends only on the (co-)variance and (cross-asset) market impact of the assets, while the speed with which these portfolios are reached depends only on the utility function of the trader. We derive partial differential equations for both the sequence of portfolios reached during the execution and the trading speed.

Appendix: Proofs of results

This appendix consists of three parts. First, we discuss mean–variance optimal strategies and prove Theorem 4.1. By extending methods from calculus of variations to the infinite-horizon setting, we show that optimal strategies exist, are unique and satisfy Bellman’s principle of optimality. In the second subsection, we show that a smooth solution to the HJB equation exists and provide some of its properties. This is achieved by first obtaining a solution of the PDE for $\tilde{a}$ and then defining $\tilde{v}$ by a transport equation with coefficient $\tilde{a}$. In the third subsection, we apply a verification argument and show that this solution of the HJB equation must be equal to the value function. Theorems 5.1–5.3 and Corollary 5.4 are direct consequences of the propositions in the last two subsections. The proofs there have a similar structure to the proofs in [30]. However, they differ in a few subtle points and we therefore provide them in full detail.

1.1 A.1 Optimal mean–variance strategies

To obtain optimal trading strategies for the infinite-horizon setting, we first show that optimal strategies exist for the setting with finite horizon $T$ (i.e., $X_{t}=0$ for $t\geq T$) and then consider the limit $T\to\infty$.

Lemma A.1

If a mean–variance optimal trading strategy exists for $X_{0}\in\mathbb {R}^{n}$ and time horizon $T\in(0,\infty]$, then this strategy is unique.

Proof

This is clear from strict convexity of the functional $X\mapsto f({\xi})+\frac{1}{2}X^{\top}\varSigma X$. □

Proposition A.2

For finite liquidation horizons $T\in(0,\infty)$, a mean–variance optimal liquidation strategy ${\xi}^{(X_{0},T)}$ exists for all initial portfolios $X_{0}\in\mathbb {R}^{n}$. The portfolio evolution $(X_{t}^{{\xi}^{(X_{0},T)}})$ is $C^{1}$ in $t$ (i.e., the optimal trading vector $t\mapsto{\xi}_{t}^{(X_{0},T)}$ is continuous). We denote the time at which the portfolio $X_{t}^{\xi}$ attains zero by

$$ T_{0}:=\inf\{t>0:X_{t}^{\xi}=0\}\in(0,T]. $$

For $t\in[0,T_{0})$, the portfolio evolution $t\mapsto X_{t}^{\xi}$ is even $C^{2}$ and fulfills the Euler–Lagrange equation

$$ \varSigma X_{t} = D^{2}f(-\dot{X}_{t})\ddot{X}_{t}. $$

The optimal trading vector ${\xi}^{(X_{0},T)}$ satisfies Bellman’s principle of optimality, i.e.,

$$ {\xi}_{t}^{(X_{0},T)}={\xi}_{0}^{(X_{t},T-t)}. $$

Furthermore, the initial trading speed ${\xi}_{0}$ is locally uniformly bounded. More precisely, for each portfolio ${\bar{X}}_{0}\in\mathbb {R}$ and each time horizon ${\bar{T}}$, there are $\delta>0$ and $C>0$ such that $|{\xi} _{0}^{(X_{0},T)}|< C$ for all $|X_{0}-{\bar{X}}_{0}|<\delta$ and $T\geq{\bar{T}}$.

Theorem 2.2 in [31] establishes the existence of a mean–variance optimal strategy for finite liquidation horizons, but not the uniform bound on ${\xi}_{0}$, which we need for our proof of Proposition A.3. We therefore present a self-contained proof of Proposition A.2 establishing this bound.

Proof of Proposition A.2

First, we observe that for any mean–variance optimal ${\xi}$, there is an a priori upper bound $K>0$ independent of $T$ such that

$$ \sup\big\{ |X_{t}^{\xi}|:t\in[0,T]\big\} < K. $$

To see this, select an arbitrary $\tilde{K}>X_{0}^{\top}\varSigma X_{0}$ and assume that $t\mapsto\frac{1}{2}X_{t}^{\top}\varSigma X_{t}$ attains $\tilde{K}$ at $T_{2}:=\min\{t>0:\frac{1}{2}X_{t}^{\top}\varSigma X_{t}\geq\tilde{K}\}$. Then

$$ \frac{\tilde{K}}{2}\leq\frac{1}{2} X_{t}^{\top}\varSigma X_{t}\leq\tilde{K} $$

for all $t\in[T_{1},T_{2}]$ with $T_{1}:=\max\{t< T_{2}:\frac{1}{2}X_{t}^{\top}\varSigma X_{t}\leq\frac {\tilde{K}}{2}\}$. We therefore have

$$\begin{aligned} \int_{T_{1}}^{T_{2}}\left(f({\xi}_{t})+\frac{1}{2}X_{t}^{\top}\varSigma X_{t}\right)\,dt \geq& \left( \min_{{\scriptstyle \tilde{X}\in{\bar{\mathcal {X}}}\ \mathrm{such}\ \mathrm{that}\atop\scriptstyle \frac{1}{2}\tilde{X}_{T_{1}}^{\top}\varSigma\tilde{X}_{T_{1}}=\frac{\tilde{K}}{2}, \frac{1}{2}\tilde{X}_{T_{2}}^{\top}\varSigma\tilde{X}_{T_{2}}=\tilde{K}}} \int_{T_{1}}^{T_{2}} f(\tilde{{\xi}}_{t})\,dt\right) \\ &{}+ (T_{2}-T_{1})\frac{\tilde{K}}{2} \end{aligned}$$

Let $\tilde{X}^{*}$ (resp., $\tilde{{\xi}}^{*}$) be a minimizer of the first term on the right-hand side. Then $\frac{2}{\tilde{K}}\tilde {X}_{T_{1}+t(T_{2}-T_{1})}^{*}$ has derivative $\frac{2(T_{2}-T_{1})}{\tilde{K}}\tilde{{\xi }}_{T_{1}+t(T_{2}-T_{1})}^{*}$ and satisfies $\frac{1}{2}\tilde{X}_{0}^{\top}\varSigma\tilde{X}_{0}=1$ and $\frac{1}{2}\tilde{X}_{1}^{\top}\varSigma\tilde{X}_{1}=2$. Due to the scaling property from (2.1), we have

$$\begin{aligned} &\min_{{\scriptstyle \tilde{X}\in{\bar{\mathcal {X}}}\ \mathrm{such}\ \mathrm{that}\atop\scriptstyle \frac{1}{2}\tilde{X}_{T_{1}}^{\top}\varSigma\tilde{X}_{T_{1}}=\frac{\tilde{K}}{2}, \frac{1}{2}\tilde{X}_{T_{2}}^{\top}\varSigma\tilde{X}_{T_{2}}=\tilde{K}}} \int_{T_{1}}^{T_{2}} f(\tilde{{\xi}}_{t})\,dt\\ &\quad = \int_{T_{1}}^{T_{2}} f(\tilde{{\xi}}_{t}^{*})\,dt \\ & \quad= (T_{2}-T_{1}) \int_{0}^{1} f\big(\tilde{{\xi }}_{T_{1}+t(T_{2}-T_{1})}^{*}\big)\,dt \\ &\quad = (T_{2}-T_{1}) \int_{0}^{1} \bigg(\frac{\tilde {K}}{2(T_{2}-T_{1})}\bigg)^{\alpha +1} f\left(\frac{2(T_{2}-T_{1})}{\tilde{K}}\tilde{{\xi}}_{T_{1}+t(T_{2}-T_{1})}^{*}\right)\,dt\\ &\quad \geq\left(\frac{1}{T_{2}-T_{1}}\right)^{\alpha}\bigg(\frac{\tilde {K}}{2}\bigg)^{\alpha+1} \min_{{\scriptstyle \tilde{X}\in{\bar{\mathcal {X}}}\ \mathrm{such}\ \mathrm{that}\atop\scriptstyle \frac{1}{2}\tilde{X}_{0}^{\top}\varSigma\tilde{X}_{0}=1,\frac{1}{2}\tilde {X}_{1}^{\top}\varSigma\tilde{X}_{1}=2}} \int_{0}^{1} f(\tilde{{\xi}}_{t})\,dt. \end{aligned}$$

Combining the previous two derivations, we obtain

$$\begin{aligned} \int_{T_{1}}^{T_{2}}\bigg(f({\xi}_{t})+\frac{1}{2}X_{t}^{\top}\varSigma X_{t}\bigg)\,dt & \geq\bigg(\frac{1}{T_{2}-T_{1}}\bigg)^{\alpha}\bigg(\frac{\tilde {K}}{2}\bigg)^{\alpha+1}\tilde{C}+ (T_{2}-T_{1})\frac{\tilde{K}}{2} \end{aligned}$$

(A.1)

with the constant

$$ \tilde{C}:=\min_{{\scriptstyle \tilde{X}\in{\bar{\mathcal {X}}}\ \mathrm{such}\ \mathrm{that}\atop\scriptstyle \frac{1}{2}\tilde{X}_{0}^{\top}\varSigma\tilde {X}_{0}=1,\frac{1}{2}\tilde{X}_{1}^{\top}\varSigma\tilde{X}_{1}=2}} \int _{0}^{1} f(\tilde{{\xi}}_{t})\,dt>0. $$

Since $\tilde{C}$ is independent of $T_{1}$ and $T_{2}$, the right-hand side of (A.1) is bounded from below by a function of $\tilde{K}$ that is increasing and unbounded. This establishes that an optimal ${\xi}$ cannot attain arbitrarily large values of $X_{t}^{\top}\varSigma X_{t}$ (resp., $\sup_{t} |X_{t}|$).

We can therefore reduce the optimization problem with unbounded $X _{t}\in\mathbb {R}^{n}$ to an optimization problem with bounded $X_{t}\in [-K,K]^{n}$. By Tonelli’s existence theorem (see e.g. [16, Theorem 2.20]), a mean–variance optimal trading strategy exists for the bounded optimization problem; by our previous considerations, this strategy is also optimal for the unbounded optimization problem $X_{t}\in\mathbb {R}^{n}$, and we denote this strategy by ${\xi}^{(X_{0},T)}$.

In order to apply theorems ensuring continuity of even smoothness of ${\xi}$, we need to show that the optimal ${\xi}={\xi}^{(X_{0},T)}$ is essentially bounded.¹³ The idea of the following proof is that if ${\xi}$ trades extremely quickly at some points in time, then the mean–variance costs of ${\xi}$ can be reduced by “smoothing” the trading speed, i.e., slowing down trading when it is fast and accelerating it when it is slow. To formalize this argument, we first observe that there are bounds $(X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}< K_{0}$ and $\int_{0}^{T} f({\xi}_{t})dt=K_{1}<\infty$, and we define

$$\mu :\mathbb{R}\ni t\mapsto {\mu }_t:={\int }_0^t{\mathbb{1}}_{\{f({\xi }_s)\ge K_2\}}ds\in \mathbb{R},$$

where $K_{2}>0$ is a large, arbitrary constant. ${\xi}$ is essentially bounded if there is a $K_{2}> 0$ with $\mu\equiv0$. We assume that $\mu \not\equiv0$ for all $K_{2}\in(0,\infty)$ and establish a contradiction. We define the time transformation

$$ {\tilde{t}}(t,s):=s\mu_{t} + \frac{T-s\mu_{T}}{T-\mu_{T}}(t-\mu_{t}). $$

For $0< s<\frac{T}{\mu_{T}}$, this transformation is a bijection ${\tilde{t}} (\cdot,s):[0,T]\to[0,T]$ satisfying ${\tilde{t}}(0,s)=0$ and ${\tilde{t}}(T,s)=T$. When using the variables ${\tilde{t}}$ and $t$ in the following, we always assume that they are connected by this bijection, i.e., that ${\tilde{t}} ={\tilde{t}}(t,s)$. We can now define a new portfolio evolution ${Y}$ depending on $s$ by

$$ {Y}^{(s)}:\mathbb {R}_{+}\ni{\tilde{t}}\to{Y}_{{\tilde{t}}}^{(s)} := X_{t}. $$

The portfolio evolution ${Y}^{(s)}$ is absolutely continuous and fulfills

$$ {\xi}_{\tilde{t}}^{(s)}:=-\frac{d}{d{\tilde{t}}}{Y}_{\tilde{t}}^{(s)}= \textstyle\begin{cases} \frac{1}{s}{\xi}_{t} & \text{for }f({\xi}_{t})\geq K_{2}, \\ \frac{T-\mu_{T}}{T-s\mu_{T}}{\xi}_{t} & \text{for }f({\xi}_{t}) < K_{2}. \end{cases} $$

Note that ${\xi}^{(1)}={\xi}$. The mean–variance costs of executing ${Y} ^{(s)}$ are given by

$$\begin{aligned} &\int_{0}^{T} \big(f({\xi}_{\tilde{t}}^{(s)}) + ({Y}_{{\tilde {t}}}^{(s)})^{\top}\varSigma{Y}_{{\tilde{t}}}^{(s)}\big)\,d{\tilde{t}}\\ &\quad = \int_{f({\xi}_{t})\geq K_{2}} \big(f({\xi}_{\tilde{t}}^{(s)}) + ({Y}_{{\tilde{t}} }^{(s)})^{\top}\varSigma{Y}_{{\tilde{t}} }^{(s)}\big)\,d{\tilde{t}} \\ &\qquad{} + \int_{f({\xi}_{t}) < K_{2}} \big(f({\xi}_{\tilde{t}}^{(s)}) + ({Y}_{{\tilde{t}} }^{(s)})^{\top}\varSigma{Y}_{{\tilde{t}} }^{(s)}\big)\,d{\tilde{t}} \\ &\quad = s \int_{f({\xi}_{t})\geq K_{2}} \bigg(f\Big(\frac{1}{s} {\xi }_{t}\Big) + (X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}\bigg)\,dt \\ &\qquad{} + \frac{T-s\mu_{T}}{T-\mu_{T}} \int_{f({\xi}_{t}) < K_{2}} \bigg(f\Big(\frac{T-\mu_{T}}{T-s\mu_{T}}{\xi}_{t}\Big) + (X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}\bigg)\,dt \\ &\quad = \left(\frac{1}{s}\right)^{\alpha}\int_{f({\xi}_{t})\geq K_{2}} f({\xi}_{t})\,dt + s \int_{f({\xi}_{t})\geq K_{2}} (X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}\,dt \\ &\qquad{} + \left(\frac{T-\mu_{T}}{T-s\mu_{T}}\right)^{\alpha}\int _{f({\xi}_{t}) < K_{2}} f({\xi}_{t})\,dt + \frac{T-s\mu_{T}}{T-\mu_{T}} \int_{f({\xi}_{t}) < K_{2}} (X_{t}^{{\xi }})^{\top}\varSigma X_{t}^{{\xi}}\,dt. \end{aligned}$$

By differentiating with respect to $s$ at $s=1$, we have

$$\begin{aligned} &\frac{d}{ds}\bigg|_{s=1}\int_{0}^{T} \big(f({\xi}_{\tilde{t}}^{(s)}) + ({Y}_{{\tilde{t}} }^{(s)})^{\top}\varSigma{Y}_{{\tilde{t}} }^{(s)}\big)\,d{\tilde{t}} \\ &\quad = - \alpha\underbrace{\int_{f({\xi}_{t})\geq K_{2}} f({\xi} _{t})\,dt}_{\geq K_{2} \mu_{T}} + \underbrace{\int_{f({\xi}_{t})\geq K_{2}} (X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}\,dt}_{\leq K_{0} \mu _{T}} \\ &\qquad{} + \alpha\frac{\mu_{T}}{T-\mu_{T}} \underbrace{\int _{f({\xi} _{t}) < K_{2}} f({\xi}_{t})\,dt}_{\leq K_{1}-K_{2}\mu_{T}} - \frac{\mu_{T}}{T-\mu_{T}} \underbrace{\int_{f({\xi}_{t}) < K_{2}} (X_{t}^{{\xi} })^{\top}\varSigma X_{t}^{{\xi} }\,dt}_{\geq0}. \end{aligned}$$

If $K_{2}$ is large enough, the right-hand side of the above equation is smaller than zero for all possible values $\mu_{T}\in(0,\frac {K_{1}}{K_{2}}]$, which contradicts the optimality of ${\xi}={\xi}^{(1)}$. This completes the proof that ${\xi}$ is essentially bounded. Note that a suitably large bound $K_{2}$ holds for all time horizons larger than $T$ and for all initial portfolios that are close to $X_{0}$, establishing the uniform boundedness of ${\xi}_{0}$.

Since ${\xi}$ is essentially bounded, we can apply the theorems of Tonelli and Weierstrass (see [16, Theorem 2.6]) and find that $t\mapsto X_{t}^{\xi}$ is $C^{1}$ everywhere and $C^{2}$ until it attains zero. Furthermore, it fulfills the Euler–Lagrange equation. Bellman’s principle of optimality for the optimal trading vector ${\xi}$ follows by the additivity of mean costs and variance of proceeds, as already noted by [6]. □

Proposition A.3

For an infinite liquidation time horizon, a mean–variance optimal liquidation strategy ${\bar{{\xi}}}^{(X_{0})}$ exists for all initial portfolios $X_{0}\in\mathbb {R}^{n}$. The portfolio evolution $(X_{t}^{{\bar {{\xi}}}^{(X _{0})}})$ is $C^{1}$ in $t$ (i.e., the optimal trading vector $t\mapsto {\bar{{\xi}}} _{t}^{(X_{0})}$ is continuous). We denote the time at which the portfolio $X_{t}$ attains zero by

$$ T_{0}:=\inf\{t>0:X_{t}^{\bar{{\xi}}}=0\}\in(0,\infty]. $$

For $t\in[0,T_{0})$, the portfolio evolution $t\mapsto X_{t}^{\xi}$ is $C^{2}$ and fulfills the Euler–Lagrange equation

$$ \varSigma X_{t} = D^{2}f(-\dot{X}_{t})\ddot{X}_{t}. $$

(A.2)

The optimal trading vector ${\bar{{\xi}}}^{(X_{0})}$ satisfies Bellman’s principle of optimality, i.e.,

$$ {\bar{{\xi}}}_{t}^{(X_{0})}={\bar{{\xi}}}_{0}^{(X_{t})}=:{\bar{{a}}}(X_{t}), $$

with a continuous vector field ${\bar{{a}}}:\mathbb {R}^{n}\ni X\mapsto {\bar{{a}}}(X)\in\mathbb {R}^{n}$.

Proof

We first introduce some notation. For a sequence $(X^{(i)}_{0},T^{(i)})\in\mathbb {R}^{n}\times\mathbb {R}$, we define

$$\begin{aligned} {\xi}^{(i)} & := {\xi}^{(X^{(i)}_{0},T^{(i)})}, \\ X^{(i)}_{t} & := X_{t}^{{\xi}^{(i)}}, \\ T_{0}^{(i)} & := \inf\{t>0:X_{t}^{(i)}=0\}\in(0,T^{(i)}]. \end{aligned}$$

In Proposition A.2, we have established the uniform boundedness of ${\xi}_{0}$. For each $X_{0}\in\mathbb {R}^{n}$, we can therefore select a sequence $(X^{(i)}_{0},T^{(i)})$ with $\lim_{i\to\infty} X_{0}^{(i)}=X_{0}$ and $\lim_{i\to\infty} T^{(i)}=\infty$ such that ${\xi}_{0}^{(i)}$ converges to $\lim_{i\to\infty}{\xi}_{0}^{(i)} =: {\xi}_{0}^{(\infty)}$. Then we define $t\mapsto X_{t}^{(\infty)}$ as the solution to the Euler–Lagrange equation with initial values $X_{0}^{(\infty)}=X_{0}$ and $\dot{X}_{0}^{(\infty)}=-{\xi}_{0}^{(\infty)}$ until $t\mapsto X_{t}^{(\infty)}$ attains zero at time $T_{0}^{(\infty )}\in(0,\infty]$. On $[T_{0}^{(\infty)},\infty)$, we define $X _{t}^{(\infty)}\equiv0$.

Let $\phi(t,X_{0},{\xi}_{0})$ be the position and trading speed $(X _{t},\dot{X}_{t})$ at time $t$ of the solution to the Euler–Lagrange equation with initial values $X_{0}$ and ${\xi}_{0}$. Since the coefficient function $D^{2} f$ in the Euler–Lagrange equation is Lipschitz-continuous for $X\neq0$, for any compact subset $[0,T]\subset[0,T_{0}^{(\infty)})$, there exists an open set $O$ with $(X_{0},{\xi}_{0})\in O\subset\mathbb {R}^{2n}$ such that $\phi$ is continuous on $[0,T]\times O$. On any compact subset of $[0,T]\times O$, this ensures uniform continuity. Since $(X ^{(i)})$ is a family of solutions to the Euler–Lagrange equation with converging initial values $(X_{0}^{(i)},{\xi}_{0}^{(i)})$, this implies uniform convergence of $(X^{(i)},\dot{X}^{(i)})$ to $(X^{(\infty )},\dot{X}^{(\infty)})$ on any compact subset $[0,T]\subset [0,T_{0}^{(\infty)})$. Therefore,

$$\begin{aligned} &\int_{0}^{\infty}\big(f({\xi}_{s}^{(\infty)}) + (X_{s}^{(\infty )})^{\top}\varSigma X_{s}^{(\infty)}\big)\,ds\\ &\quad = \lim_{T\to T_{0}^{(\infty)}} \int_{0}^{T} \big(f({\xi}_{s}^{(\infty )}) + (X_{s}^{(\infty)})^{\top}\varSigma X_{s}^{(\infty)}\big)\,ds \\ &\quad = \lim_{T\to T_{0}^{(\infty)}} \lim_{i\to\infty} \int_{0}^{T} \big(f({\xi} _{s}^{(i)}) + (X_{s}^{(i)})^{\top}\varSigma X_{s}^{(i)}\big)\,ds \\ &\quad \leq\lim_{i\to\infty} \int_{0}^{\infty}\big(f({\xi}_{s}^{(i)}) + (X_{s}^{(i)})^{\top}\varSigma X_{s}^{(i)}\big)\,ds. \end{aligned}$$

This establishes that $X^{(\infty)}$ is “at least as good” as the limit of the finite-horizon strategies $X^{(i)}$. In Proposition A.6, we show that $X^{(\infty)}\in{\bar{\mathcal {X}}}$. We now show that no strategy can be better than this limit. Let $X^{[\infty]}\in {\bar{\mathcal {X}}}$ be a deterministic admissible strategy with $X_{0}^{[\infty]}=X _{0}$ and finite mean–variance cost

$$ \int_{0}^{\infty}\big(f({\xi}_{s}^{[\infty]})+(X_{s}^{[\infty]})^{\top}\varSigma X_{s}^{[\infty]}\big)\,ds < \infty. $$

Then $X_{t}^{[\infty]}$ converges to zero as $t$ tends to infinity. We define a sequence of trading strategies $X^{[i]}$ that liquidate the portfolio $X^{(i)}_{0}$ by time $T^{(i)}>2$ via

$$ X_{t}^{[i]}:= \textstyle\begin{cases} X_{t}^{[\infty]}+(1-t)(X^{(i)}_{0}-X_{0}) & \text{for }0 \leq t\leq1, \\ X_{t}^{[\infty]} & \text{for }1 < t < T^{(i)}-1, \\ (T^{(i)}-t)X_{T^{(i)}-1}^{[\infty]} & \text{for }T^{(i)}-1 \leq t \leq T^{(i)}, \\ 0 & \text{for } t > T^{(i)}. \end{cases} $$

We then have

$$\begin{aligned} &\lim_{i\to\infty} \int_{0}^{\infty}\big(f({\xi}_{s}^{(i)}) + (X_{s}^{(i)})^{\top}\varSigma X_{s}^{(i)}\big)\,ds\\ &\quad \leq\lim_{i\to\infty} \int_{0}^{\infty}\big(f({\xi}_{s}^{[i]}) + (X_{s}^{[i]})^{\top}\varSigma X_{s}^{[i]}\big)\,ds \\ &\quad = \lim_{T\to\infty} \lim_{i\to\infty} \int_{0}^{T} \big(f({\xi} _{s}^{[i]}) + (X_{s}^{[i]})^{\top}\varSigma X_{s}^{[i]}\big)\,ds \\ &\quad = \int_{0}^{\infty}\big(f({\xi}_{s}^{[\infty]}) + (X_{s}^{[\infty ]})^{\top}\varSigma X_{s}^{[\infty]}\big)\,ds. \end{aligned}$$

Hence $X^{(\infty)}$ is mean–variance optimal. Because it is unique by Lemma A.1, we see that ${\xi}_{0}^{(i)}$ converges to the same vector ${\xi}_{0}^{\infty}$ for any sequence $(X ^{(i)}_{0},T^{(i)})$. Therefore ${\xi}_{0}^{(\infty)}$ depends continuously on $X_{0}$. The validity of the Euler–Lagrange equation carries over by construction, and Bellman’s principle of optimality follows again by the additivity of mean costs and variance of proceeds. □

The next proposition establishes a special form of the identity established by [9] and rediscovered by Hilbert in 1900; see also [12, Chap. 3, Sect. 17].

Proposition A.4

The vector field ${\bar{{a}}}$ fulfills

$$ \frac{f({\bar{{a}}}(X))}{X^{\top}\varSigma X} = \frac{1}{2\alpha }\quad\textit{for all }X\in\mathbb {R}^{n}\backslash\{0\}. $$

Proof

Let $(X_{t})$ be a mean–variance optimal strategy. Then

$$\begin{aligned} \frac{d}{dt}\left(f(-\dot{X}_{t})+\frac{1}{2}X_{t}^{\top}\varSigma X_{t}\right) & = -\nabla f(-\dot{X}_{t})\ddot{X}_{t} + X_{t}^{\top}\varSigma\dot{X }_{t} \\ & = -\nabla f(-\dot{X}_{t})\ddot{X}_{t}+(\ddot{X}_{t})^{\top}D^{2}f(-\dot{X}_{t})\dot{X}_{t} \end{aligned}$$

(A.3)

$$\begin{aligned} & = \frac{d}{dt}\big(-\nabla f(-\dot{X}_{t})\dot{X}_{t}\big) \\ & = \frac{d}{dt} \big((\alpha+1) f(-\dot{X}_{t})\big), \end{aligned}$$

(A.4)

where (A.3) follows by the Euler–Lagrange equation (A.2) and (A.4) by the scaling property (2.1) which implies

$$ \nabla f({a}){a}=\lim_{s\to0} \frac{f((1+s){a})-f({a})}{s} = (\alpha+1) f({a}). $$

(A.5)

Hence

$$ -\alpha f\big({\bar{{a}}}(X_{0})\big)+\frac{1}{2} X_{0}^{\top}\varSigma X_{0} = \lim_{t\to0}\bigg(-\alpha f\big({\bar{{a}}}(X_{t})\big)+\frac{1}{2} X_{t}^{\top}\varSigma X_{t}\bigg) = 0. $$

The desired equality follows immediately. □

Finally, we show that the mean–variance value function fulfills the dynamic programming PDE.

Proposition A.5

The mean–variance value function

$$ X_{0} \mapsto{\bar{v}}(X_{0}):=\inf_{{\bar{{\xi}}}\in{\bar{\mathcal {X}}}} \int_{0}^{\infty}\bigg(f({\bar{{\xi}}} _{s}) + \frac{1}{2} (X_{s}^{{\bar{{\xi}}}})^{\top}\varSigma X_{s}^{{\bar{{\xi }}}}\bigg) \,ds $$

is $C^{1}$ and fulfills

$$ \nabla f\big({\bar{{a}}}(X)\big) = {\bar{v}}_{X}(X). $$

(A.6)

Proof

The mean–variance value function is convex because of the convexity of the mapping $X\mapsto f({\xi})+\frac{1}{2}X^{\top}\varSigma X$. The function ${\bar{v}}$ is therefore necessarily differentiable at $X_{0}\in\mathbb {R}^{n}$ if it is bounded from above by a smooth function $\tilde{v}$ that touches ${\bar{v}}$ at $X_{0}$, i.e., $\tilde{v}(X_{0})={\bar{v}}(X_{0})$. Such a function $\tilde{v}$, however, can be constructed as

$$ \tilde{v}(X) = \int_{0}^{\infty}\big(f({\xi}^{X}_{t})+(X_{t}^{{\xi }^{X}})^{\top}\varSigma X_{t}^{{\xi}^{X}}\big) \,dt $$

with

$$ {\xi}^{X}_{t} := {\bar{{\xi}}}_{t}^{(X_{0})} + M_{t}(X-X_{0}), $$

where

$$ M_{t}:=\big({\bar{{\xi}}}_{t}^{(X_{0}+e_{1})}-{\bar{{\xi}}}^{(X_{0})}_{t}, {\bar{{\xi}}}_{t}^{(X_{0}+e_{2})}-{\bar{{\xi}}}_{t}^{(X_{0})}, \dots, {\bar{{\xi}}}_{t}^{(X_{0}+e_{n})}-{\bar{{\xi}}}_{t}^{(X_{0})}\big) \in \mathbb {R}^{n\times n} $$

and $e_{i}$ is the $i$th unit vector. Therefore, ${\bar{v}}$ is differentiable. By the dynamic programming principle, we have that for any absolutely continuous path $X:\mathbb {R}_{+}\to \mathbb {R}^{n}$,

$$ {\bar{v}}(X_{0})\leq{\bar{v}}(X_{t})+\int_{0}^{t}\bigg(f(-\dot {X}_{s})+\frac{1}{2}X_{s}^{\top}\varSigma X_{s}\bigg)\,ds, $$

with equality for the optimal strategy $X^{\bar{{\xi}}}$. Since ${\bar{v}}$ is differentiable, this implies

$$ 0\leq{\bar{v}}_{X}(X_{0})\dot{X}_{0} + f(-\dot{X}_{0}) + \frac{1}{2} X_{0}^{\top}\varSigma X_{0}. $$

The right-hand side therefore attains its minimum at the optimal $\dot {X}_{0}=-{\bar{{a}}}(X_{0})$ and therefore

$$ \nabla f\big({\bar{{a}}}(X_{0})\big) = {\bar{v}}_{X}(X_{0}). $$

This establishes (A.6) and that the mean–variance cost ${\bar{v}}$ is $C^{1}$. □

Proposition A.6

For any $X_{0}\in\mathbb {R}^{n}$, the deterministic mean–variance optimal trading strategy ${\bar{{\xi}}}={\bar{{\xi}}}^{(X_{0})}$ satisfies

$$ \lim_{t\to\infty} (X_{t}^{{\bar{{\xi}}}})^{\top}\varSigma X_{t}^{{\bar {{\xi}}}} t \ln\ln t = 0. $$

(A.7)

It is hence an admissible trading strategy, i.e., ${\bar{{\xi}}}\in {\bar{\mathcal {X}}}\subset \mathcal {X}$.

For the proof, we need the following lemma.

Lemma A.7

Let ${Y}_{0}=rX_{0}$ and let $X$ and ${Y}$ be the corresponding mean–variance optimal strategies. Then we have that

$$ {Y}_{t} = r X_{bt}\quad \textit{with }b:=r^{\frac{1-\alpha}{1+\alpha}}. $$

Proof of Lemma A.7

Let us define

$$\begin{aligned} \hat{X}_{t} & := \frac{1}{r} {Y}_{\frac{t}{b}},\\ \hat{{Y}}_{t} & := r X_{bt}. \end{aligned}$$

Then $\hat{X}$ and $\hat{{Y}}$ are deterministic strategies with $\hat {X}_{0}=X_{0}$ and $\hat{{Y}}_{0}={Y}_{0}$, and we obtain

$$\begin{aligned} {\bar{v}}(X_{0}) & \leq\int_{0}^{\infty}\left(f(-\dot{\hat{X}}_{s}) + \frac{1}{2} \hat{X}_{s}^{\top}\varSigma\hat{X}_{s}\right) \,ds \\ & = \left(\frac{1}{rb}\right)^{\alpha+1}b\int_{0}^{\infty}f(-\dot{{Y} }_{s})\,ds + \left(\frac{1}{r}\right)^{2} b \int_{0}^{\infty}\frac{1}{2} {Y}_{s}^{\top}\varSigma{Y}_{s} \,ds \\ & = r^{-\frac{3\alpha+1}{\alpha+1}}{\bar{v}}({Y}_{0}) \\ & \leq r^{-\frac{3\alpha+1}{\alpha+1}} \int_{0}^{\infty}\left(f(-\dot{\hat{{Y}}}_{s}) + \frac{1}{2} \hat {{Y}}_{s}^{\top}\varSigma\hat{{Y}}_{s}\right) \,ds \\ & = r^{-\frac{3\alpha+1}{\alpha+1}}\left( \left(rb\right)^{\alpha+1}\frac{1}{b}\int_{0}^{\infty}f(-\dot{X}_{s})\,ds + r^{2} \frac{1}{b} \int_{0}^{\infty}\frac{1}{2} X_{s}^{\top}\varSigma X_{s} \,ds\right) \\ & = {\bar{v}}(X_{0}). \end{aligned}$$

All the inequalities above are thus equalities, and hence $\hat{X}$ and $\hat{{Y}}$ are optimal. The lemma follows since the optimal strategies are unique. □

Proof of Proposition A.6

It is clear that ${\bar{{\xi}}}^{(X_{0})}$ satisfies the conditions of Sect. 2.¹⁴ To see that it is admissible, i.e., that ${\bar{{\xi}}}^{(X_{0})}\in \mathcal {X}$, the only thing left to prove is (A.7). First, we observe that by Lemma A.7, it is sufficient to prove this equation for $X_{0}$ with $X_{0}^{\top}\varSigma X_{0}=1$. Let us first define

$$ \tau_{0}:=\sup_{X_{0}\ \mathrm{with}\ X_{0}^{\top}\varSigma X_{0}=1} \max\bigg\{ t>0:(X_{t}^{{\bar{{\xi}}}^{(X_{0})}})^{\top}\varSigma X_{t}^{{\bar{{\xi}}}^{(X_{0})}}\geq\frac{1}{2}\bigg\} . $$

This $\tau_{0}$ is the time it takes at most until $X_{0}^{\top}\varSigma X_{0}$ is reduced from 1 to $\frac{1}{2}$. By Lemma A.7, we obtain that

$$ \tau_{1}:=\sup_{X_{0}\ \mathrm{with}\ X_{0}^{\top}\varSigma X_{0}=\frac{1}{2}} \max\bigg\{ t>0:(X_{t}^{{\bar{{\xi}}}^{(X_{0})}})^{\top}\varSigma X_{t}^{{\bar{{\xi}}}^{(X_{0})}}\geq\frac{1}{4}\bigg\} = 2^{\frac {1-\alpha}{1+\alpha}}\tau_{0}, $$

or more generally,

$$ \tau_{k}:=\sup_{X_{0}\ \mathrm{with}\ X_{0}^{\top}\varSigma X_{0}=(\frac{1}{2})^{k}} \max\bigg\{ t>0:(X_{t}^{{\bar{{\xi}}}^{(X_{0})}})^{\top}\varSigma X_{t}^{{\bar{{\xi}}}^{(X_{0})}}\geq\left(\frac{1}{2}\right )^{k+1}\bigg\} = 2^{k\frac{1-\alpha}{1+\alpha}}\tau_{0}. $$

Let $X_{0}\in\mathbb {R}^{n}$ with $X_{0}^{\top}\varSigma X_{0}=1$. Then for all $t \ge \sum_{0}^{k} \tau_{i}$, we have that

$$ (X_{t}^{{\bar{{\xi}}}^{(X_{0})}})^{\top}\varSigma X_{t}^{{\bar{{\xi }}}^{(X_{0})}}\leq\left(\frac{1}{2}\right)^{k+1}. $$

For $\alpha\geq1$, we have $\tau_{k}\leq\tau_{0}$; so $(X_{t}^{{\bar {{\xi}}}})^{\top}\varSigma X_{t}^{{\bar{{\xi}}}}$ is bounded from above by an exponential function. For $0<\alpha<1$, we see that $(X_{t}^{{\bar{{\xi}}}})^{\top}\varSigma X_{t}^{{\bar{{\xi}}}}$ is bounded from above by $K(t+1)^{\frac {\alpha+1}{\alpha-1}}$ for a $K>0$. In both cases, we see that (A.7) holds. □

1.2 A.2 Existence and characterization of a smooth solution of the HJB equation

As a first step, we observe that $\lim_{M\to\infty}u(M)<\infty$ due to the boundedness of the risk aversion, and we can thus assume without loss of generality that

$$ \lim_{M\to\infty}u(M)=0. $$

Proposition A.8

There exists a smooth ($C^{2,4}$) solution of

$$ \tilde{a}_{Y} = -\frac{2\alpha+1}{\alpha+1} \tilde{a}^{\alpha}\tilde {a}_{M}+ \frac{\alpha (\alpha-1)}{\alpha+1} \left(\frac{\tilde{a}_{M}}{\tilde{a}}\right )^{2} + \frac {\alpha}{\alpha+1}\frac{\tilde{a}_{MM}}{\tilde{a}} $$

(A.8)

with initial value

$$\tilde{a}(0,M) = A(M)^{\frac{1}{\alpha+1}}. $$

The solution satisfies

$$\tilde{a}_{\mathrm{min}}:=\inf_{M\in\mathbb{R}}A(M)^{\frac{1}{\alpha+1}} \leq\tilde{a}(Y,M) \leq\sup_{M\in\mathbb{R}}A(M)^{\frac{1}{\alpha+1}}=:\tilde{a}_{\mathrm{max}}. $$

The function $\tilde{a}$ in Proposition A.8 is $C^{2,4}$ in the sense that it has a continuous derivative $\frac{\partial ^{i+j}}{\partial Y^{i} \partial M^{j}}\tilde{a}(Y,M)$ if $2i+j\leq4$. In particular, $\tilde{a}_{YMM}$ and $\tilde{a}_{MMM}$ exist and are continuous.

The statement in Proposition A.8 follows from the following auxiliary theorem from the theory of parabolic partial differential equations. We do not establish the uniqueness of $\tilde{a}$ directly in the preceding proposition. However, it follows from Proposition A.16.

Theorem A.9

(Auxiliary theorem: solution of Cauchy problem)

There exists a smooth ($C^{2,4}$) solution

$$ g: \mathbb {R}_{+}\times\mathbb {R}\ni(t,x) \mapsto g(t,x)\in\mathbb {R} $$

for the parabolic partial differential equation ¹⁵

$$ g_{t}-\frac{d}{dx} \kappa(x,t,g,g_{x})+\theta(x,t,g,g_{x})=0 $$

(A.9)

with initial value condition

$$ g(0,x)=\psi_{0}(x) $$

if all of the following conditions are satisfied:

• $\psi_{0}(x)$ is smooth ($C^{4}$) and bounded;

• $\kappa$ and $\theta$ are smooth ($C^{3}$ respectively $C^{2}$);

• There are constants $b_{1}$ and $b_{2}\geq0$ such that for all $x$ and $u$,

  $$ \left(\theta(x,t,u,0) - \frac{\partial \kappa}{\partial x}(x,t,u,0)\right)u \geq-b_{1} u^{2} -b_{2}; $$

• For all $M>0$, there are constants $\mu_{M}\geq\nu_{M} > 0$ such that for all $x$, $t$, $u$ and $p$ that are bounded in modulus by $M$,

  $$ \nu_{M}\leq\frac{\partial \kappa}{\partial p}(x,t,u,p) \leq\mu_{M} $$

  and

  $$ \left(|\kappa| + \left|\frac{\partial \kappa}{\partial u}\right |\right)(1+|p|) +\left|\frac{\partial \kappa}{\partial x}\right| + |\theta| \leq \mu_{M} (1+|p|)^{2}. $$

Proof

The theorem is a direct consequence of Theorem 8.1 in [24, Chap. V]. In the following, we outline the last step of its proof because we use it for the proof of subsequent propositions.

The conditions of Theorem A.9 guarantee the existence of solutions $g_{N}$ of (A.9) on the strip $\mathbb {R} _{+}\times[-N,N]$ with boundary conditions

$$ g_{N}(0,x)=\psi_{0}(x)\quad \mbox{for all }x\in[-N,N] $$

and

$$ g_{N}(t,\pm N)=\psi_{0}(\pm N)\quad \mbox{for all }t\in\mathbb {R}_{+}. $$

These solutions converge as $N$ tends to infinity. □

Proof of Proposition A.8

We want to apply Theorem A.9 and set

$$\begin{aligned} \kappa(x,t,u,p) & :=h_{1}(u)p, \\\theta(x,t,u,p) & := h_{2}(u) p - h_{3}(u) p^{2} + h_{1}'(u)p^{2}, \\\psi_{0}(x) & := A(M)^{\frac{1}{\alpha+1}}, \end{aligned}$$

with smooth functions $h_{1},h_{2},h_{3}:\mathbb {R}\to\mathbb {R}$. With

$$\begin{aligned} h_{1}(u) & = \frac{\alpha}{(\alpha+1)u}, \quad h_{2}(u) = \frac{2\alpha+1}{(\alpha+1)}u^{\alpha}, \quad h_{3}(u) = \frac{\alpha(\alpha-1)}{(\alpha+1)u^{2}}, \end{aligned}$$

(A.10)

Equation (A.9) becomes (A.8) by relabeling the coordinates from $t$ to $Y$ and from $x$ to $M$. All conditions of Theorem A.9 are fulfilled if we take $h_{1}$, $h_{2}$ and $h_{3}$ to be smooth nonnegative functions bounded away from zero and infinity and fulfilling (A.10) for $\tilde{a} _{\mathrm{min}}\leq u\leq\tilde{a}_{\mathrm{max}}$. With these functions, there exists a smooth solution to

$$ g_{t} = - h_{2}(g) g_{x} + h_{3}(g) g_{x}^{2} + h_{1}(g) g_{xx}. $$

(A.11)

We now show that this solution $g$ also fulfills

$$ g_{t} = -\frac{2\alpha+1}{\alpha+1} g^{\alpha}g_{x} + \frac{\alpha (\alpha -1)}{\alpha+1} \left(\frac{g_{x}}{g}\right)^{2} + \frac{\alpha }{\alpha +1}\frac{g_{xx}}{g} $$

(A.12)

by using the maximum principle to show that $\tilde{a}_{\mathrm{min}}\leq g\leq\tilde{a} _{\mathrm{max}}$. (This is enough because (A.11) and (A.12) coincide whenever $g$ takes values in a range where $h_{1}, h_{2}, h_{3}$ satisfy (A.10), and that range condition translates for $g$ into the preceding inequality.) First, assume that there is a $(t_{0},x_{0})$ such that $g(t_{0},x_{0})> \tilde{a}_{\mathrm{max}}$. Then there are $N>0$ and $\gamma>0$ such that also $\tilde {g}_{N}(t_{0},x_{0}):=g_{N}(t_{0},x_{0})e^{-\gamma t_{0}}>\tilde{a}_{\mathrm{max}}$ with $g_{N}$ as constructed in the proof of Theorem A.9. Then $\max _{t\in[0,t_{0}],x\in[-N,N]}\tilde{g}_{N}(t,x)$ is attained at an interior point $(t_{1},x_{1})$, i.e., $0< t_{1}\leq t_{0}$ and $-N < x_{1} < N$. We thus have

$$\begin{aligned} \tilde{g}_{N,t}(t_{1},x_{1}) & \ge0,\\ \tilde{g}_{N,x}(t_{1},x_{1}) & = 0,\\ \tilde{g}_{N,xx}(t_{1},x_{1}) & \le0. \end{aligned}$$

We furthermore have that

$$\begin{aligned} \tilde{g}_{N,t} & = e^{-\gamma t} g_{N,t} - \gamma e^{-\gamma t} g_{N} \\ & = -e^{-\gamma t} h_{2}(g_{N}) g_{N,x} + e^{-\gamma t} h_{3}(g_{N})g_{N,x}^{2} + e^{-\gamma t} h_{1}(g_{N})g_{N,xx} -\gamma e^{-\gamma t} g_{N} \\ & = - h_{2}(g_{N}) \tilde{g}_{N,x} + h_{3}(g_{N})\tilde{g}_{N,x} {g}_{N,x} + h_{1}(g_{N})\tilde{g}_{N,xx} -\gamma\tilde{g}_{N} \end{aligned}$$

and therefore that

$$ \tilde{g}_{N} (t_{1},x_{1}) \leq0. $$

This, however, contradicts $\tilde{g}_{N}(t_{1},x_{1})\geq\tilde {g}_{N}(t_{0},x_{0})\geq\tilde{a}_{\mathrm{max}} > 0$.

By a similar argument, we can show that if there is a point $(t_{0},x_{0})$ such that $g(t_{0},x_{0})<\tilde{a}_{\mathrm{min}}$, then the interior minimum $(t_{1},x_{1})$ of a suitably chosen function $\tilde{g}_{N}:=e^{-\gamma t}(g_{N}-\tilde{a}_{\mathrm{max}})<0$ satisfies $\tilde {g}_{N}(t_{1},x_{1})\geq0$ and thus causes a contradiction. □

Proposition A.10

There exists a $C^{2,4}$-solution $\tilde{w}:\mathbb{R}_{+}\times \mathbb{R}\to\mathbb {R}$ of the transport equation

$$ \tilde{w}_{Y} = - \tilde{a}^{\alpha}\tilde{w}_{M} $$

(A.13)

with initial value

$$ \tilde{w}(0,M)= u(M). $$

The solution satisfies

$$ 0 \geq\tilde{w}(Y,M) \geq u(M-\tilde{a}_{\mathrm{max}}^{\alpha}Y) $$

and is increasing in $M$ and decreasing in $Y$.

Proof

The proof uses the method of characteristics. Consider the function

$$ P: \mathbb {R}_{+}\times\mathbb {R}\ni(Y,S) \mapsto P(Y,S)\in\mathbb {R} $$

satisfying the ODE

$$ P_{Y}(Y,S) = \tilde{a}\big(Y,P(Y,S)\big)^{\alpha}$$

(A.14)

with initial value condition $P(0,S)=S$. Since $\tilde{a}^{\alpha}$ is smooth and bounded, a solution of the above ODE exists for each fixed $S$. For every $Y$, $P(Y,\cdot)$ is a diffeomorphism mapping ℝ onto ℝ that has the same regularity as $\tilde{a}$, i.e., belongs to $C^{2,4}$. We define

$$ \tilde{w}(Y,M)=u(S) \quad\text{iff}\quad P(Y,S)=M. $$

Then $\tilde{w}$ is a $C^{2,4}$-function satisfying the initial value condition. By definition, we have

$$\begin{aligned} 0 & = \frac{d}{dY}\tilde{w}\big(Y,P(Y,S)\big) \\ & = \tilde{w}_{M}\big(Y,P(Y,S)\big)P_{Y}(Y,S) + \tilde{w}_{Y}\big(Y,P(Y,S)\big) \\ & = \tilde{w}_{M}\big(Y,P(Y,S)\big) \tilde{a}\big(Y,P(Y,S)\big)^{\alpha}+ \tilde{w} _{Y}\big(Y,P(Y,S)\big). \end{aligned}$$

Therefore, $\tilde{w}$ fulfills the desired partial differential equation. Since $\tilde{a}\leq\tilde{a}_{\mathrm{max}}$, we know that $P_{Y}\leq \tilde{a}_{\mathrm{max}}^{\alpha}$ and hence $P(Y,S)\leq S+Y \tilde{a}_{\mathrm{max}}^{\alpha}$ and thus $\tilde{w}(Y,M)\geq u(M- \tilde{a}_{\mathrm{max}}^{\alpha}Y)$.

The monotonicity statements in the proposition follow because the solutions of the ODE (A.14) do not cross and since $\tilde{a}$ is positive. □

Proposition A.11

The function $w(X,M):=\tilde{w}({\bar{v}}(X),M)$ has continuous derivatives up to $w_{XMM}$ and $w_{MMMM}$, and it solves the HJB equation

$$\min_{{a}\in\mathbb {R}^{n}} \bigg( -\frac{1}{2} w_{MM} X^{\top}\varSigma X + w_{M}f({a}) + w_{X} {a}\bigg)=0. $$

The unique minimum is attained at

$$ {a}(X,M):=\tilde{a}\big({\bar{v}}(X),M\big){\bar{{a}}}(X). $$

(A.15)

Note that $w$ is not necessarily everywhere twice differentiable in $X$; the single asset case with $\alpha<1$ is a counterexample (see (4.4)).

Proof

Assume for the moment that

$$ \tilde{a}^{\alpha+1}=-\frac{\tilde{w}_{MM}}{\tilde{w}_{M}}. $$

(A.16)

Then with $Y={\bar{v}}(X)$, we get

$$\begin{aligned} 0 &= -\frac{1}{2} X^{\top}\varSigma X \tilde{w}_{M}\left(\frac{\tilde {w}_{MM}}{\tilde{w} _{M}} + \tilde{a}^{\alpha+1}\right) \\& = - \frac{1}{2} X^{\top}\varSigma X \tilde{w}_{M}\left(\frac{\tilde {w}_{MM}}{\tilde{w} _{M}} + \frac{2 \alpha f({\bar{{a}}})}{X^{\top}\varSigma X}\tilde {a}^{\alpha+1}\right) \end{aligned}$$

(A.17)

$$\begin{aligned} & = -\frac{1}{2} \tilde{w}_{MM} X^{\top}\varSigma X - \alpha \tilde{w}_{M}f({a} ) \end{aligned}$$

(A.18)

$$\begin{aligned} & = \inf_{{a}\in\mathbb {R}^{n}}\bigg( -\frac{1}{2} w_{MM} X^{\top}\varSigma X + w_{M}f({a}) + w_{X} {a}\bigg). \end{aligned}$$

(A.19)

Equation (A.17) holds because of Theorem 4.1, (A.18) by the scaling property (2.1) of $f$, and (A.19) again because of the scaling property of $f$ as in (A.5). Note that the minimizer ${a}$ in (A.15) is unique since $\nabla f$ is injective due to the convexity of $f$.

We now show that (A.16) is fulfilled for all $M$ and $Y={\bar{v}}(X)$. First, observe that it holds for $Y=0$. For general $Y$, consider the two equations

$$\begin{aligned} \begin{aligned} \frac{d}{dY}\tilde{a}^{\alpha+1} & = -(2\alpha+1)\tilde {a}^{2\alpha}\tilde{a}_{M}+ \alpha(\alpha-1)\tilde{a}^{\alpha-2}\tilde{a}_{M}^{2} + \alpha\tilde {a}^{\alpha-1} \tilde{a} _{MM}, \\-\frac{d}{dY} \frac{\tilde{w}_{MM}}{\tilde{w}_{M}} & = \tilde {a}^{\alpha}\frac{d}{dM}\frac{\tilde{w}_{MM}}{\tilde{w}_{M}} + \alpha\tilde {a}^{\alpha -1}\tilde{a}_{M}\frac{\tilde{w}_{MM}}{\tilde{w}_{M}} + \alpha(\alpha-1)\tilde{a}^{\alpha-2}\tilde{a}_{M}^{2} + \alpha \tilde{a}^{\alpha-1} \tilde{a} _{MM}. \end{aligned} \end{aligned}$$

(A.20)

The first of these holds because of (A.8) and the second because of (A.13). Now we have

$$ \frac{d}{dY}\bigg(\tilde{a}^{\alpha+1} + \frac{\tilde {w}_{MM}}{\tilde{w}_{M} }\bigg) = -\tilde{a}^{\alpha}\frac{d}{dM}\bigg(\tilde{a}^{\alpha+1}+\frac {\tilde{w}_{MM }}{\tilde{w}_{M}}\bigg) - \alpha\tilde{a}^{\alpha-1}\tilde{a}_{M}\bigg(\tilde{a}^{\alpha +1}+\frac{\tilde{w}_{M M}}{\tilde{w}_{M}}\bigg). $$

Hence, the function $g(Y,M):=\tilde{a}^{\alpha+1}+\frac{\tilde{w}_{MM }}{\tilde{w}_{M}}$ satisfies the linear PDE

$$ g_{Y} = -\tilde{a}^{\alpha}g_{M}-\alpha\tilde{a}^{\alpha-1}\tilde{a}_{M}g $$

with initial value condition $g(0,M)=0$. One obvious solution to this PDE is the function $g(Y,M)\equiv0$. By the method of characteristics, this is the unique solution to the PDE since $\tilde{a}$ and $\tilde{a}_{M}$ are smooth and hence locally Lipschitz. □

The next auxiliary lemma will prove useful in the sequel.

Lemma A.12

There are positive constants $c_{1}$, $c_{2}$, $c_{3}$, $c_{4}$ and $b$ such that

$$\begin{aligned} \textstyle\begin{array}{rcl} u(M) \geq& w(X,M) & \geq u(M) \exp\big(b {\bar{v}}(X)\big),\\ 0 \leq& w_{M}(X,M) & \leq c_{1} + c_{2} \exp\big(-c_{3} M + c_{4} {\bar{v}}(X)\big) \end{array}\displaystyle \end{aligned}$$

(A.21)

for all $(X,M)\in\mathbb {R}^{n}\times\mathbb {R}$.

Proof

The left-hand side of the first inequality follows by the boundary condition for $w$ and the monotonicity of $w$ with respect to $X$ as established in Proposition A.10. Since the risk aversion of $u$ is bounded from above by $\tilde {a}_{\mathrm{max}}^{\alpha+1}$, we have

$$ u(M-\varDelta) \geq u(M) e^{\tilde{a}_{\mathrm{max}}^{\alpha+1} \varDelta }\quad \text{for }\varDelta\geq0 $$

(A.22)

and thus by Proposition A.10

$$ w(X, M) \ge u\big(M- \tilde{a}_{\mathrm{max}}^{\alpha}{\bar{v}}(X)\big) \ge u(M) e^{\tilde{a}_{\mathrm{max}}^{2\alpha+1} {\bar{v}}(X)} $$

which establishes the right-hand side of the first inequality with $b=\tilde{a}_{\mathrm{max}}^{2\alpha+1}$.

For the second inequality, we show the equivalent inequality

$$ 0\leq\tilde{w}_{M}(Y,M)\leq c_{1} + c_{2} \exp(-c_{3} M+ c_{4} Y). $$

The left-hand side follows since $\tilde{w}$ is increasing in $M$ by Proposition A.10. For the right-hand side, note that also the “risk aversion” of $\tilde{w}$ is bounded by $\tilde{a} _{\mathrm{max}}^{\alpha+1}$ due to (A.16). Hence

$$ \tilde{w}(Y,M_{0}) \ge\tilde{w}(Y,M) + \frac{\tilde {w}_{M}(Y,M)}{\tilde{a} _{\mathrm{max}}^{\alpha+1}}\big(1-e^{-\tilde{a}_{\mathrm{max}}^{\alpha+1}(M _{0}-M)}\big). $$

Since

$$ \lim_{M_{0}\to\infty} \tilde{w}(Y,M_{0}) \leq\lim_{M_{0}\to\infty} u(M_{0}) = 0, $$

we have

$$ 0 \ge\tilde{w}(Y,M) + \frac{\tilde{w}_{M}(Y,M)}{\tilde{a}_{\mathrm{max}}^{\alpha+1}} $$

and thus

$$ \tilde{w}_{M}(Y,M)\le- \tilde{w}(Y,M)\tilde{a}_{\mathrm{max}}^{\alpha+1} \le- u(M- \tilde{a}_{\mathrm{max}}^{\alpha}Y)\tilde{a}_{\mathrm{max}}^{\alpha+1}. $$

Since $u$ is bounded by an exponential function, we obtain the desired bound on $\tilde{w}_{M}$. □

1.3 A.3 Verification argument

We now connect the PDE results from Sect. A.2 with the optimal stochastic control problem introduced in Sect. 2. For any admissible strategy ${\xi}\in\mathcal {X}$ and $k\in{\mathbb {N}}$, we define

$$ \tau^{\xi}_{k}:=\inf\bigg\{ t\ge0:\int_{0}^{t} f({\xi}_{s}) \,ds\ge k\bigg\} . $$

We proceed by first showing that $u(M^{\xi})$ and $w(X^{\xi}, M^{\xi})$ fulfill local supermartingale inequalities. Thereafter we show that $w(X_{0},M_{0}) \geq\lim_{t\to\infty} \mathbb {E}[u(M_{t}^{\xi} )]$ with equality for ${\xi}=\hat{{\xi}}$.

Lemma A.13

For any admissible strategy ${\xi}$, the expected utility $t\mapsto\mathbb {E}[ u(M^{\xi}_{t})]$ is decreasing in  $t$. Moreover, we have $\mathbb {E}[ u(M^{\xi}_{t\wedge \tau ^{\xi}_{k}}) ]\ge\mathbb {E}[ u(M^{\xi}_{t}) ]$.

Proof

Since $M^{\xi}-M_{0}$ is the difference of the true martingale $(\int_{0}^{t} (X_{s}^{{\xi}})^{\top}\sigma \,d{B}_{s})$ and the increasing process $(\int _{0}^{t} f({\xi}_{s}) \,ds)$, it satisfies for $s\leq t$ the supermartingale inequality $\mathbb {E}[ M^{\xi}_{t} | \mathcal{F}_{s} ]\le M^{\xi}_{s}$ (even though it may fail to be a supermartingale due to the possible lack of integrability). Hence $t\mapsto\mathbb {E}[ u(M^{\xi}_{t}) ]$ is decreasing according to Jensen’s inequality.

For the second assertion, we write $\tau_{k}:=\tau^{\xi}_{k}$ and have

$$\begin{aligned} \mathbb {E}[\,u(M^{\xi}_{t\wedge\tau_{k}})]&= \mathbb {E}\bigg[u\bigg( M_{0}+\int _{0}^{t\wedge\tau_{k}}(X_{s}^{{\xi}})^{\top}\sigma\,d{B}_{s}-\int _{0}^{t\wedge\tau_{k}}f({\xi} _{s})\,ds \bigg)\bigg] \\ &\geq\mathbb {E}\bigg[u\bigg(M_{0}+\int_{0}^{t}(X_{s}^{{\xi}})^{\top}\sigma\,d{B}_{s}-\int _{0}^{t\wedge\tau_{k}}f({\xi}_{s})\,ds\bigg)\bigg] \\ &\geq\mathbb {E}[\,u(M^{\xi}_{t})], \end{aligned}$$

(A.23)

where (A.23) follows by dominated convergence because $u$ is bounded from below by an exponential function, the integral of $f({\xi})$ is bounded by $k$, and the stochastic integrals are uniformly bounded from below by $\inf_{s\le K t}W_{s}$, where $W$ is the DDS-Brownian motion of $\int(X_{s}^{{\xi}})^{\top}\sigma\,d{B}_{s}$ and $K$ is an upper bound for $(X^{{\xi}})^{\top}\varSigma X^{{\xi}}$. Finally, the term in (A.23) is clearly larger than or equal to $\mathbb {E}[ u(M^{\xi}_{t}) ]$. □

Lemma A.14

For any admissible strategy ${\xi}$, $w(X^{\xi}, M^{\xi})$ is a local supermartingale with localizing sequence $(\tau^{\xi}_{k})$.

Proof

We use a verification argument which is similar to the ones in [31] and [30]. For $T>t\geq0$, Itô’s formula yields that

$$\begin{aligned} &w(X^{\xi}_{T},{M}^{\xi}_{T}) - w(X^{\xi}_{t},{M}^{\xi}_{t}) \\ &\quad= \int_{t}^{T} w_{M}(X^{\xi}_{s},{M}^{\xi}_{s}) (X_{s}^{{\xi}})^{\top}\sigma \,d{B}_{s} \\ &\qquad{}- \int_{t}^{T}\bigg(w_{M}f({\xi}_{s}) + w_{X}{\xi}_{s} - \frac{1}{2}(X_{s}^{{\xi}})^{\top}\varSigma X_{s}^{{\xi}}w_{MM} \bigg) (X^{\xi }_{s},{M}^{\xi}_{s}) \,ds. \end{aligned}$$

(A.24)

By Proposition A.11, the $ds$-integral is nonnegative and we obtain

$$ w(X^{\xi}_{t},{M}^{\xi}_{t}) \ge w(X^{\xi}_{T}, M_{T}^{\xi})-\int_{t}^{T} w_{M}(X^{\xi}_{s},M^{\xi}_{s}) (X_{s}^{{\xi}})^{\top}\sigma \,d{B}_{s}. $$

(A.25)

We show next that the stochastic integral in (A.25) is a local martingale with localizing sequence $(\tau_{k}):=(\tau^{\xi}_{k})$. For some constant $C_{1}$ depending on $t$, $k$, $|\sigma|$, $M_{0}$ and on the upper bound of $|X^{\xi}|$, we have for $s\le t\wedge\tau_{k}$

$$\begin{aligned} {M}^{\xi}_{s} & = M_{0}+(X_{s}^{{\xi}})^{\top}\sigma{B}_{s} + \int_{0}^{s}\big({\xi}_{q}^{\top}\sigma{B}_{q} - f({\xi}_{q})\big) \,dq \ge-C_{1}\Big(1+ \sup_{q\le t} |{B}_{q}| \Big). \end{aligned}$$

Using Lemma A.12, we see that for $s\le{t\wedge\tau_{k}}$,

$$ 0\le w_{M}(X^{\xi}_{s},M^{\xi}_{s}) \le c_{1} + c_{2}\exp\bigg(c_{3} C_{1}\Big(1+ \sup_{q\le t} |{B}_{q}|\Big)+c_{4} K^{2} \bigg), $$

(A.26)

where $K$ is the upper bound of ${\bar{v}}(X^{\xi})$. Since $\sup _{q\le t} |{B}_{q}|$ has exponential moments of all orders, the martingale property of the stochastic integral in (A.25) follows. Taking conditional expectations in (A.25) thus yields the desired supermartingale property

$$ w(X^{\xi}_{t\wedge\tau_{k}},M^{\xi}_{t\wedge\tau_{k}})\ge{\mathbb {E}}[ w(X^{\xi}_{T\wedge\tau_{k}},M^{\xi}_{T\wedge\tau_{k}})|\mathcal {F}_{t} ]. $$

(A.27)

The integrability of $w(X^{\xi}_{t\wedge\tau_{k}},M^{\xi}_{t\wedge \tau_{k}})$ follows from Lemma A.12 and (A.22) in a similar way as in (A.26). □

Lemma A.15

There is an adapted strategy $\hat{\xi}$ fulfilling

$$ \hat{{\xi}}_{t} = {a}(X_{t}^{\hat{{\xi}}},M_{t}^{\hat{{\xi}}}) $$

(A.28)

with ${a}$ as defined in (A.15). This $\hat{\xi}$ is admissible and satisfies $\int_{0}^{\infty}f(\hat{\xi}_{t}) dt< K$ for some constant $K$. Furthermore, $w(X^{\hat{\xi}}, M^{\hat{\xi} })$ is a martingale and

$$ w(X_{0},M_{0}) = \lim_{t\to\infty}{\mathbb {E}}[ u(M^{\hat{\xi} }_{t}) ] \le v_{2}(X_{0},M_{0}) := \sup_{{\xi}\in\mathcal {X}} \lim_{t\to\infty} \mathbb {E} [u(M^{{\xi}}_{t})]. $$

(A.29)

Proof

Consider the stochastic differential equation

$$ d\left( \textstyle\begin{array}{c}s_{t} \\ M_{t} \end{array}\displaystyle \right) = \left( \textstyle\begin{array}{l} \tilde{a}({\bar{v}}(X_{s_{t}}^{\bar{{\xi}}}),M_{t}) \,dt \\ -\tilde{a}({\bar{v}}(X_{s_{t}}^{\bar{{\xi}}}),M_{t})^{\alpha+1} f({\bar{{a}}}(X_{s_{t}}^{\bar{{\xi}}})) \,dt + (X_{s_{t}}^{\bar{{\xi}}})^{\top}\sigma \,dB_{t} \end{array}\displaystyle \right) $$

(A.30)

with initial condition $s_{0}=0$. The functions $\tilde{a}$ and ${\bar {v}}$ are differentiable, $X_{s}^{\bar{{\xi}}}$ is differentiable in $s$, and by the Beltrami identity (4.3), we have

$$ f\big({\bar{{a}}}(X_{s}^{\bar{{\xi}}})\big)=\frac{(X_{s}^{{\bar {{\xi}}}})^{\top}\varSigma X_{s}^{{\bar{{\xi}}}}}{2\alpha} $$

which establishes that $f({\bar{{a}}}(X_{s}^{\bar{{\xi}}}))$ is differentiable in $s$. Hence, (A.30) satisfies local boundedness and Lipschitz conditions and hence has a solution. We can now set $\hat{X}_{t}:=X_{s_{t}}^{\bar{{\xi}}}$; the resulting stochastic process $\hat{X}$ is absolutely continuous, and by setting $\hat{{\xi} }_{t}:=-\dot{\hat{X}}_{t}$, we obtain a solution of Eq. (A.28) since

$$\hat{{\xi}}_{t}=-\dot{X}_{s_{t}}^{\bar{{\xi}}}\dot{s}_{t} = {\bar {{a}}}(\hat{X}_{t})\tilde{a}\big({\bar{v}}(\hat{X}_{t}),M_{t}\big)={a}(\hat{X}_{t},M_{t}). $$

We observe that $\hat{{\xi}}$ is admissible if $\int_{0}^{\infty}f(\hat {\xi} _{t}) dt< K$ for some constant $K$; the conditions (3.1) and (3.2) are clearly satisfied due to Proposition A.6 and the lower bound on $\tilde{a}$ (Proposition A.8). The upper bound for $\int_{0}^{\infty}f(\hat{\xi}_{t}) dt$ can be derived by writing

$$\begin{aligned} \int_{0}^{\infty}f(\hat{{\xi}}_{t})\,dt & = \int_{0}^{\infty}f\Big(\tilde {a}\big({\bar{v}}(X _{t}^{\hat{{\xi}}}),M_{t}^{\hat{{\xi}}}\big) {\bar{{a}}}(X_{t}^{\hat {{\xi}}})\Big)\,dt\\ &= \int_{0}^{\infty}\tilde{a}^{\alpha+1}\big({\bar{v}}(X_{t}^{\hat {{\xi}}}),M _{t}^{\hat{{\xi}}}\big) f\big({\bar{{a}}}(X_{t}^{\hat{{\xi}}})\big)\,dt \\ & \leq\tilde{a}_{\mathrm{max}}^{\alpha+1} \int_{0}^{\infty}f\big({\bar {{a}}}(X_{t}^{\hat {\xi}})\big)\,dt \leq\frac{\tilde{a}_{\mathrm{max}}^{\alpha+1}}{\tilde{a}_{\mathrm{min}}} {\bar{v}}(X_{0}). \end{aligned}$$

Next, with the choice ${\xi}=\hat{\xi}$, the $ds$-integral in (A.24) vanishes and we get equality in (A.27). Since $\tau^{\hat{\xi}}_{K}=\infty$, this proves the martingale property of $w(X^{\hat{\xi}}, M^{\hat{\xi}})$. Furthermore, we obtain from (A.21) that

$$ u(M^{\hat{\xi}}_{t}) \geq w(X^{\hat{\xi}}_{t},M^{\hat{\xi}}_{t}) \geq u(M^{\hat{\xi}}_{t}) \exp\big(b {\bar{v}}(X^{\hat{\xi }}_{t})\big). $$

Since ${\bar{v}}(X_{t}^{\hat{\xi}})$ uniformly converges to zero as $t$ tends to infinity, we obtain (A.29). □

Proposition A.16

We have $v_{2}=w$, and $\hat{\xi}$ of Lemma  A.15 is the a.s. unique optimal strategy to achieve $v_{2}$.

Proof

By Lemma A.15, we already have $w\leq v_{2}$. We now show that $v_{2}\leq w$. Let ${\xi}$ be any admissible strategy such that

$$ \lim_{t\to\infty}\mathbb {E}[u(M_{t}^{\xi})] > -\infty. $$

(A.31)

By Lemmas A.14 and A.12, we have for all $k$, $t$ and $(\tau_{k}):=(\tau^{\xi}_{k})$ that

$$\begin{aligned} w(X_{0},M_{0}) \ge\mathbb {E}[w(X_{t\wedge\tau_{k}}^{\xi}, M_{t\wedge \tau_{k}}^{\xi})] \geq\mathbb {E}\big[u(M_{t\wedge\tau_{k}}^{\xi})\exp\big({b {\bar {v}}(X_{t\wedge \tau_{k}}^{\xi})}\big)\big]. \end{aligned}$$

As in the proof of Lemma A.13, one shows that

$$\begin{aligned} \liminf_{k\to\infty}\mathbb {E}\big[u(M_{t\wedge\tau_{k}}^{\xi})\exp \big({b {\bar{v}}(X_{t\wedge\tau_{k}}^{\xi})}\big)\big] & \ge\liminf_{k\to\infty}\mathbb {E}\big[u(M_{t}^{\xi})\big({b {\bar{v}}(X _{t\wedge\tau_{k}}^{\xi})}\big)\big] \\ & = \mathbb {E}\big[u(M_{t}^{\xi})\exp\big(b {\bar{v}}(X_{t}^{\xi })\big)\big]. \end{aligned}$$

Hence,

$$ w(X_{0},M_{0}) \ge\mathbb {E}[u(M_{t}^{\xi})] + \mathbb {E}\Big[u(M _{t}^{\xi})\Big(\exp\big({b {\bar{v}}(X_{t}^{\xi})}\big)-1\Big)\Big]. $$

Let us assume for a moment that the second expectation on the right attains values arbitrarily close to zero. Then

$$ w(X_{0}, M_{0}) \geq\lim_{t\to\infty} \mathbb {E}[u(M_{t}^{\xi})]. $$

Taking the supremum over all admissible strategies ${\xi}$ gives $w\geq v_{2}$. The optimality of $\hat{\xi}$ follows from Lemma A.15, and its uniqueness from the fact that the functional ${\xi}\mapsto\mathbb {E}[u(M_{t}^{\xi})]$ is strictly concave since $u$ is concave and increasing and $M_{t}^{\xi}$ is concave.

We now show that $\mathbb {E}[u(M_{t}^{\xi})(\exp({b {\bar {v}}(X_{t}^{\xi} )})-1)]$ attains values arbitrarily close to zero. First we observe that

$$ 0 \geq u(M) \geq c_{5} u_{MM}(M) $$

for a constant $c_{5}>0$, due to the boundedness of the risk aversion of $u$, and that

$$ \exp\big({b {\bar{v}}(X_{t}^{\xi})}\big)-1 \leq c_{6} b {\bar {v}}(X_{t}^{\xi}), $$

due to the bound on $X_{t}^{\xi}$. Since $X_{t}^{\xi}$ is uniformly bounded, we see that for every $\epsilon _{1}>0$, there is an $\epsilon_{2}>0$ such that we have the uniform bound

$$ {\bar{v}}(X_{t}^{\xi})< \epsilon_{1} + \epsilon_{2} (X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}. $$

Combining the last three inequalities, we obtain

$$\begin{aligned} 0 & \geq\mathbb {E}\Big[u(M_{t}^{\xi})\Big(\exp\big({b {\bar {v}}(X_{t}^{\xi} )}\big)-1\Big)\Big] \\ & \geq b c_{6} \epsilon_{1} \mathbb {E}[u(M_{t}^{\xi})] + b c_{5} c_{6} \epsilon_{2} \mathbb {E}[(X_{t}^{{\xi}})^{\top}\varSigma X_{t}^{{\xi}}u_{MM}(M_{t}^{\xi} )]. \end{aligned}$$

(A.32)

Let us now assume that the second expectation of (A.32) attains values arbitrarily close to zero. Then for each $\epsilon_{1}>0$, there is a ${\tilde{t}}\in(0,\infty)$ such that

$$ 0 \geq\mathbb {E}\Big[u(M_{{\tilde{t}}}^{\xi})\Big(\exp\big({b {\bar{v}}(X_{{\tilde{t}} }^{\xi})}\big)-1\Big)\Big] \geq b c_{6} \epsilon_{1} \lim_{t\to\infty}\mathbb {E}[u(M_{t}^{\xi})]. $$

Sending $\epsilon_{1}$ to zero yields that $\mathbb {E}[u(M_{t}^{\xi })(\exp ({b {\bar{v}}(X_{t}^{\xi})})-1)]$ attains values arbitrarily close to zero, since $\lim_{t\to\infty}\mathbb {E}[u(M_{t}^{\xi})]$ is bounded by assumption (see (A.31)).

We finish the proof by showing that the second expectation of (A.32) attains values arbitrarily close to zero. By Lemma A.13 and the same line of reasoning as in the proof of Lemma A.14, we have for all $k$, $t$ and $(\tau_{k}):=(\tau^{\xi}_{k})$ that

$$\begin{aligned} -\infty& < \lim_{s\to\infty} \mathbb {E}[u(M_{s}^{\xi})] \le\mathbb {E}[u(M _{t}^{\xi})] \le\mathbb {E}[u({M}^{\xi}_{t\wedge\tau_{k}})] \\ & = u(M_{0}) + \mathbb {E}\left[\int_{0}^{t\wedge\tau_{k}} u_{M}({M}^{\xi} _{s}) (X_{s}^{{\xi}})^{\top}\sigma \,d{B}_{s}\right] \\ & \quad{} - \mathbb {E}\bigg[\int_{0}^{t\wedge\tau_{k}}\bigg( u_{M}f({\xi}_{s}) - \frac{1}{2} (X_{s}^{{\xi}})^{\top}\varSigma X_{s}^{{\xi}} u_{MM} \bigg) ({M}^{\xi }_{s}) \,ds\bigg] \\ & = u(M_{0}) - \mathbb {E}\bigg[\int_{0}^{t\wedge\tau_{k}}\bigg(u_{M} f({\xi}_{s}) - \frac{1}{2}(X_{s}^{{\xi}})^{\top}\varSigma X_{s}^{{\xi}}u_{MM} \bigg) ({M}^{\xi}_{s}) \,ds\bigg]. \end{aligned}$$

Sending $k$ and $t$ to infinity yields

$$\int_{0}^{\infty}\mathbb {E}\big[ (X_{s}^{{\xi}})^{\top}\varSigma X_{s}^{{\xi }}u_{MM}(M^{\xi}_{s})\big]\, ds > -\infty $$

which concludes the proof. □

Proposition A.17

We have $v=w$, and $\hat{\xi}$ of Lemma  A.15 is the a.s. unique optimal strategy.

Proof

For any admissible strategy ${\xi}$, the martingale

$$ \int_{0}^{t} (X^{\xi}_{s})^{\top}\sigma \,d{B}_{s} $$

is uniformly integrable due to the requirement in (3.1). Therefore,

$$ \mathbb {E}[u(M^{\xi}_{t})] \ge\mathbb {E}[u(M^{\xi}_{\infty})] $$

follows as in the proof of Lemma A.13. Hence, Proposition A.16 yields

$$ \mathbb {E}[u(M_{\infty}^{\xi})] = \lim_{t\to\infty} \mathbb {E}[u(M_{t}^{\xi})] \leq v_{2}(X_{0},M_{0})\leq w(X_{0},M_{0}). $$

Taking the supremum over all admissible strategies ${\xi}$ gives $v\leq w$. The converse inequality follows from Lemma A.14, since $\hat{\xi}$ is admissible. □

Proof of Theorem 5.7

Fix $N>0$ and let $g_{i}$ denote the function $\tilde{g}_{N}$ constructed in the proof of Proposition A.8 when the parabolic boundary condition is given by $\tilde{g}_{N}(Y,M)=A_{i}(M)^{\frac{1}{\alpha+1}}$ for $Y=0$ or $|M|= N$, where $i\in\{0,1\}$. The result follows if we can show that $h:=g_{1}-g_{0}\ge0$. A straightforward computation shows that $h$ solves the linear PDE

$$\begin{aligned} h_{Y} =& \frac{2\alpha+1}{\alpha+1}(g_{1}^{\alpha}g_{1,M} - g_{0}^{\alpha}g_{0,M}) + \frac{\alpha(\alpha-1)}{\alpha+1}\bigg(\Big(\frac{g_{1,M }}{g_{1}}\Big)^{2}-\frac{g_{0,M}}{g_{0}}\bigg)\\ &{}+ \frac{\alpha}{\alpha+1}\left(\frac{g_{1,MM}}{g_{1}}-\frac {g_{0,MM}}{g_{0}}\right) \\ = & \frac{2\alpha+1}{\alpha+1}\left(g_{1}^{\alpha}h_{M} + \frac {g_{1}^{\alpha}- g_{0}^{\alpha}}{g_{1}-g_{0}}g_{0,M} h \right)\\ &{}+ \frac{\alpha(\alpha-1)}{\alpha+1}\frac{g_{0}^{2} g_{1,M} h_{M} + g_{0}^{2} g_{0,M} h_{M}- g_{0} g_{0,M}^{2} h -g_{1} g_{0,M }^{2} h}{g_{0}^{2} g_{1}^{2}} \\ &{} + \frac{\alpha}{\alpha+1}\frac{g_{0} h_{MM} -g_{0,M M} h}{g_{0} g_{1}} \\ = & \frac{1}{2}b_{1}h_{MM}+b_{2}h_{M}+Vh, \end{aligned}$$

where the coefficients $b_{1}$ and $b_{2}$ and the potential $V$ are given by

$$\begin{aligned} b_{1} &= \frac{\alpha}{\alpha+1}\frac{1}{g_{1}}, \\ b_{2} &= \frac{2\alpha+1}{\alpha+1}g_{1}^{\alpha}+ \frac{\alpha(\alpha-1)}{\alpha+1}\frac{g_{1,M} + g_{0,M}}{g_{1}^{2}}, \\ V &= \frac{2\alpha+1}{\alpha+1}\frac{g_{1}^{\alpha}- g_{0}^{\alpha}}{g_{1}-g_{0}}g_{0,M} - \frac{\alpha(\alpha-1)}{\alpha+1}\frac{(g_{0}+g_{1}) g_{0,M }^{2}}{g_{0}^{2} g_{1}^{2}} - \frac{\alpha}{\alpha+1}\frac{g_{0,MM}}{g_{0} g_{1}}. \end{aligned}$$

The parabolic boundary condition of $h$ is

$$ h(Y,M)=A_{1}^{\frac{1}{\alpha+1}} - A_{0}^{\frac{1}{\alpha+1}} =:\psi (M)\quad\text{for }Y=0\text{ or }|M|= N. $$

The functions $b_{1}$, $b_{2}$, $V$ and $\psi$ are smooth and (at least locally) bounded on the set $(0,\infty)\times[-N,N]$, and $b_{1}$ is bounded away from zero. Next, take $T>0$ and $M \in(-N,N)$, and let $Z$ be the solution of the stochastic differential equation

$$ dZ_{t}=\sqrt{b_{1}(T-t,Z_{t})} dB_{t}+b_{2}(T-t,Z_{t}) dt ,\quad Z_{0}=M, $$

which is defined up to the time

$$ \tau:=\inf\{t\ge0:|Z_{t}|=N\text{ or }t=T\}. $$

By a standard Feynman–Kac argument, $h$ can then be represented as

$$ h(T,M)=\mathbb {E}\bigg[ \psi(Z_{\tau})\exp\bigg(\int_{0}^{\tau}V(T-t,Z_{t}) dt\bigg) \bigg]. $$

Hence $h\ge0$ as $\psi\ge0$ by assumption. □

Proof of Theorem 5.6

In Theorem 5.7, take $u^{0}(x):=u(x)$ and $u^{1}(x):=u(x+r)$. If $u$ exhibits IARA, then $A^{1}\ge A^{0}$ if $r>0$ and hence $\tilde{a}^{1}\ge\tilde{a}^{0}=\tilde{a}$. But we clearly have $\tilde{a}^{1}(X,M)=\tilde{a} (X,M+r)$. The result for decreasing $A$ follows by taking $r<0$. □