Bisimulation for Stochastic Systems


Nathanaël Fijalkow


Research Fellow @ Alan Turing Institute
Affiliated to Warwick

Short Talks at the Turing, Thursday 2nd March, 2017

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Probabilistic Bisimulation

(due to Larsen, Skou 91)

Bisimulation for Non-Deterministic Systems


Bisimulation game: Duplicator claims that $p \equiv q$ We say that $p$ and $q$ are bisimilar if Duplicator can play forever


Coinductive definition: If $p \equiv q$, then for all $a$ in $A$, for all $p' \in \Delta(p,a)$, there exists $q' \in \Delta(q,a)$ such that $p' \equiv q'$

Bisimulation for Labelled Markov Chains


Labelled Markov Chains: $\Delta : Q \times A \to \mathcal{D}(Q)$

Coinductive definition: If $p \equiv q$, then for all $a$ in $A$, for all union of equivalence classes $C$ of $\equiv$, $$\Delta(p,a,C) = \Delta(q,a,C)$$


Equivalent game: Duplicator claims that $p \equiv q$
  • Spoiler: pick $a$ in $A$ and $C \subseteq Q$ such that $\Delta(p,a,C) \neq \Delta(q,a,C)$
  • Duplicator: $p'$ in $C$, $q'$ not in $C$ and claim that $p' \equiv q'$
Bisimulation game: Duplicator claims that $p \equiv q$ Duplicator claims they are bisimilar Duplicator claims they are bisimilar Spoiler picks a set of green states Duplicator claims they are bisimilar Spoiler picks a set of green states Duplicator picks a green state and a non-green state Duplicator claims they are bisimilar Spoiler picks a set of green states Duplicator picks a green state and a non-green state Spoiler's move was stupid! Duplicator claims they are bisimilar Spoiler picks a set of green states Duplicator claims they are bisimilar Spoiler picks a set of green states Duplicator picks a green state and a non-green state

25 years of research on Probabilistic Bisimulation


How to prove that two systems are not bisimilar?


Find a formula satisfying one but not the other!

Logical Characterisation

\[ \phi\ =\ \top\ \mid\ \phi \wedge \phi\ \mid\ \langle a \rangle_{> p} \phi\ \] Two Theorems:
- Desharnais, Edalat, Panangaden 1998: the state space is a Polish space, the transition function is measurable and the set of labels is countable
- F., Klin, Panangaden 2017: the state space is a Polish space and the transition function is continuous

Two states are bisimiliar if, and only if, they satisfy the same formulas of this logic