A continuous-time asset market game with short-lived assets

Abstract

We propose a continuous-time game-theoretic model of an investment market with short-lived assets. The first goal of the paper is to obtain a stochastic equation which determines the wealth processes of investors and to provide conditions for the existence of its solution. The second goal is to show that there exists a strategy such that the logarithm of the relative wealth of an investor who uses it is a submartingale regardless of the strategies of the other investors, and the relative wealth of any other essentially different strategy vanishes asymptotically. This strategy can be considered as an optimal growth portfolio in the model.

Appendix

The first part of this appendix provides a known condition for a semimartingale to be a submartingale, in a form convenient for our applications in the proof of Theorem 5.1. The next two parts assemble several facts about the Lebesgue decomposition and Lebesgue derivatives of $\sigma$-finite measures, and prove auxiliary results for random measures generated by predictable nondecreasing càdlàg processes.

1.1 A.1 Submartingality conditions

A scalar semimartingale $Z$ with $Z_{0}=0$ is called a $\sigma$-submartingale if there exists a nondecreasing sequence of predictable sets $\Pi _{n} \in \mathcal {P}$ such that $Z_{t}^{\Pi _{n}} := \int _{0}^{t}I_{s}(\Pi _{n}) d Z_{s}$ is a submartingale for each $n$ and $\bigcup _{n}\Pi _{n}= \Omega \times \mathbb{R}_{+}$. Suppose the triplet $(B^{h},C,\nu )$ of predictable characteristics of $Z$ with respect to a truncation function $h(z)$ admits the representation

$$ B^{h} = b^{h}\boldsymbol {\cdot }G,\quad C=c\boldsymbol {\cdot }G,\quad \nu = K\otimes G, $$

where $b^{h}=(b_{t}^{h})_{t\ge 0}$, $c=(c_{t})_{t\ge 0}$ are predictable processes, $K=(K_{t}(dz))_{t\ge 0}$ is a transition kernel and $G=(G_{t})_{t\ge 0}$ is a nondecreasing predictable càdlàg process. Then $Z$ is a $\sigma$-submartingale if and only if $(P\otimes G)$-a.e. on $\Omega \times \mathbb{R}_{+}$,

$$ \int _{\{|z|>1\}} |z| K_{t}(dz) < \infty \qquad \text{and}\qquad \mathfrak{d}_{t} := b_{t}^{h} + \int _{\mathbb{R}} \bigl(z - h(z)\bigr) K_{t}(dz) \ge 0 $$

(see Karatzas and Kardaras [27, Proposition 11.2], Kallsen [26, Lemma 3.1]). The predictable process $\mathfrak{d}$ is called the drift rate of $Z$ with respect to $G$. One can see that it does not depend on the choice of the truncation function $h$ (see Jacod and Shiryaev [25, Proposition II.2.24]).

Observe that if we have for all $t\ge 0$ that

$$ \int _{\mathbb{R}}|z| K_{t}(dz)< \infty , $$

then $\mathfrak{d}_{t} = b^{0}_{t} + \int _{\mathbb{R}}z K_{t}(dz)$, where $b_{t}^{0} = b_{t}^{h} - \int _{\mathbb{R}}h(z) K_{t}(dz)$ is a well-defined predictable process which does not depend on the choice of $h$. From this, we obtain the following result which is used in the proof of Theorem 5.1.

Proposition A.1

Suppose $Z$ is a nonpositive semimartingale. Then $Z$ is a submartingale if $(P\otimes G)$-a.e.,

$$ \int _{\{z< 0\}} z K_{t}(dz) >- \infty \qquad \textit{and}\qquad \mathfrak{d}_{t} = b^{0}_{t} + \int _{\mathbb{R}}z K_{t}(dz) \ge 0. $$

(A.1)

In particular, if (A.1) is satisfied, then the process $\mathfrak{d}$ is $G$-integrable and the compensator of $Z$ is

$$ A_{t} = \mathfrak{d}\boldsymbol {\cdot }G_{t}. $$

(A.2)

Formula (A.2) follows from observing that for a nonpositive semimartingale, we have $\int _{\{z>0\}} z K_{t}(dz)<\infty$ for $t\ge 0$ since $K_{t}(\{z : z > -Z_{t-}\}) = 0$.

1.2 A.2 Lebesgue decomposition of $\sigma$-finite measures

Let $(\Omega ,\mathcal {F})$ be a measurable space. First recall the following known result, which can be found (in a slightly different form) e.g. in Bogachev [10, Chap. 3.2].

Proposition A.2

Let $P,\widetilde {P}$ be $\sigma$-finite measures on $(\Omega ,\mathcal {F})$. Then there exist a measurable function $Z \ge 0$ ($P$-a.s. and $\widetilde {P}$-a.s.) and a set $\Gamma \in \mathcal {F}$ such that

$$ \widetilde {P} [A] = \int _{A} Z dP+ \widetilde {P} [A\cap \Gamma ] \qquad \textit{for any $A\in \mathcal {F}$} $$

(A.3)

and

$$ P[\Gamma ] = 0. $$

(A.4)

Such a $Z$ is $P$-a.s. unique and $\Gamma$ is $\widetilde {P}$-a.s. unique, i.e., if $Z'$ and $\Gamma '$ also satisfy the above properties, then $Z=Z'$ $P$-a.s., and $\widetilde {P} [\Gamma \triangle \Gamma '] = 0$ (where we denote by $\Gamma \triangle \Gamma ' = (\Gamma \setminus \Gamma ')\cup ( \Gamma '\setminus \Gamma )$ the symmetric difference of the two sets).

The function $Z$ – the Lebesgue derivative of $\widetilde {P}$ with respect to $P$ – is denoted in this paper by ${d\widetilde {P}}/{dP}$. If $\widetilde {P} \ll P$, the Lebesgue derivative coincides with the Radon–Nikodým derivative and one can take $\Gamma = \emptyset$. When it is necessary to emphasise that the set $\Gamma$ is related to $\widetilde {P}$ and $P$, we use the notation $\Gamma _{\widetilde {P} / P}$.

In an explicit form, $Z$ and $\Gamma$ can be constructed as follows. Let $Q$ be any $\sigma$-finite measure on $(\Omega ,\mathcal {F})$ such that $P\ll Q$, $\widetilde {P} \ll Q$ (for example, $Q=P+\widetilde {P}$). Then

$$ Z = \frac {d{\widetilde {P}}}{d{ Q}}\left (\frac{dP}{d Q} \right )^{-1} I_{\left\{\frac{dP}{d Q}>0\right\}}, \qquad \Gamma = \left \{\omega :\frac{dP}{d Q}( \omega ) = 0\right \}, $$

where the derivatives are in the Radon–Nikodým sense.

By approximating a measurable function with simple functions, it follows from (A.3) that for any ℱ-measurable function $f\ge 0$, we have

$$ \int _{\Omega}f d\widetilde {P} = \int _{\Omega}f \frac{d\widetilde {P}}{d P}d P+ \int _{\Omega}f I_{\Gamma} d \widetilde {P} $$

(A.5)

(where the integrals may assume the value $+\infty$).

The following proposition contains facts about Lebesgue derivatives that are used in the paper.

Proposition A.3

Let $P,\widetilde {P} ,Q$ be $\sigma$-finite measures on $(\Omega ,\mathcal {F})$. Then the following statements are true:

(a) Suppose $Q$ is representable in the form $Q[A] = \int _{A} f dP+ \int _{A} \widetilde{f} d\widetilde {P}$, where $f,\widetilde{f}\ge 0$ are measurable functions and $\widetilde{f} = 0$ $P$-a.s. Then

$$ \frac {d{P}}{d{ Q}} = \frac{1}{f} I_{\{f>0,\widetilde{f}=0\}}, \qquad \Gamma _{P/Q} = \{f=0, \widetilde{f}=0\}. $$

(b) If $R$ is a $\sigma$-finite measure such that $R \ll P$ and $R \ll Q$, then

$$ \frac {d{\widetilde {P}}}{d{ P}} = \frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}} \qquad \textit{$R $-a.s.} $$

(A.6)

(c) If $R$ is as in (b), then $dQ/dP> 0$ $R$-a.s. and $dP/dQ>0$ $R$-a.s.

Proof

(a) is obtained by straightforward verification of (A.3) and (A.4).

(b) Observe that for any $A\in \mathcal {F}$, we have

$$ \begin{aligned} \widetilde {P} [A] &= \int _{A} \frac {d{\widetilde {P}}}{d{ Q}} dQ+ \widetilde {P} [A\cap \Gamma _{ \widetilde {P} /Q} ] \\ &= \int _{A}\frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}}d P+ \int _{\Omega} I_{A\cap \Gamma _{Q/P}} \frac {d{\widetilde {P}}}{d{ Q}} dQ+ \widetilde {P} [A\cap \Gamma _{ \widetilde {P} /Q} ] \\ &= \int _{A}\frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}}d P+ \widetilde {P} [A\cap (\Gamma _{Q/P}\cup \Gamma _{ \widetilde {P} /Q}) ], \end{aligned} $$

(A.7)

where we applied (A.5) to obtain the second equality, and the third follows by expressing the second integral in the second line from the equality

$$ \widetilde {P} [A \cap \Gamma _{Q/P}] = \int _{\Omega} I_{A \cap \Gamma _{Q/P}} \frac {d{\widetilde {P}}}{d{ Q}}d Q+ \widetilde {P} [A\cap \Gamma _{Q/P}\cap \Gamma _{ \widetilde {P} /Q} ]. $$

Suppose for $A=\{\frac {d{\widetilde {P}}}{d{ P}} > \frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}}\}$ that $R [A]>0$. Then we also have $R [A']>0$ for $A'=A\cap (\Gamma _{Q/P}\cup \Gamma _{\widetilde {P} /Q} \cup \Gamma _{\widetilde {P} /P})^{c}$ because $R [\Gamma _{Q/P}] = R [\Gamma _{\widetilde {P} /Q}] = R [\Gamma _{\widetilde {P} /P}]=0$. Consequently, $P[A']>0$. But this leads to a contradiction between the decomposition (A.3) and the equality (A.7) for $\widetilde {P} [A']$ since according to them, we should have

$$ \int _{A'} \frac {d{\widetilde {P}}}{d{ P}} dP= \int _{A'} \frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}} dP, $$

which is impossible due to the choice of $A$. Hence $R [\frac {d{\widetilde {P}}}{d{ P}} > \frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}}] = 0$. In the same way, we show that $R [\frac {d{\widetilde {P}}}{d{ P}} < \frac {d{\widetilde {P}}}{d{ Q}} \frac {d{Q}}{d{ P}}] = 0$.

(c) follows from (A.6) if one takes $\widetilde {P} = P$. □

1.3 A.3 Lebesgue decomposition of nondecreasing predictable processes

Let $(\Omega ,\mathcal {F},(\mathcal {F}_{t})_{t\ge 0}, P)$ be a filtered probability space satisfying the usual assumptions, and $\mathcal {P}$ the predictable $\sigma$-algebra on $\Omega \times \mathbb{R}_{+}$. For a scalar nondecreasing càdlàg predictable process $G$, denote by $P\otimes G$ the measure on $\mathcal {P}$ defined as

$$ (P\otimes G)[A] = E[I(A)\boldsymbol {\cdot }G_{\infty}],\qquad A\in \mathcal {P}. $$

(A.8)

Observe that $P\otimes G$ is $\sigma$-finite on $\mathcal {P}$. Indeed, this can be shown by considering the predictable stopping times $\tau _{n} = \inf \{t\ge 0 : G_{t} \ge n\}$. The stochastic intervals $A_{n} = [\!\![0,\tau _{n}[\!\![\, := \{(\omega ,t) : t< \tau _{n}(\omega )\}$ are predictable, i.e., $A_{n}\in \mathcal {P}$, and we have $(P\otimes G)[A_{n}]\le n$ and $\bigcup _{n} A_{n} = \Omega \times \mathbb{R}_{+}$.

Proposition A.4

(a) For any scalar nondecreasing càdlàg predictable processes $G,\widetilde{G}$, there exist a predictable process $\xi \ge 0$ and a set $\Gamma \in \mathcal {P}$ such that up to $P$-indistinguishability,

$$ \widetilde{G} = \widetilde{G}_{0} + \xi \boldsymbol {\cdot }G + I_{\Gamma} \boldsymbol {\cdot }\widetilde{G} \qquad \textit{and}\qquad I_{\Gamma} \boldsymbol {\cdot }G = 0. $$

(A.9)

(b) A predictable process $\xi \ge 0$ and a set $\Gamma \in \mathcal {P}$ satisfy (A.9) if and only if $\xi$ is a version of the Lebesgue derivative $d(P\otimes \widetilde{G})/d(P\otimes G)$ and $\Gamma$ is the corresponding set from the Lebesgue decomposition.

We denote any $(P\otimes G)$-version of such a process $\xi$ by $d{\widetilde{G}}/d{G}$ or $d{\widetilde{G}_{t}}/d{G_{t}}$ and call it a predictable Lebesgue derivative of $\widetilde{G}$ with respect to $G$. When it is necessary to emphasise that the set $\Gamma$ is related to $\widetilde{G}$ and $G$, we use the notation $\Gamma _{\widetilde{G}/ G}$.

Proof

Without loss of generality, assume $\widetilde{G}_{0}=0$.

(a) Let $\xi =d(P\otimes \widetilde{G})/d(P\otimes G)$ and $\Gamma$ be the corresponding set from the Lebesgue decomposition. Define the process

$$ \widetilde{G}' = \xi \boldsymbol {\cdot }G + I_{\Gamma} \boldsymbol {\cdot }\widetilde{G}. $$

We have to show that $\widetilde{G}' = \widetilde{G}$. Since $\widetilde{G}'$ and $\widetilde{G}$ are càdlàg, it is enough to show that $\widetilde{G}'_{t} = \widetilde{G}_{t}$ a.s. for any $t\ge 0$, and this is equivalent to having

$$ E[\widetilde{G}'_{t} I_{B} ] = E[\widetilde{G}_{t}I_{B} ] \qquad \text{for any $B\in \mathcal {F}_{t}$}. $$

(A.10)

Let $M$ be the bounded càdlàg martingale given by $M_{u} = E[I_{B}\mid \mathcal {F}_{u}]$. We have

$$ E[\widetilde{G}'_{t} I_{B} ] = E[\widetilde{G}'_{t} M_{t}] = E[M_{-} \boldsymbol {\cdot }\widetilde{G}'_{t}], $$

and similarly

$$ E[\widetilde{G}_{t} I_{B} ]= E[M_{-} \boldsymbol {\cdot }\widetilde{G}_{t}], $$

(A.11)

where we used the following fact: If $A$ is a nondecreasing càdlàg predictable process and $M$ is a bounded càdlàg martingale, then for any stopping time $\tau$, we have $E[M_{\tau}A_{\tau}] = E[M_{-}\boldsymbol {\cdot }A_{\tau}]$. The latter result can be found in Jacod and Shiryaev [25, Lemma I.3.12] for the case $E[A_{\infty}] < \infty$, from which our case follows by a localisation procedure.

Finally, from the definition of $\widetilde{G}'$ and the Lebesgue decomposition of the measure $P\otimes \widetilde{G}$, it follows that the measures $P\otimes \widetilde{G}$ and $P\otimes \widetilde{G}'$ coincide. Hence for any nonnegative $\mathcal {P}$-measurable function $f$, we have $E[f\boldsymbol {\cdot }\widetilde{G}_{t}] = E[f\boldsymbol {\cdot }\widetilde{G}_{t}']$, which finishes the proof by (A.10) and (A.11).

(b) In view of the construction in (a), it only remains to show that if $\xi ,\Gamma$ satisfy (A.9), then $\xi$ is the Lebesgue derivative and $\Gamma$ is the corresponding predictable set. This follows from straightforward verification of properties (A.3) and (A.4). □