On the Market Selection Hypothesis in a Mean Reverting Environment

Abstract

We investigate the market selection hypothesis in a mean reverting environment. We consider three models varying the endowment process and agents’ beliefs and we show that with a constant relative risk aversion utility, controlling for the discount factor, agents with incorrect beliefs about the level of the endowment process cannot survive.

Appendix

Let $X(t)$ be a real irreducible Markov process on ${\mathbb {R}}_+$, with an invariant measure $\nu$ on ${\mathbb {R}}$ and $X(0)=x$. Let $p(s,t,x,dy)$ be its transition kernel and ${\mathcal {A}}$ its generator. We need the following definitions.

Definition 2

(Small Set). A subset $C$ of ${\mathbb {R}}$ is a small set if there exist an $\epsilon >0$, a probability measure on $[0,+\infty )$ with density $m(t)$ which admits a mean and probability measure $\pi$ on ${\mathbb {R}}$ such that for each $x \in {\mathbb {R}}$ and for each $A \in {\mathcal {B}}({\mathbb {R}})$

$$\begin{aligned} \int _0^{+\infty } p(0,t,x,A)\, m(t)\, dt \ge \epsilon \, {\mathbb {I}}_{C}(x) \pi (A), \end{aligned}$$

(44)

where

$$\begin{aligned} p(0,t,x,A)=\int _A p(0,t,x,dy) \end{aligned}$$

and ${\mathbb {I}}_{C}(x)$ denotes the indicator function of the set $C$.

Definition 3

(Lyapunov function). $V(x)$ is a Lyapunov function if there exist a “test” function $W(x):{\mathbb {R}} \rightarrow [1,\infty )$, a small set $C \subset {\mathbb {R}}$ and two constants $\delta >0,\; b<+\infty$ such that

$$\begin{aligned} {\mathcal {A}}V(x) \le -\delta W(x) + b \, {\mathbb {I}}_C(x) \end{aligned}$$

(45)

We can now state the following result, see Kontoyiannis and Meyn (2005).

Lemma 1

(“Generalized Ergodic”). Given a Lyapunov function $V(x)$ and a “test” function $W(x)$ which satisfy (45), for each function $F(x):{\mathbb {R}} \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \sup \frac{\vert F(x) \vert }{W(x)} < +\infty , \end{aligned}$$

(46)

we have that

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \int _0^t F(X(t)) = \int F(x) \, \nu (dx) = {\mathbb {E}}_\nu \big [F(x)\big ] \quad a.s. \end{aligned}$$

We are now able to prove the main result used in our analysis.

Theorem 1

Given a Ornstein–Uhlenbeck process $X(t)$

$$\begin{aligned} dX(t)=-b(X(t)-a)+\sigma dZ(t), \end{aligned}$$

(47)

the following result holds true

$$\begin{aligned} \lim _{t \rightarrow +\infty } \frac{1}{t} \int _0^t X^2(s)ds = \int x^2 \nu (dx) = a^2+\frac{\sigma ^2}{2b} \quad a.s. \end{aligned}$$

(48)

where $\nu$ is the invariant measure of $X(t)$.

Proof

Thanks to the Ergodic Theorem for Markov processes we know that for each continuous and bounded function $f: {\mathbb {R}} \rightarrow {\mathbb {R}}$ we have

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t f(X(s))\,ds=\int f(x)\,\nu (dx) =E_\nu \big [f(x)\big ] \quad a.s. \end{aligned}$$

In our case, $f(x)=x^2$ so we need a generalization of the result to a larger class of functions. The generalization is provided by Lemma 1.

In fact we can prove (48) applying Lemma 1 with $F(x)=x^2$. By the law of $X(t)$ we have

$$\begin{aligned} p(0,t,x,dy)=\frac{1}{\sqrt{ \frac{\pi \sigma ^2}{2b} (1-e^{-2bt})}} \exp \bigg \{ -\frac{b(y-\mu (t))^2}{\sigma ^2 (1-e^{-2bt})} \bigg \}, \end{aligned}$$

where $\mu (t)={\mathbb {E}}[X(t)]=a+(x-a) e^{-bt}$.

We must now find a Lyapunov function $V(x)$, a “test” function $W(x)$, a “small set” $C$ and two constants $\delta >0,\; b<\infty$ such that (45) and (46) are satisfied with $F(x)=x^2$. The idea is to try with a “test” function $W(x)$ which is a simple translation of $F(x)$, assuring that (46) is respected and $W(x) \ge 1$ for each $x$ (as requested in the definition for a Lyapunov function). Then let us suppose that

$$\begin{aligned} W(x)=x^2+1. \end{aligned}$$

(49)

Using the expression for Ornstein–Uhlenbeck generator

$$\begin{aligned} {\mathcal {A}}(x)=\frac{\sigma ^2}{2}\frac{d^2}{dx^2}-\alpha x \frac{d}{dx} \end{aligned}$$

in (45) we obtain

$$\begin{aligned} \frac{\sigma ^2}{2}V''(x)-\alpha x V'(x) \le -\delta W(x)+b \, {\mathbb {I}}_C(x), \end{aligned}$$

which, with (49), leads to the following condition for the Lyapunov function $V(x)$

$$\begin{aligned} \frac{\sigma ^2}{2}V''(x)-\alpha x V'(x) \le -\delta (x^2+1) +b \, {\mathbb {I}}_C(x). \end{aligned}$$

Let us suppose $V(x)=\frac{x^2}{2}$, we obtain

$$\begin{aligned} \frac{\sigma ^2}{2}-\alpha x^2 \le -\delta x^2 -\delta + b \, {\mathbb {I}}_C(x) \end{aligned}$$

which is satisfied taking $\delta < \alpha$, $b \ge \delta + \frac{\sigma ^2}{2}$ and $C=[-k,k]$ where $k= \sqrt{\frac{\sigma ^2+2 \delta }{2(\alpha -\delta )}}$.

We have found our candidates for the Lyapunov function ($V(x)$), the “test” function ($W(x)$), the set $C$, the constants $\delta$ and $b$. We have now to prove that $C=[-k,k]$ is a small set. Let $x$ be $\in [-k,k]$ and $t^*$ such that $1-e^{-2bt} \ge \frac{1}{2}$ for each $t \ge t^*$.

For each $t \ge t^*$ we have

$$\begin{aligned} \vert y-\mu (t) \vert ^2&\le y^2 + \mu ^2(t) \\&\le y^2+a^2+ 2\vert a \vert \vert x-a \vert e^{-bt} + \vert x-a \vert ^2 e^{-2bt} \nonumber \\&\le y^2 +a^2 + 2\vert a \vert \vert x-a \vert + \vert x-a \vert ^2 \nonumber \\&\le y^2 +4a^2 +4k \vert a \vert + k^2. \end{aligned}$$

Let $h=4a^2 +4k \vert a \vert + k^2$. We obtain

$$\begin{aligned} p(0,t,x,dy)&= \frac{1}{\sqrt{ \frac{\pi \sigma ^2}{2b} (1-e^{-2bt})}} \exp \bigg \{ -\frac{b(y-\mu (t))^2}{\sigma ^2 (1-e^{-2bt})} \bigg \} \\&\ge \frac{1}{\sqrt{ \frac{\pi \sigma ^2}{2b} (1-e^{-2bt})}} \exp \bigg \{ -\frac{2b(y^2+h)}{\sigma ^2} \bigg \}. \end{aligned}$$

For each probability density $m(t)$ on $[0,+\infty )$ then we have that

$$\begin{aligned}&\int _0^{+\infty } p(0,t,x,A) \, m(t) dt \ge \int _{t^*}^{+\infty } p(0,t,x,A)\, m(t)\, dt \\&\quad \ge \int _{t^*}^{+\infty } \left( \int _{A} \frac{1}{\sqrt{ \frac{\pi \sigma ^2}{2b} (1-e^{-2bt})}} \exp \bigg \{ -\frac{2b(y^2+h)}{\sigma ^2} \bigg \} dy \right) \, m(t)\, dt \\&\quad =\int _{A} \left( \int _{t^*}^{+\infty } \frac{1}{\sqrt{ \frac{\pi \sigma ^2}{2b} (1-e^{-2bt})}} \exp \bigg \{ -\frac{2b(y^2+h)}{\sigma ^2} \bigg \}\, m(t) dt \right) dy = \epsilon \, \pi (A), \end{aligned}$$

where

$$\begin{aligned} \pi (A)&= \frac{1}{\sqrt{\frac{\pi \sigma ^2}{2b}}} \int _A \exp \bigg \{ -\frac{2by^2}{\sigma ^2} \bigg \} dy, \end{aligned}$$

(50)

$$\begin{aligned} \epsilon&= c_m \sqrt{\frac{\pi \sigma ^2}{2b}} \exp \bigg \{-\frac{2b h}{\sigma ^2} \bigg \}, \end{aligned}$$

(51)

with

$$\begin{aligned} c_m = \int _{t^*}^{+\infty } \frac{1}{\sqrt{ \frac{\pi \sigma ^2}{2b} (1-e^{-2bt})}} m(t) dt. \end{aligned}$$

It is easy to recognize that $\pi$ is a gaussian probability measure. We require $c_m$ to be bounded, this can be achieved choosing $m(t)$ as a $\chi ^2$ density random variable.

Condition (44) is then verified with $\epsilon$ and $\pi$ given by (51) and (50). We can now apply Lemma 1: (48) follows considering for the invariant measure $\nu$ of the Ornstein–Uhlenbeck process $X(t)$, thanks to the fact that the expression for the variance is $\sigma ^2 / 2b$, and the one for mean is $a$. $\square$

Theorem 2

Let $Y(t)$ be an Ornstein–Uhlenbeck process with zero long run mean, mean reversion and volatility parameters $k$ and $\sigma _Y$, that is

$$\begin{aligned} dY(t)=-kY(t)+\sigma _Y dZ(t) \end{aligned}$$

then the process $X(t)=\int \limits _0^t Y^2(s)ds -mt$ is such that a.s.

Proof

By Theorem 1

$$\begin{aligned} \lim _{t\rightarrow +\infty }\frac{1}{t}\int _0^t Y^2(s)ds= \frac{\sigma _Y^2}{2k} \quad a.s. \end{aligned}$$

then, if $t \rightarrow +\infty$, we have

$$\begin{aligned} X(t) \sim \bigg ( \frac{\sigma _Y^2}{2k}-m \bigg )t. \end{aligned}$$

$\square$

Theorem 3

Let $(\Omega , {\mathcal {F}},\{ {\mathcal {F}}_t\}, {\mathbb {P}})$ be a stochastic base where a Brownian motion $Z(t)$ is defined and let $\{ {\mathcal {F}}_t\}$ be its natural filtration.

Define the translated Brownian motion

$$\begin{aligned} Y(t)=Z(t)+x, \quad x \in {\mathbb {R}}. \end{aligned}$$

and the process $Z_{{\mathbb {Q}}}(t)$ such that

$$\begin{aligned} dZ^{{\mathbb {Q}}}(t) = d Y(t) - \alpha ( \Pi - Y(t))dt = d Z(t) - \alpha (\Pi -x-Z(t))dt. \end{aligned}$$

Then $Z_{{\mathbb {Q}}}(t)$ is a $\mathbb {Q}$-Brownian motion where

$$\begin{aligned} \frac{d{\mathbb {Q}}}{d{\mathbb {P}}} =\exp \Bigg \{ \int _0^t \alpha (\Pi -x-Z(s)) dZ(s) -\frac{1}{2} \int _0^t \alpha ^2 (\Pi -x-Z(s))^2 ds \Bigg \} \end{aligned}$$

and $Y(t)$ is a $\mathbb {Q}$ Ornstein–Uhlenbeck process with long run mean $\Pi$, mean reversion parameter $\alpha$ and initial value $x$.

Proof

The change of measure $\frac{d{\mathbb {Q}}}{d{\mathbb {P}}}$ follows directly from Girsanov Theorem with $\theta (t)=\alpha (\Pi -x-Z(t))$. Note that

$$\begin{aligned} dZ_{{\mathbb {Q}}}(t) = d Y(t) - \alpha ( \Pi - Y(t))dt \end{aligned}$$

implies

$$\begin{aligned} d Y(t) = \alpha ( \Pi - Y(t))dt + dZ_{{\mathbb {Q}}}(t), \end{aligned}$$

where we recognize a Ornstein–Uhlenbeck dynamics with long run mean $\Pi$ and mean reversion parameter $\alpha$. $\square$