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Relative-entropy decrease and convergence for a finite detailed-balance chain

Physical scope

A finite, time-homogeneous continuous-time rate model satisfying detailed balance; convergence is asserted only when its positive-rate graph is connected, and no general continuum or non-Markovian relaxation theorem is claimed.

Statement

Let $S$ have $N \geq 2$ states, positive detailed-balance law $π$ , and connected undirected rate graph as in Finite detailed-balance rate and transition models. For every strictly positive initial probability vector $p (0)$ , the master equation has a unique strictly positive probability solution $p (t)$ for all $t \geq 0$ . The relative entropy $D (p (t) ∥ π) = \sum_{i} p_{i} (t) \log (p_{i} (t) / π_{i})$ is nonincreasing, and $p_{i} (t) \to π_{i}$ for every state. No monotonicity of Shannon entropy $- \sum_{i} p_{i} \log p_{i}$ is asserted.

Facts & Assumptions

Given: A finite state set of size $N \geq 2$ , finite rates, a positive normalized detailed-balance vector $π$ , and an initially strictly positive normalized vector $p (0)$ .

[F1]

Finite detailed-balance rate and transition models: $q_{i j} \geq 0$ for $i \neq j$ , $q_{i i} = - \sum_{j \neq i} q_{i j}$ , and ${\dot{p}}_{i} = \sum_{j \neq i} (p_{j} q_{j i} - p_{i} q_{i j})$ .

[F2]

Finite detailed-balance rate and transition models: $a_{i j} : = π_{i} q_{i j} = π_{j} q_{j i}$ is positive on graph edges, and the graph is connected.

[F3]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\log (x y) = \log x + \log y$ , $\log (1 / x) = - \log x$ , and $\log x \leq 0$ for $0 < x \leq 1$ .

[F4]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : for $x > 0$ , $(\log x)^{'} = 1 / x$ , and the ordinary chain and finite-product rules differentiate the entropy sum.

[F5]

The exponential function is smooth and $(\exp)^{'} = \exp$ and The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : differentiate $e^{c t}$ and infer monotonicity on an interval from a nonpositive derivative.

[F6]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the finite Cauchy--Schwarz inequality bounds a sum of at most $N - 1$ edge increments by $(N - 1)$ times their squared sum.

Proof

Given: [F1]--[F6] and $p_{i} (0) > 0$ .

1.1F1F5givenalgebra

Put $c = 1 + \max_{i} \sum_{j \neq i} q_{i j}$ and $P = I + Q / c$ , where $Q = (q_{i j})$ . Then $P$ has nonnegative entries and row sums one. The series $p (t) = e^{- c t} \sum_{n = 0}^{\infty} (c t)^{n} p (0) P^{n} / n!$ and its differentiated series converge uniformly on bounded time intervals; its coefficients are nonnegative and sum to one, so it is a probability vector, starts at $p (0)$ , and differentiates to $\dot{p} = p Q$ . Its $n = 0$ term gives $p_{i} (t) \geq e^{- c t} p_{i} (0) > 0$ . For uniqueness, the difference $z$ of two solutions satisfies $z^{'} = z Q$ ; with $C = \max_{i} \sum_{j} ∣ q_{i j} ∣$ and $U (t) = \int_{0}^{t} ∥ z (s) ∥_{1} d s$ , one has $U^{'} (t) = ∥ z (t) ∥_{1} \leq C U (t)$ . Thus $(e^{- C t} U (t))^{'} \leq 0$ and $U (0) = 0 \leq U (t)$ force $U = 0$ , hence $z = 0$ .

2.1F1F2F3F4F5step 1.1

With $r_{i} = p_{i} / π_{i}$ , detailed balance gives ${\dot{p}}_{i} = \sum_{j \neq i} a_{i j} (r_{j} - r_{i})$ . By [F4], differentiating the finite entropy sum gives $D^{'} = \sum_{i} {\dot{p}}_{i} (\log r_{i} + 1) = \sum_{i} {\dot{p}}_{i} \log r_{i}$ , since $\sum_{i} {\dot{p}}_{i} = 0$ . Pairing the $i, j$ terms gives $D^{'} = - \sum_{{i, j} \in E} a_{i j} (r_{i} - r_{j}) (\log r_{i} - \log r_{j}) \leq 0$ : if $r_{i} \geq r_{j}$ , then $\log r_{i} - \log r_{j} = - \log (r_{j} / r_{i}) \geq 0$ by [F3], and the reversed order gives the same product sign. The mean value theorem [F5] on each compact time interval makes $D$ nonincreasing.

2.2F1F2step 1.1

Set $u_{i} = r_{i} - 1$ , so $\sum_{i} π_{i} u_{i} = 0$ , and $V (t) = \sum_{i} π_{i} u_{i}^{2}$ . Since ${\dot{u}}_{i} = π_{i}^{- 1} \sum_{j \neq i} a_{i j} (u_{j} - u_{i})$ , differentiating and pairing the two orientations of each edge gives $V^{'} = 2 \sum_{i} u_{i} \sum_{j \neq i} a_{i j} (u_{j} - u_{i}) = - 2 \sum_{{i, j} \in E} a_{i j} (u_{i} - u_{j})^{2}$ .

3.1F2F6step 2.2algebra

Let $a_{*} = \min_{{i, j} \in E} a_{i j} > 0$ . Fix a root state and choose one simple path from it to each state; this is only a finite family of choices, obtainable successively, and each path has length at most $N - 1$ . Telescoping and [F6] give $(u_{i} - u_{r})^{2} \leq (N - 1) \sum_{{j, k} \in E} (u_{j} - u_{k})^{2}$ . The zero-mean identity gives $u_{r} = - \sum_{i} π_{i} (u_{i} - u_{r})$ ; expanding $\sum_{i} π_{i} [(u_{i} - u_{r}) - \sum_{j} π_{j} (u_{j} - u_{r})]^{2} \geq 0$ yields $u_{r}^{2} \leq \sum_{i} π_{i} (u_{i} - u_{r})^{2}$ . Since $(x + y)^{2} \leq 2 x^{2} + 2 y^{2}$ and $\sum_{i} π_{i} = 1$ , it follows that $V \leq 4 (N - 1) \sum_{E} (u_{i} - u_{j})^{2} \leq 4 (N - 1) a_{*}^{- 1} \sum_{E} a_{i j} (u_{i} - u_{j})^{2}$ .

4.1F5step 2.2step 3.1∎

Thus $V^{'} \leq - 2 λ V$ with $λ = a_{*} / [4 (N - 1)] > 0$ . By [F5], the derivative of $e^{2 λ t} V (t)$ is nonpositive and that function is nonincreasing, so $V (t) \leq e^{- 2 λ t} V (0)$ . Since $π_{i} u_{i} (t)^{2} \leq V (t)$ and $p_{i} - π_{i} = π_{i} u_{i}$ , each $p_{i} (t) \to π_{i}$ .

Relative-entropy decrease and convergence for a finite detailed-balance chain

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources