Aging of the Metropolis dynamics on the random energy model

Abstract

We study the Metropolis dynamics of the simplest mean-field spin glass model, the random energy model. We show that this dynamics exhibits aging by showing that the properly rescaled time change process between the Metropolis dynamics and a suitably chosen ‘fast’ Markov chain converges in distribution to a stable subordinator. The rescaling might depend on the realization of the environment, but we show that its exponential growth rate is deterministic.

Appendix: Extremal characterization of mean hitting time

In this appendix we give the proof of the formula (4.1) which gives a lower bound on the mean hitting time of a set when starting from stationarity. This formula is a continuous-time version of (a half of) Proposition 3.2 from [16]. This proposition, as well as the underlying result [2, Proposition 3.41], is stated for a continuous-time Markov chain whose waiting times are mean-one exponential random variables. We were not able to find analogous statements for general continuous-time Markov chains in the literature, so we provide short proofs here, for the sake of completeness.

We start by introducing some notation. Let Y be a reversible continuous-time Markov chain on a finite state space $\mathcal {S}$ with transition rates $q_{xy}$ and invariant probability measure $\nu _x$, denote by $P_{\nu }$ and $P_x$ the laws of Y started stationary and from x respectively, and by $E_{\nu }$, $E_x$ the corresponding expectations. Define the conductances as $c_{xy}=\nu _x q_{xy} = \nu _y q_{yx}$. Let $q_x=\sum _y q_{xy}$ and $c_x = \sum _y c_{xy}$. The transition probability from x to y is $p_{xy}=\frac{q_{xy}}{q_x}=\frac{c_{xy}}{c_x}$. In the same way as in Sect. 2, we define the hitting time $H_x$ and the return time $H^+_x$ to x by Y, and similarly $H_A$ and $H^+_A$ for sets $A\subset \mathcal {S}$.

A function g on $\mathcal {S}$ is called harmonic in x, if $\sum _{y}g(y)p_{xy} = g(x)$. For $x\in \mathcal {S}$ and $B\subset \mathcal {S}{\setminus }\{x\}$, the equilibrium potential $g^{\star }_{x,B}$ is defined as the unique function on $\mathcal {S}$ that is harmonic on $(x\cup B)^c$, 1 on x and 0 on B. It is well known that

$$\begin{aligned} g^{\star }_{x,B}(y) = P_y[H_x \le H_B]. \end{aligned}$$

For a function $g:\mathcal {S}\rightarrow \mathbb {R}$, the Dirichlet form is defined as

$$\begin{aligned} D(g,g) = \frac{1}{2} \sum _{z\in \mathcal {S}} \sum _{y\sim z} \nu _z q_{zy} (g(z)-g(y))^2, \end{aligned}$$

(8.9)

where $y\sim z$ means that y and z are neighbors in the sense that $q_{zy}>0$.

The following proposition is the required generalization of Proposition 3.2 of [16].

Proposition 8.5

For every $x\in \mathcal {S}$ and $B\subset \mathcal {S}{\setminus }\{x\}$

$$\begin{aligned} \frac{1}{E_{\nu }[H_x]} \le D(g^{\star }_{x,B},g^{\star }_{x,B}) \nu (B)^{-2} = c_x P_x[H^+_x> H_B]\nu (B)^{-2}. \end{aligned}$$

(8.10)

To prove this proposition we will need a lemma which is a generalization of [2, Proposition 3.41] giving the extremal characterization of the mean hitting time.

Lemma 8.6

For every $x\in \mathcal {S}$,

$$\begin{aligned} \frac{1}{E_{\nu }[H_x]} = \inf \left\{ D(g,g):~g:\mathcal {S}\rightarrow \mathbb {R},~g(x)=1, \sum _{y \in \mathcal {S}}\nu _y g(y)=0\right\} . \end{aligned}$$

(8.11)

Proof

The proof follows the lines of [2] with some minor changes to fit into the setting of general continuous-time chains.

We first show that there is a minimizing function g that equals $g(y) = \frac{Z_{yx}}{Z_{xx}}$, where

$$\begin{aligned} Z_{yx} = \int _0^{\infty } \left( P_y[Y_t=x]-\nu _x\right) \,{\mathrm d}t. \end{aligned}$$

To this end, we introduce the Lagrange multiplier $\gamma$ and consider g as the minimizer of $D(g,g) + \gamma \sum _z \nu _z g(z)$ with $g(x)=1$. The contribution to this of g(y) for $y\ne x$ is

$$\begin{aligned} \sum _{z\sim y} \nu _y q_{yz} (g(y)-g(z))^2 + \gamma \nu _y g(y), \end{aligned}$$

which is minimized if

$$\begin{aligned} 2\sum _{z\sim y} \nu _y q_{yz} (g(y)-g(z)) + \gamma \nu _y =0. \end{aligned}$$

From this we get for all $y\in \mathcal {S}$, by introducing the term including the parameter $\beta$ for the case $y=x$, that

$$\begin{aligned} g(y) = \sum _{z\sim y} \frac{q_{yz}}{q_y} g(z) - \frac{\gamma }{2} \frac{1}{q_y} + \frac{\beta }{q_y} \mathbf {1}_{\{y=x\}}. \end{aligned}$$

Multiplying by $q_y$ and $\nu _y$, and summing over all $y\in \mathcal {S}$,

$$\begin{aligned} \sum _{y} \sum _{z\sim y} \nu _y q_{yz} g(y) = \sum _y \sum _{z\sim y} \nu _y q_{yz} g(z) - \frac{\gamma }{2} + \beta \nu _x. \end{aligned}$$

By reversibility $\nu _y q_{yz} = \nu _z q_{zy}$, so the term on the left and the first term on the right are identical, which gives $\frac{\gamma }{2} = \beta \nu _x$. Thus there is a minimizing g such that

$$\begin{aligned} g(y) = \frac{\beta }{q_y} \left( \mathbf {1}_{\{y=x\}}-\nu _x\right) + \sum _{z\sim y} \frac{q_{yz}}{q_y} g(z). \end{aligned}$$

(8.12)

We now show that up to the factor $\beta$ the function $y\mapsto Z_{yx}$ satisfies the same relation. Indeed, by the strong Markov property at the time $J_1$ of the first jump of Y, which under $P_y$ is an exponential random variable with mean $\frac{1}{q_y}$,

$$\begin{aligned} \begin{aligned} Z_{yx}&= \int _0^{\infty }\left( \int _0^{J_1} \left( \mathbf {1}_{\{y=x\}}-\nu _x\right) {\mathrm d}t +\sum _{z\sim y} \frac{q_{yz}}{q_y} \int _0^{\infty } \left( P_z[Y_t=x] -\nu _x\right) {\mathrm d}t\right) P_y({\mathrm d}J_1) \\&= \frac{1}{q_y}\left( \mathbf {1}_{\{y=x\}}-\nu _x\right) + \sum _{z\sim y} \frac{q_{yz}}{q_y} Z_{zx}. \end{aligned} \end{aligned}$$

The function $g(y)= \frac{Z_{yx}}{Z_{xx}}$ thus satisfies the constrains of the variational problem in (8.11) and fulfills (8.12) with $\beta = 1/Z_{xx}$. It is thus the minimizer of this variational problem.

Moreover, by [2, Lemmas 2.11 and 2.12], we have $Z_{xx}=E_{\nu }[H_x] \nu _x$ and $\nu _x E_y[H_x] = Z_{xx}-Z_{yx}$. Denoting $h(y)=E_y[H_x]$ and using these equalities, we obtain

$$\begin{aligned} D(g,g) = \frac{1}{E_{\nu }[H_x]^2}D(h,h) = \frac{1}{E_\nu [H_x]}, \end{aligned}$$

where for the last equality we used $D(h,h) = E_{\nu }[H_x]$, by e.g. [1, Lemma 6]. This completes the proof. $\square$

With this lemma the proof of Proposition 8.5 follows the lines of [16].

Proof of Proposition 8.5

To prove the inequality in (8.10), it is sufficient to modify the function $g^{\star }_{x,B}$ so that it becomes admissible for the variational problem in Lemma 8.6. Write $g^{\star }$ for $g^{\star }_{x,B}$ and define $\tilde{g}$ on $\mathcal {S}$ as

$$\begin{aligned} \tilde{g} (z)=\frac{g^{\star }(z)-\sum _{y\in \mathcal {S}}\nu _y g^{\star }(y)}{1-\sum _{y\in \mathcal {S}}\nu _y g^{\star }(y)}. \end{aligned}$$

Then $\tilde{g}$ equals 1 on x and $\sum _{z\in \mathcal {S}}\nu _z \tilde{g}(z)=0$. Hence, by Lemma 8.6,

$$\begin{aligned} \frac{1}{E_{\nu }[H_x]} \le D(\tilde{g},\tilde{g}) = D(g^{\star },g^{\star }) \left( 1-\sum _{y\in \mathcal {S}}\nu _y g^{\star }(y) \right) ^{-2}. \end{aligned}$$

But $g^{\star }$ is non-negative, bounded by 1 and non-zero only on $B^c$, therefore $\sum _{y\in \mathcal {S}}\nu _y g^{\star }(y) \le \nu (B^c)$, the first part of Proposition 8.5 follows.

To prove the equality in (8.10), we show that

$$\begin{aligned} D(g^{\star }_{x,B},g^{\star }_{x,B}) = P_x[H^+_x > H_B]c_x. \end{aligned}$$

(8.13)

Indeed, let again $g^{\star }=g^{\star }_{x,B}$. If $g^{\star }$ is harmonic in z, the second sum in the Dirichlet form (8.9) is

$$\begin{aligned} \sum _{y\sim z} c_{zy}(g^{\star }(z)-g^{\star }(y))^2 = \sum _{y\sim z} c_{zy} (g^{\star }(y)^2 - g^{\star }(z)^2). \end{aligned}$$

This shows that the contribution to the Dirichlet form of every edge that connects two vertices in which $g^{\star }$ is harmonic or zero vanishes. Therefore $D(g^{\star },g^{\star })$ reduces to

$$\begin{aligned} D(g^{\star },g^{\star })&=\frac{1}{2}\left( \sum _{y\sim x} c_{xy}(1-g^{\star }(y))^2 + \sum _{y\sim x} c_{xy}(1-g^{\star }(y)^2)\right) \\&= \sum _{y\sim x} c_{xy}(1-g^{\star }(y))\\&= c_x \sum _{y\sim x} p_{xy} P_y[H_x>H_B]\\&= c_x P_x[H_x^+ > H_B]. \end{aligned}$$

This proves (8.13) and thus the proposition. $\square$