Optimal bubble riding with price-dependent entry: a mean field game of controls with common noise

Abstract

In this paper we further extend the optimal bubble riding model proposed in Tangpi and Wang (Optimal bubble riding: a mean field game with varying entry times, 2022) by allowing for price-dependent entry times. Agents are characterized by their individual entry threshold that represents their belief in the strength of the bubble. Conversely, the growth dynamics of the bubble is fueled by the influx of players. Price-dependent entry naturally leads to a mean field game of controls with common noise and random entry time, for which we provide an existence result. The equilibrium is obtained by first solving discretized versions of the game in the weak formulation and then examining the measurability property in the limit. In this paper, the common noise comes from two sources: the price of the asset which all agents trade, and also the the exogenous bubble burst time, which we also discretize and incorporate into the model via progressive enlargement of filtration.

Two auxiliary results

For a càdlàg process Y, denote by $M^Y_t = \sup _{0 \le s \le t}Y_s$ its running maximum. Recall from (1) and the price dynamics of the N-player game that the bubble trend function b naturally depends on $F_p(M^P_t)$, which is not Lipschitz in $M^P$. In general, the dynamics of asset price in the bubble phase is not well-posed. However, as the bubble is fueled by players’ entry, b should be increasing in $F_p(M^P_t)$, hence also increasing in $M^P_t$ at each time $t \in [0, T]$ since $F_p$ is a CDF. This monotonicity property of the path-dependent SDE (2) restores unique solvability.

Proposition A.1

The following path-dependent SDE

$$\begin{aligned} X_t = x + \int _0^t{\tilde{b}}(s, M^X_s, X_s)ds + \sigma _0 B_t \end{aligned}$$

(50)

has a unique strong solution satisfying $\mathbb {E}[\Vert X\Vert _\infty ^2] < \infty$ if for each fixed $t\in [0,T]$:

• (1) There exists $C > 0$ such that for all $\varvec{x}\in C([0, T]; \mathbb {R})$:

  $$\begin{aligned} \left| {\tilde{b}}\left( t, M^{\varvec{x}}_t, \varvec{x}_t\right) \right| \le C\left( 1 + M^{|\varvec{x}|}_t\right) . \end{aligned}$$

• (2) ${\tilde{b}}(t, \cdot , \cdot )$ is increasing (not necessarily strictly) in each argument.

Proof

We adapt the proof of [11, Theorem 4.1]. The first condition guarantees a weak solution satisfying the integrability condition that is unique in law (see [42, Proposition 5.3.6 and Remark 5.3.8]). By the well-known result of Yamada and Watanabe [64], we only need to show pathwise uniqueness. Suppose X and Y are two solutions on the same probability space with respect to the same Brownian motion B. Observing that $X - Y$ is absolutely continuous, by Tanaka’s formula we get

$$\begin{aligned} \begin{aligned} X_t \vee Y_t&= Y_t + (X_t - Y_t)^+ = Y_t + \int _0^t\mathbb {1}_{\{X_s> Y_s\}}d(X_s - Y_s) \\&= x + \sigma _0 B_t + \int _0^t\mathbb {1}_{\{X_s> Y_s\}}{\tilde{b}}(s, M^X_s, X_s)ds + \int _0^t\mathbb {1}_{\{X_s \le Y_s\}}{\tilde{b}}(s, M^Y_s, Y_s)ds.\\ Y_t \vee X_t&= x + \sigma _0 B_t + \int _0^t\mathbb {1}_{\{Y_s > X_s\}}{\tilde{b}}(s, M^Y_s, Y_s)ds + \int _0^t\mathbb {1}_{\{Y_s \le X_s\}}{\tilde{b}}(s, M^X_s, X_s)ds. \end{aligned} \end{aligned}$$

(51)

We can equate the above expressions for all t, implying that for almost every t we have

$$\begin{aligned} \mathbb {1}_{\{X_t = Y_t\}} \left( {\tilde{b}}(t, M^Y_t, Y_t) - {\tilde{b}}(t, M^X_t, X_t)\right) = 0. \end{aligned}$$

(52)

We now show that if $X_s > Y_s$, then $M^X_s \ge M^Y_s$. Define

$$\begin{aligned} s_0 :=\sup \{u \in [0, s]: X_u = Y_u\}. \end{aligned}$$

The case is trivial if $s_0 = 0$.

On the event $\{s_0 > 0\}$, continuity of X and Y implies that $X_t > Y_t$ for all $t \in (s_0, s]$. Suppose $M^X_s < M^Y_s$, then there must exist $s^* \in [0, s_0)$ where $Y_{s^*} = M^Y_s > M^X_s \ge X_{s^*}$. Then define

$$\begin{aligned} s_1 :=\inf \{u \in [s^*,s_0]: X_u = Y_u\}. \end{aligned}$$

By continuity again, $Y_t > X_t$ for all $t \in [s^*, s_1)$. By definition of $s^*$, we must also have $M^Y_{t} > M^X_{t}$ for all $t \in [s^*, s_1)$. Monotonicity of ${\tilde{b}}$ leads to a contradiction

$$\begin{aligned} 0 > X_{s^*} - Y_{s^*} = \int _{s^*}^{s_1}{\tilde{b}}(t, M^Y_t, Y_t) - {\tilde{b}}(t, M^X_t, X_t)dt\ge 0. \end{aligned}$$

Therefore, $M^X_s \ge M^Y_s$ and in particular, $M^{X\vee Y}_s = M^{X}_s$. We can then rewrite (51) as

$$\begin{aligned} X_t \vee Y_t&= x + \sigma _0 B_t + \int _0^t{\tilde{b}}(s, M^{X \vee Y}_s, X_s \vee Y_s)ds \\&\quad + \int _0^t\mathbb {1}_{\{X_s = Y_s\}}\left( {\tilde{b}}(s, M^{Y}_s, Y_s) - {\tilde{b}}(s, M_s^{X \vee Y}, X_s \vee Y_s)\right) ds\\&= x + \sigma _0 B_t + \int _0^t{\tilde{b}}(s, M^{X \vee Y}_s, X_s \vee Y_s)ds \\&\quad + \int _0^t\mathbb {1}_{\{\{X_s = Y_s\} \cap \{M^{X}_s > M^{Y}_s\}\}}\left( {\tilde{b}}(s, M^{Y}_s, Y_s) - {\tilde{b}}(s, M_s^{X \vee Y}, X_s \vee Y_s)\right) ds. \end{aligned}$$

where the last line vanishes by (52). Therefore, $X \vee Y$ also satisfies (50). Similarly, one can show $X \wedge Y$ is also a solution. Then by uniqueness of law, we have $\mathbb {E}[|X - Y|] = \mathbb {E}[X\vee Y - X\wedge Y] = 0$ which leads to pathwise-uniqueness and completes the proof. The integrability property easily follows from Grönwall’s inequality. $\square$

The following measure theoretic result is probably well known. We give a proof since we could not find a directly citable reference.

Lemma A.2

Let $(S, \Sigma , \mu )$ be a complete measurable space. A function $f: S \times \mathbb {R}\rightarrow \mathbb {R}$ is jointly measurable if for all $x \in \mathbb {R}$:

• (1) $f(\cdot , x)$ is measurable.

• (2) $f(\cdot , x_n)$ converges to $f(\cdot , x)$ in $\mu$-measure for any increasing sequence $x_n \uparrow x$.

Proof

First let $E \subseteq \mathbb {R}$ be any closed set and let $X = \{x_m\}_{m \ge 1}$ be a countable, dense subset of $\mathbb {R}$. For $\varepsilon > 0$, denote by ${\mathcal {O}}_{\epsilon }(E)$ the open set $\{x \in \mathbb {R}: \inf _{e \in E}|x - e| < \varepsilon \}$. We claim that for $\mu$-almost every $s \in S$ and any $x \in \mathbb {R}$, $f(s, x) \in E$ if and only if for each $n \in \mathbb {N}$, there is $x_m \in X \cap (x - \frac{1}{n}, x]$ such that $f(s, x_m) \in {\mathcal {O}}_{\frac{1}{n}}(E)$. Note that we can always approximate any $x \in \mathbb {R}$ by an increasing sequence $\{x_{m_k}\}_{k \ge 1}$ with elements in X such that the functions $f(\cdot , x_{m_k})$ converge $\mu$-almost everywhere to $f(\cdot , x)$. The claim follows almost immediately. Denoting by $f^{-1}$ the preimage of f, joint measurability is proved by writing

$$\begin{aligned} f^{-1}(E) = \bigcap _{n = 1}^\infty \bigcup _{m = 1}^{\infty }\left\{ s \in S:f(s, x_m) \in {\mathcal {O}}_{\frac{1}{n}}(E)\right\} \times \left[ x_m, x_m + \frac{1}{n}\right) . \end{aligned}$$

$\square$