Mean Field Games of Timing and Models for Bank Runs

Abstract

The goal of the paper is to introduce a set of problems which we call mean field games of timing. We motivate the formulation by a dynamic model of bank run in a continuous-time setting. We briefly review the economic and game theoretic contributions at the root of our effort, and we develop a mathematical theory for continuous-time stochastic games where the strategic decisions of the players are merely choices of times at which they leave the game, and the interaction between the strategic players is of a mean field nature.

Appendices

Appendix A: Proof of Proposition 6.2

To prove Theorem 6.2, we need a preliminary results, borrowed from previous works of the authors. Recall that $\Rightarrow$ denotes convergence in law.

Proposition A.1

(Proposition C.1 of [12]) Let X and Y be random variables defined on a common probability space, taking values in some Polish spaces E and F. If the law of X is nonatomic, and if F is (homeomorphic to) a convex subset of a locally convex space, then there exists a sequence of continuous functions $\phi _n : E \rightarrow F$ such that $(X,\phi _n(X)) \Rightarrow (X,Y)$.

Proposition 6.2 extends Proposition A.1 to the dynamic setting, and this is where the role of compatibility is the clearest. This is contained in the third author’s PhD thesis [27, Proposition 2.1.6], which itself was implicitly present in the proof of [12, Lemma 3.11], though we include the proof for the sake of completeness.

1.1 Proof of Proposition 6.2

The proof is an inductive application of Proposition A.1. First, in light of the assumption that the law of $Y_1$ is nonatomic, Proposition A.1 allows us to find a sequence of continuous functions $h^j_1 : {\mathcal Y}\rightarrow {\mathcal X}$ such that $(Y_1,h^j_1(Y_1)) \Rightarrow (Y_1,X_1)$ as $j\rightarrow \infty$. Let us show that in fact $(Z,h^j_1(Y_1))$ converges to $(Z,X_1)$. Let $\phi : {\mathcal Z}\rightarrow {\mathbb R}$ be bounded and measurable, and let $\psi : {\mathcal X}\rightarrow {\mathbb R}$ be continuous. Note that Z and $X_1$ are conditionally independent given $Y_1$, by assumption. Now use Lemma 6.1 to get

$$\begin{aligned} \lim _{j\rightarrow \infty }{\mathbb E}[\phi (Z)\psi (h^j_1(Y_1))]&= \lim _{j\rightarrow \infty }{\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\right| Y_1\right] \psi (h^j_1(Y_1))\right] \\&= {\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\right| Y_1\right] )\psi (X_1)\right] \\&= {\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\right| Y_1\right] {\mathbb E}\left[ \left. \psi (X_1)\right| Y_1\right] \right] \\&= {\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\psi (X_1)\right| Y_1\right] \right] \\&= {\mathbb E}\left[ \phi (Z)\psi (X_1)\right] . \end{aligned}$$

The class of functions of the form ${\mathcal Z}\times {\mathcal X}\ni (z,x) \mapsto \phi (z)\psi (x)$, where $\phi$ and $\psi$ are as above, is convergence determining (see, e.g., [16, Proposition 3.4.6(b)]), and we conclude that $(Z,h^j_1(Y_1)) \Rightarrow (Z,X_1)$.

We proceed inductively as follows. Abbreviate $Y^n := (Z_1,\ldots ,Z_n)$ for each $n=1,\ldots ,N$, noting $Y^N=Y$, and similarly define $X^n$. Suppose we are given $1 \le n < N$ and continuous functions $g^j_k : {\mathcal Y}^k \rightarrow {\mathcal X}$, for $k \in \{1,\ldots ,n\}$ and $j \ge 1$, satisfying

$$\begin{aligned} \lim _{j\rightarrow \infty }(Z,g^j_1(Y^1),\ldots ,g^j_n(Y^n)) = (Z,X^n), \end{aligned}$$

(A.1)

where convergence is in distribution, as usual. We will show that there exist continuous functions $h^i_k : {\mathcal Y}^k \rightarrow {\mathcal X}$ for each $k \in \{1,\ldots ,n+1\}$ and $i \ge 1$ such that

$$\begin{aligned} \lim _{i\rightarrow \infty }(Z,h^i_1(Y^1),\ldots ,h^i_{n+1}(Y^{n+1})) = (Z,X_1,\ldots ,X_{n+1}). \end{aligned}$$

(A.2)

By Proposition A.1 there exists a sequence of continuous functions $\hat{g}^j : ({\mathcal Y}^{n+1} \times {\mathcal X}^n) \rightarrow {\mathcal X}$ such that

$$\begin{aligned} \lim _{j\rightarrow \infty }(Y^{n+1},X^n,\hat{g}^j(Y^{n+1},X^n)) = (Y^{n+1},X^n,X_{n+1}) = (Y^{n+1},X^{n+1}). \end{aligned}$$

(A.3)

We claim now that

$$\begin{aligned} \lim _{j\rightarrow \infty }(Z,X^n,\hat{g}^j(Y^{n+1},X^n)) = (Z,X^n,X_{n+1}). \end{aligned}$$

(A.4)

Indeed, let $\phi$, $\psi _n$, and $\psi$ be bounded measurable functions on ${\mathcal Z}$, ${\mathcal X}^n$, and ${\mathcal X}$, respectively, with $\psi _n$ and $\psi$ continuous. Use the conditional independence of Z and $(Y^{n+1},X^{n+1})$ given $Y^{n+1}$ along with (A.3) and Lemma 6.1 to get

$$\begin{aligned}&\lim _{j\rightarrow \infty }{\mathbb E}[\phi (Z)\psi _n(X^n)\psi (\hat{g}^j(Y^{n+1},X^n))]\\&\quad = \lim _{j\rightarrow \infty }{\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\right| Y^{n+1}\right] \psi _n(X^n)\psi (\hat{g}^j(Y^{n+1},X^n))\right] \\&\quad = {\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\right| Y^{n+1}\right] \psi _n(X^n)\psi (X_{n+1})\right] \\&\quad = {\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z)\right| Y^{n+1}\right] {\mathbb E}\left[ \left. \psi _n(X^n)\psi (X_{n+1})\right| Y^{n+1}\right] \right] \\&\quad = {\mathbb E}\left[ {\mathbb E}\left[ \left. \phi (Z) \psi _n(X^n)\psi (X_{n+1})\right| Y^{n+1}\right] \right] \\&\quad = {\mathbb E}\left[ \phi (Z) \psi _n(X^n)\psi (X_{n+1})\right] . \end{aligned}$$

Again, the class of functions of the form ${\mathcal Z}\times {\mathcal X}^n \times {\mathcal X}\ni (z,x,x') \mapsto \phi (z)\psi _n(x)\psi (x')$, where $\phi$, $\psi _n$, and $\psi$ are as above, is convergence determining, and (A.4) follows.

By continuity of $\hat{g}^j$, the limit (A.1) implies that, for each j,

$$\begin{aligned} \lim _{k\rightarrow \infty }&(Z,g^k_1(Y^1),\ldots ,g^k_n(Y^n),\hat{g}^j(Y^{n+1},g^k_1(Y^1),\ldots ,g^k_n(Y^n))) \nonumber \\&= (Z,X_1,\ldots ,X_n,\hat{g}^j(Y^{n+1},X_1,\ldots ,X_n)) \nonumber \\&= (Z,X^n,\hat{g}^j(Y^{n+1},X^n)). \end{aligned}$$

(A.5)

Combining the two limits (A.4) and (A.5), we may find a subsequence $j_k$ such that

$$\begin{aligned} \lim _{k\rightarrow \infty }&(Z,g^{j_k}_1(Y^1),\ldots ,g^{j_k}_n(Y^n),\hat{g}^k(Y^{n+1},g^{j_k}_1(Y^1),\ldots ,g^{j_k}_n(Y^n))) = (Z,X^n,X_{n+1}). \end{aligned}$$

Define $h^k_\ell := h^{j_k}_\ell$ for $\ell =1,\ldots ,n$ and $h^k_{n+1}(Y^{n+1}) := \hat{g}^k(Y^{n+1},g^{j_k}_1(Y^1),\ldots ,g^{j_k}_n(Y^n))$ to complete the induction. $\square$

Appendix B: The Lattice of Stopping Times

In this section, we prove that the set ${\mathcal S}$ of stopping times defined in Sect. 5 is a complete lattice. Recall that ${\mathcal S}$ is defined as the set of (equivalence classes of a.s. equal) random times $\tau$ defined on the probability space $\Omega ^{\mathrm {com}} \times \Omega ^{\mathrm {ind}}$, which are stopping times with respect to the filtration ${\mathbb F}^{\text {sig}}$. Recall that the essential supremum of a family $\Phi$ of random variables is defined as the minimal (with respect to a.s. order) random variable exceeding a.s. each element of ${\mathbb T}$:

Theorem B.1

(Theorem A.33 of [17]) Let $\Phi$ be a set of real-valued random variables. Then there exists a unique (up to a.s. equality) random variable $Z = {{\mathrm{ess\,sup}}}\Phi$ such that $Z \ge X$ a.s. for each $X \in \Phi$, and also $Z \le Y$ a.s. for every random variable Y satisfying $Y \ge X$ a.s. for every $X \in \Phi$. Moreover, there exists a countable set $\Phi _0 \subset \Phi$ such that $Z = \sup _{X \in \Phi _0}X$ a.s.

Proof

Existence and uniqueness is stated in [17, Theorem A.33], and the proof therein constructs the desired $\Phi _0$. $\square$

The essential infimum is defined analogously, or simply by ${{\mathrm{ess\,inf}}}\Phi = -{{\mathrm{ess\,sup}}}(-\Phi )$.

Theorem B.2

The set ${\mathcal S}$ is a complete lattice.

Proof

Fix a set $\Phi \subset {\mathcal S}$. Define $Z = {{\mathrm{ess\,sup}}}\Phi$ and find a countable set $\{\tau _n : n \ge 1\} \subset \Phi$ such that $Z = \sup _n\tau _n$ a.s. Define $\sigma _n = \max _{k=1,\ldots ,n}\tau _k$, so that $\sigma _n$ is an increasing sequence of stopping times with $\sigma _n \uparrow Z$ a.s. The increasing limit of a sequence of stopping times is again a stopping time [13, Theorem IV.55(b)], so $Z \in {\mathcal S}$.

A similar argument applies to show that the essential infimum of $\Phi$ is also a stopping time, and the only difference is that this step crucially uses the right-continuity of the filtration $\overline{{\mathbb F}}^{\text {sig}}$; indeed, while the supremum of a sequence of stopping times is always a stopping time, the infimum of a sequence of stopping times is only guaranteed to be a stopping time if the underlying filtration is right-continuous [13, Theorem IV.55(c)]. $\square$