Multidimensional investment problem

Sören Christensen; Paavo Salminen

doi:10.1007/s11579-017-0195-y

Mathematics and Financial Economics

January 2018, Volume 12, Issue 1, pp 75–95 | Cite as

Multidimensional investment problem

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Sören ChristensenEmail author
Paavo Salminen

Article

First Online: 04 August 2017

Received: 26 August 2016

Accepted: 22 July 2017

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Abstract

In this paper we demonstrate that the Riesz representation of excessive functions is a useful and enlightening tool to study optimal stopping problems. After a short general discussion of the Riesz representation we concretize to geometric Brownian motions. After this, a classical investment problem, also known as exchange-of-baskets-problem, is studied. It is seen that the boundary of the stopping region in this problem can be characterized as a unique solution of an integral equation arising immediately from the Riesz representation of the value function. The two-dimensional case is studied in more detail and a numerical algorithm is presented.

Keywords

Geometric Brownian motion Convex set Resolvent kernel Duality Integral representation of excessive function Optimal investment problem American put option Integral equation

JEL Classification

C61 G11

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Notes

Acknowledgements

Paavo Salminen thanks the Mathematisches Seminar at Christian-Albrechts-Universität for the hospitality and the support during the stay in Kiel. Paavo Salminen’s research was supported in part by a grant from Svenska kulturfonden via Stiftelsernas professorspool, Finland.

A Appendix

A.1 Proof of Proposition 1

Proof

Let X be a d-dimensional geometric Brownian motion with parameters ${\varvec{\mu }}$ and ${\varvec{R}}$ started at $\mathbf{1}$ as introduced in Sect. 2.1, and denote the Lebesgue density of $X_t$ by $f_{X_t}$.

In the following, we understand each operation (as multiplication, division etc.) componentwise. Since 1/X is a geometric Brownian motion a short calculation yields

$$\begin{aligned} f_{X_t}(1/\mathbf{y})=f_{1/X_t}(\mathbf{y})\exp \left( 2\sum _{i=1}^d\log (y_i)\right) \end{aligned}$$

and

$$\begin{aligned} f_{1/X_t}(\mathbf{y})=f_{X_t}(\mathbf{y})\exp \left( -2\,{\varvec{ \overline{\mu }}}\,{\varvec{\Sigma }}^{-1}\log (\mathbf{y})^{tr}\right) \end{aligned}$$

where $\varvec{ \overline{\mu }}=\big ((\mu _i-\frac{1}{2}a_i^2)/a_i\big )_{i=1}^d$. Let m be the measure on ${\mathbb R}_+^d$ with Lebesgue density

$$\begin{aligned} h(\mathbf{y})=\exp \left( -\sum _{i=1}^d\log (y_i)+2\,{\varvec{\overline{\mu }}}\,{\varvec{\Sigma }}^{-1}\log ( \mathbf{y})^{tr}\right) , \end{aligned}$$

then for all nonnegative and measurable functions f and g, it holds that

$$\begin{aligned}&\int _{{\mathbb R}_+^d} f(\mathbf{x})G_rg(\mathbf{x})m(\mathbf{dx})\\&\quad =\int _0^\infty \mathrm{e}^{-rt}\int _{{\mathbb R}_+^d}\int _{{\mathbb R}_+^d} f(\mathbf{x})g (\mathbf{x}{} \mathbf{z})h(\mathbf{x})f_{X_t}(\mathbf{z})\mathbf{dx} \mathbf{dz}dt\\&\quad =\int _0^\infty \mathrm{e}^{-rt}\int _{{\mathbb R}_+^d}\int _{{\mathbb R}_+^d} f(\mathbf{w}/\mathbf{z})g(\mathbf{w})h(\mathbf{w}/\mathbf{z})f_{X_t}(\mathbf{z}) \exp \left( -\sum _{i=1}^d\log (z_i)\right) \mathbf{dw} \mathbf{dz}dt\\&\quad =\int _0^\infty \mathrm{e}^{-rt}\int _{{\mathbb R}_+^d}\int _{{\mathbb R}_+^d} f(\mathbf{y}{} \mathbf{w})g(\mathbf{w})h(\mathbf{y}{} \mathbf{w})f_{X_t}(1/\mathbf{y})\exp \left( -\sum _{i=1}^d\log (y_i)\right) \mathbf{dy} \mathbf{dw}dt\\&\quad =\int _0^\infty \mathrm{e}^{-rt}\int _{{\mathbb R}_+^d}\int _{{\mathbb R}_+^d} f(\mathbf{z}{} \mathbf{y})g(\mathbf{z})h(\mathbf{z}{} \mathbf{y})f_{1/X_t}(\mathbf{y}) \exp \left( \sum _{i=1}^d\log (y_i)\right) \mathbf{dz} \mathbf{dy}dt\\&\quad =\int _0^\infty \mathrm{e}^{-rt}\int _{{\mathbb R}_+^d}\int _{{\mathbb R}_+^d} f(\mathbf{z}{} \mathbf{y})g(\mathbf{z})h(\mathbf{z})h(\mathbf{y})f_{X_t}(\mathbf{y})\times \\&\qquad \times \exp \big (-2\,{\varvec{\overline{\mu }}}\,{\varvec{\Sigma }}^{-1} \log (\mathbf{y})^{tr}\big )\exp \left( \sum _{i=1}^d\log (y_i)\right) \mathbf{dz} \mathbf{dy}dt\\&\quad =\int _0^\infty \mathrm{e}^{-rt}\int _{{\mathbb R}_+^d}\int _{{\mathbb R}_+^d} f(\mathbf{z} \mathbf{y})g(\mathbf{z})h(\mathbf{z})f_{X_t}(\mathbf{y})\mathbf{dz} \mathbf{dy}dt\\&\quad =\int _{{\mathbb R}_+^d} G_rf(\mathbf{z})g(\mathbf{z})m(\mathbf{dz}). \end{aligned}$$

$\square $

A.2 Proof of Proposition 3

Proof

By noting that convex sets have a Lipschitz surfaces there exists (see, e.g. [30, Appendix D]) a sequence $(u_j)_{j\in {\mathbb {N}}}$ of $C^2$-functions such that $u_j\rightarrow u$ uniformly on compacts and $\mathcal{G}u_j\rightarrow \mathcal{G}u$ uniformly on compact subsets of ${\mathbb R}_+^d{\setminus } \partial D$. More precisely, this sequence is given by convolution with a sequence of $C^\infty $-functions $\eta _j,\;j\in {\mathbb {N}},$ with compact support:

$$\begin{aligned} u_j(\mathbf{x})=u*\eta _j(\mathbf{x})=\int _{{\mathbb R}^d}u(\mathbf{x}-\mathbf{y})\eta _j(\mathbf{y})\mathbf{dy}, \end{aligned}$$

where $*$ denotes convolution. Furthermore, by the proof of [30, Appendix D] it is immediate that $(\mathcal{G}-r)u_j=((\mathcal{G}-r)u)*\eta _j.$ Therefore, the boundedness assumption on $(\mathcal{G}-r)u$ implies that the sequence $(\mathcal{G}-r)u_j,\;j\in {\mathbb {N}}$, is uniformly locally bounded around $\partial S$. Now, take a sequence $(K_n)_{n\in {\mathbb {N}}}$ of compact subsets of ${\mathbb R}_+^d$ such that $\tau _{n}:=\inf \{t\ge 0: X_t\not \in K_n\}\rightarrow \infty $ as $n\rightarrow \infty $. By Dynkin’s formula

$$\begin{aligned} u_j(\mathbf{x})={\mathbb {E}}_\mathbf{x}\left( \mathrm{e}^{-r\tau _n}u_j(X_{\tau _n})\right) -{\mathbb {E}}_\mathbf{x}\left( \int _0^{\tau _n}\mathrm{e}^{-rs}(\mathcal{G}-r)u_j(X_s)ds\right) . \end{aligned}$$

Letting $j\rightarrow \infty $ and keeping in mind that the Green measure is absolutely continuous with respect to the Lebesgue measure, we obtain

$$\begin{aligned} u(\mathbf{x})={\mathbb {E}}_\mathbf{x}\left( \mathrm{e}^{-r\tau _n}u(X_{\tau _n})\right) -{\mathbb {E}}_\mathbf{x}\left( \int _0^{\tau _n}\mathrm{e}^{-rs}(\mathcal{G}-r)u(X_s)ds\right) \end{aligned}$$

by dominated convergence and the convention that $(\mathcal{G}-r)u(\mathbf{x})=0$ (or any other value) on the null set $\partial S$. Letting $n\rightarrow \infty $ and using the boundedness of u, we have

$$\begin{aligned} \int G_r(\mathbf{x},\mathbf{y})\sigma (\mathbf{dy})&=u(\mathbf{x})={\mathbb {E}}_\mathbf{x}\left( \int _0^{\infty }\mathrm{e}^{-rs}(-\mathcal{G}+r)u(X_s)ds\right) \\ {}&=\int G_r(\mathbf{x},\mathbf{y})(r-\mathcal{G})u(\mathbf{y})m(\mathbf{dy}) \end{aligned}$$

and, by the uniqueness of the integral representation,

$$\begin{aligned} \sigma (\mathbf{dy})=(r-\mathcal{G})u(\mathbf{y})m(\mathbf{dy}). \end{aligned}$$

$\square $

A.3 Proof of Lemma 2

Proof

Let $\mathbf{x_0}\in \partial S$ and ${\varvec{\zeta }}\in {\mathbb R}^d$ with Euclidian norm 1. Then $V(\mathbf{x_0})=g(\mathbf{x_0})$ and $V\ge g$. Therefore, for $\epsilon $ such that $\mathbf{x_0}+\epsilon {\varvec{\zeta }}\in {\mathbb R}_+^d$ it holds

$$\begin{aligned} V(\mathbf{x_0}+\epsilon {\varvec{\zeta }})-V(\mathbf{x_0})\ge g(\mathbf{x_0}+\epsilon {\varvec{\zeta }})-g(\mathbf{x_0}), \end{aligned}$$

which yields

$$\begin{aligned} \liminf _{\epsilon \rightarrow 0}\frac{1}{\epsilon }(V(\mathbf{x_0} +\epsilon {\varvec{\zeta }})-V(\mathbf{x_0}))&\ge \liminf _{\epsilon \rightarrow 0} \frac{1}{\epsilon }(g(\mathbf{x_0}+\epsilon {\varvec{\zeta }})-g(\mathbf{x_0}))\\&=D_{\varvec{\zeta }} g(\mathbf{x_0})=-\sum _{i=1}^d\zeta _i. \end{aligned}$$

On the other hand, let $\tau ^*=\inf \{t\ge 0:X_t\in S\}$ denote the optimal stopping time and write

$$\begin{aligned} \tau _\epsilon =\inf \{t\ge 0:(\mathbf{x_0}+\epsilon {\varvec{\zeta }})X_t\in S\}. \end{aligned}$$

Then (using componentwise multiplication of vectors and the notation $\mathbf{1}=(1,\ldots ,1)$)

$$\begin{aligned} V(\mathbf{x_0}+\epsilon {\varvec{\zeta }})={\mathbb {E}}_{\mathbf{x_0}+\epsilon {\varvec{\zeta }}}\left( \mathrm{e}^{-r\tau ^*}g(X_{\tau ^*})\right) ={\mathbb {E}}_{\mathbf{1}}\left( \mathrm{e}^{-r\tau _\epsilon }g((\mathbf{x_0}+\epsilon {\varvec{\zeta }})X_{\tau _\epsilon })\right) \end{aligned}$$

and—since $\mathbf{x_0}\in S$—

$$\begin{aligned} V(\mathbf{x_0})\ge {\mathbb {E}}_{\mathbf{1}}\left( \mathrm{e}^{-r\tau _\epsilon }g(\mathbf{x_0}X_{\tau _\epsilon })\right) . \end{aligned}$$

By Lemma 1, the set

$$\begin{aligned} F:=\{\mathbf{x}\in (0,\infty )^d:\mathbf{x}\le \mathbf{x_0}\}, \end{aligned}$$

is a subset of S, hence—keeping the continuity of the sample paths in mind—

$$\begin{aligned} \tau _\epsilon \le \inf \{t\ge 0:(\mathbf{x_0}+\epsilon {\varvec{\zeta }})X_t\in F\}\rightarrow 0 \text{ under } {\mathbb {P}}_{\mathbf{1}} \text{ as } \epsilon \rightarrow 0. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{1}{\epsilon }(V(\mathbf{x_0}+\epsilon {\varvec{\zeta }})-V(\mathbf{x_0}))&\le \frac{1}{\epsilon } {\mathbb {E}}_{\mathbf{1}}\left( \mathrm{e}^{-r\tau _\epsilon }(g((\mathbf{x_0} +\epsilon {\varvec{\zeta }})X_{\tau _\epsilon })-g(\mathbf{x_0}X_{\tau _\epsilon }))\right) \\&=\frac{1}{\epsilon } {\mathbb {E}}_{\mathbf{1}}\left( \mathrm{e}^{-r\tau _\epsilon } (-\epsilon \sum _i\zeta _iX^{(i)}_{\tau _\epsilon })\right) \\&\rightarrow -\sum _i\zeta _i=D_{\varvec{\zeta }} g(\mathbf{x_0}) \text{ as } \epsilon \rightarrow 0. \end{aligned}$$

This proves that the value function V is differentiable on $\partial S$. Furthermore, $V\in C^1$ on $(0,\infty )^d{\setminus } \partial S$, and by Lemma 1, V is convex. Since differentiable convex functions are $C^1$ (see [6, Theorem 2.2.2]), we obtain the first claim. For the second one, note that $(\mathcal{G}-r)V=0$ on $S^c$ and $(\mathcal{G}-r)V(\mathbf{y})=(\mathcal{G}-r)g(\mathbf{y})=-\left( rK+\sum _{i=1}^d(\mu _i-r)y_i\right) $ for $\mathbf{y}\in int(S)$, which gives the local boundedness around $\partial S.$ $\square $

A.4 Proof of Lemma 3

Proof

By the definition of the Green kernel and the strong Markov property it holds that for all $\mathbf{x}\in {\mathbb R}_+^d$ and all stopping times $\tau $

$$\begin{aligned} w(\mathbf{x})&=\int G_r(\mathbf{x},\mathbf{y})s(\mathbf{y})m(\mathbf{dy})={\mathbb {E}}_\mathbf{x} \left( \int _0^\infty \mathrm{e}^{-rt}s(X_t)dt\right) \\&={\mathbb {E}}_\mathbf{x}\left( \int _\tau ^\infty \mathrm{e}^{-rt}s(X_t)dt\right) +{\mathbb {E}}_\mathbf{x}\left( \int _0^\tau \mathrm{e}^{-rt}s(X_t)dt\right) \\&={\mathbb {E}}_\mathbf{x}\left( \mathrm{e}^{-r\tau }{\mathbb {E}}_\mathbf{x}\left( \int _\tau ^\infty \mathrm{e}^{-r(t-\tau )}s(X_t)dt|{\mathcal {F}}_\tau \right) \right) +{\mathbb {E}}_\mathbf{x}\left( \int _0^\tau \mathrm{e}^{-rt}s(X_t)dt\right) \\&={\mathbb {E}}_\mathbf{x}\left( \mathrm{e}^{-r\tau }{\mathbb {E}}_{X_\tau }\left( \int _0^\infty \mathrm{e} ^{-rt}s(X_t)dt\right) \right) +{\mathbb {E}}_\mathbf{x}\left( \int _0^\tau \mathrm{e}^{-rt}s(X_t)dt\right) \\&={\mathbb {E}}_\mathbf{x}\left( \mathrm{e}^{-r\tau }w(X_\tau )\right) +{\mathbb {E}}_\mathbf{x} \left( \int _0^\tau \mathrm{e}^{-rt}s(X_t)dt\right) . \end{aligned}$$

$\square $

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Authors and Affiliations

Sören Christensen
- 1
Email author
Paavo Salminen
- 2

1.Department of Mathematics, Research Group Statistics and Stochastic ProcessesUniversity of HamburgHamburgGermany
2.Faculty of Science and EngineeringÅbo Akademi UniversityÅboFinland

Multidimensional investment problem

Abstract

Keywords

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Notes

Acknowledgements

References

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