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Formalizations.Probability.Kemeny

Kemeny's constant #

For a finite Markov chain with transition matrix P, a stationary distribution π and finite expected hitting times m s t (m t t = 0, m s t = 1 + Σ_k P s k · m k t for s ≠ t), the expected time Σ_t π t · m s t to reach a π-random target does not depend on s (Theorem of the Day). Proof: Kac's formula π_t (1 + Σ_k P_tk m_kt) = 1 from stationarity turns f(s) = Σ_t π_t m_st into a harmonic function f = P f, and the hitting-time equations force a harmonic function to be constant (minimum principle on the set where f is maximal).

Main results #

References #

theorem Kemeny.kac {ι : Type u_1} [Fintype ι] [DecidableEq ι] (P : Matrix ι ι ℝ) (π : ι → ℝ) (hπ : ∀ (k : ι), ∑ s : ι, π s * P s k = π k) (hπ1 : ∑ s : ι, π s = 1) (m : ι → ι → ℝ) (hmt : ∀ (t : ι), m t t = 0) (hm : ∀ (s t : ι), s ≠ t → m s t = 1 + ∑ k : ι, P s k * m k t) (t : ι) :
π t * (1 + ∑ k : ι, P t k * m k t) = 1
theorem Kemeny.harmonic {ι : Type u_1} [Fintype ι] [DecidableEq ι] (P : Matrix ι ι ℝ) (π : ι → ℝ) (hπ : ∀ (k : ι), ∑ s : ι, π s * P s k = π k) (hπ1 : ∑ s : ι, π s = 1) (m : ι → ι → ℝ) (hmt : ∀ (t : ι), m t t = 0) (hm : ∀ (s t : ι), s ≠ t → m s t = 1 + ∑ k : ι, P s k * m k t) (s : ι) :
∑ t : ι, π t * m s t = ∑ k : ι, P s k * ∑ t : ι, π t * m k t
theorem Kemeny.exists_const {ι : Type u_1} [Fintype ι] [DecidableEq ι] (P : Matrix ι ι ℝ) (hP0 : ∀ (s k : ι), 0 ≤ P s k) (hP1 : ∀ (s : ι), ∑ k : ι, P s k = 1) (π : ι → ℝ) (hπ : ∀ (k : ι), ∑ s : ι, π s * P s k = π k) (hπ1 : ∑ s : ι, π s = 1) (m : ι → ι → ℝ) (hmt : ∀ (t : ι), m t t = 0) (hm : ∀ (s t : ι), s ≠ t → m s t = 1 + ∑ k : ι, P s k * m k t) :
∃ (K : ℝ), ∀ (s : ι), ∑ t : ι, π t * m s t = K