Randomisation in Automata on Infinite Trees

We study ﬁnite automata running over inﬁnite binary trees. A run of such an automaton over an input tree is a tree labeled by control states of the automaton: the labelling is built in a top-down fashion and should be consistent with the transitions of the automaton. A branch in a run is accepting if the ω -word obtained by reading the states along the branch satisﬁes some acceptance condition (typically an ω -regular condition such as a Büchi or a parity condition). Finally, a tree is accepted by the automaton if there exists a run over this tree in which every branch is accepting. In this paper, we consider two relaxations of this deﬁnition introducing a qualitative aspect. First, we relax the notion of accepting run by allowing a negligible set (in the sense of measure theory) of non-accepting branches. In this qualitative setting, a tree is accepted by the automaton if there exists a run over this tree in which almost every branch is accepting. This leads to a new class of tree languages, qualitative tree languages . This class enjoys many good properties: closure under union and intersection (but not under complement), emptiness is decidable in polynomial time. A dual class, positive tree languages , is deﬁned by requiring that an accepting run contains a non-negligeable set of branches. The second relaxation, is to replace the existential quantiﬁcation (a tree is accepted if there exists some accepting run over the input tree) by a probabilistic quantiﬁcation (a tree is accepted if almost every run over the input tree is accepting). For the run, we may use either classical acceptance or qualitative acceptance. In particular, for the latter, we exhibit a tight connection with partial observation Markov decision processes. Moreover, if we additionally restrict to the Büchi condition, we show that it leads to a class of probabilistic automata on inﬁnite trees enjoying a decidable emptiness problem. To our knowledge, this is the ﬁrst positive result for a class of probabilistic automaton over inﬁnite trees.


INTRODUCTION
Roughly speaking a finite automaton on infinite trees is a finite memory machine that takes as input an infinite node-labelled binary tree and processes it in a top-down fashion as follows.It starts at the root of the tree in its initial state, and picks (possibly nondeterministically) two successor states, one per son, according to the current control state, the letter at the current node and the transition relation.Then the computation proceeds in parallel from both sons, and so on.Hence a run of the automaton on an input tree is a labelling of this tree by control states of the automaton, that should satisfy the local constrains imposed by the transition relation.A branch in a run is accepting if the ω-word obtained by reading the states along the branch satisfies some acceptance condition (typically an ω-regular condition such as a Büchi or a parity condition).Finally, a tree is accepted by the automaton if there exists a run over this tree in which every branch is accepting.An ω-regular tree language is a tree language accepted by some tree automaton equipped with a parity condition.
Finite automata on infinite trees were originally introduced by Rabin in [Rabin 1969] to prove the decidability of the monadic second order logic (MSOL) over the full binary tree.Indeed, Rabin proved that for any MSOL formula, one can construct a tree automaton that accepts a non empty language if and only if the original formula holds at the root of the full binary tree.These automata were also successfully used by Rabin in [Rabin 1972] to solve Church's synthesis problem [Church 1962], which ask for constructing a circuit based on a formal specification (typically a expressed in monadic second order logic) describing the desired input/output behaviour.His approach was to represent the set of all possible behaviour of a circuit by an infinite tree (directions are used to code the input while node labels along a branch code the output) and to reduce the synthesis problem to emptiness of a tree automaton accepting all those trees coding circuits satisfying the specification.
Since then, automata on infinite trees and their variants have been intensively studied and found many applications, in particular in logic.Connections between automata on infinite trees and logic are discussed in the excellent surveys [Thomas 1997;Vardi and Wilke 2007].
A fundamental result of Rabin is that ω-regular tree languages form a Boolean algebra [Rabin 1969].The hard part in this proof is the complementation, and since the publication of this result in 1969, it has been a challenging problem to simplify the proof.A much simpler one was obtained in [Gurevich and Harrington 1982] making use of two-player games on graphs for checking membership of a tree in the language accepted by the automaton: the first player (called Automaton) builds a run on the input tree while the second player (called Pathfinder ) tries to exhibit a rejecting branch in the run.Beyond this result, the tight connection between automata and games is one of the main tools in automata theory (see e.g.[Grädel et al. 2002;Löding 2011]).
In this article, we consider variations of the classical model of tree automata over infinite trees.These variations involve, in two different ways, probabilities and preserve the fruitful connection with game theory.
In the first part of this paper, we consider a relaxed notion of an accepting run.While the usual definition requires for a run to be accepting that all branches in it satisfy the acceptance condition, we allow a negligible set of non-accepting branches.In this qualitative setting, a tree is accepted by the automaton if there exists a run over this tree in which almost every (in the usual sense from measure theory) branch is accepting.With the parity condition, this leads to a new class of tree languages that we call qualitative tree languages.We show that this class enjoys many desirable properties: closure under union and intersection (but not under complement), emptiness is decidable in polynomial time (contrasting with the fact that no polynomial algorithm is known for the emptiness test of standard parity tree automata).We also prove that there exists a strong connection between automata accepting qualitative tree languages and Markov decision processes, which play here a similar role as two-player games for usual tree automata.We also discuss the positive setting, where a run is accepting if the set of accepting branches in it has a strictly positive measure.
The idea of allowing a certain amount of rejecting branches in a run was already considered in [Beauquier et al. 1991;Beauquier and Niwiński 1995], where it was required that the number of accepting branches in a run belong to a specified set of cardinals Γ.In particular, they proved that if Γ consists of all cardinals greater than some γ, then one obtains a regular tree language.Qualitative tree languages as defined in this article are not captured by the work of [Beauquier and Niwiński 1995].Indeed, our classes are incomparable with regular tree languages.
In the second part of this paper, we investigate probabilistic automata on infinite trees.Acceptance by an automaton is based on existential quantification: an input is accepted if there exists an accepting run over it.Probabilistic automata are an alternative way to define acceptance.On finite words they have been introduced by Rabin in [Rabin 1963].Compared with the standard setting, the non-deterministic guesses are replaced by random choices (according to some probabilistic distribution depending on the control state and the input letter).Hence, the set of transitions is replaced by a probability distribution over the set of all transitions which induces a probability measure on the set of runs of the automaton and acceptation is defined using a threshold 0 ă λ ă 1 on the probability of a run to be accepting.In contrast to the non-deterministic setting, the emptiness problem for probabilistic automata on finite words is undecidable [Paz 1971].
The probabilistic model was recently extended to infinite words in [Baier and Größer 2005] by Baier and Größer1 and studied in more details in [Baier et al. 2008;Chadha et al. 2009;Chatterjee et al. 2009].In addition to the threshold criterion, two additional semantics were considered : almost-sure and probable which respectively corresponds to a probability 1 or ą 0 for a run to be accepting 2 .Surprisingly the class of languages defined by Büchi automata with the probable semantics is closed under complement, which implies that it coincides with the class of languages defined by co-Büchi automata with the almost-sure semantics 3 .The emptiness problem for Büchi automata with the almost-sure semantics as well as for co-Büchi automata with the probable semantics are decidable.However the emptiness problem for Büchi automata with the probable semantics as well as for co-Büchi automata with the almost-sure semantics are undecidable.Of course, emptiness is undecidable when considering a threshold semantics.See [Baier et al. 2012] for a very rich overview of this topic.
In this paper, we consider probabilistic automata on infinite trees.We first focus on the almost-sure semantics, i.e. a tree is accepted if almost every run over it is accepting, and later discuss the probable semantics, i.e. a tree is accepted if the set of accepting runs on it has a (strictly) positive measure.Of course each semantics can be used in combination with the acceptance criteria on runs: the classical one (all branches are accepting), the qualitative one (almost all branches are accepting) and the positive one (there is a non negligible set of accepting branches).For all these combinations, we establish that the definition makes sense (i.e.we prove measurability of the set of accepting runs).
For the qualitative criterion on runs combined with the almost-sure semantics, as well as for the probable criterion on runs combined with the positive semantics, we prove that there exists a strong connection with partial observation Markov decision processes.This condition is independent of the acceptance condition on branches (Büchi, co-Büchi, parity. . .).In particular, for the Büchi (resp.co-Büchi) acceptance condition on branches, probabilistic automata on infinite trees with the qualitative criterion on runs combined with the almost-sure semantics (resp.with the positive criterion on runs combined with the probable semantics) enjoy a decidable emptiness problem.To our knowledge, this is the first positive result for a class of probabilistic automata over infinite trees.On the negative side, we derive from undecidability results on probabilistic automata on infinite words similar results for probabilistic automata on infinite trees.Note that whilst being immediate for the classical criterion on runs such a reduction is not as simple for the qualitative and positive criteria.
The paper is organised as follows.In Section 2 we introduce general notations and definitions.In Section 3 we define the class of qualitative tree languages, present their properties (closure properties, decidability properties) and their tight connections with Markov decision processes, and establish that they are incomparable with regular tree languages.We also discuss possible variants of the model (leading the notion of positive tree languages).In Section 4 we focus on the probabilistic model.We first justify the validity of the definition and give some examples.Then we provide connections with partial observation Markov decision processes from which we derive positive results for the emptiness problem on a subclass.Finally we discuss variants of the probabilistic model.Section 5 summarises the contributions of the paper.
the complement of L, namely L.
ACM Transactions on Computational Logic, Vol.V, No. N, Month 20YY.

Words and Trees
An alphabet A is a finite set of letters.In the sequel A ˚denotes the set of finite words over A, and A ω the set of infinite words over A. The empty word is written ε; the length of a word u is denoted by |u|.Let u be a finite word and v be a (possibly infinite) word.Then u ¨v denotes the concatenation of u and v; the word u is a prefix of v, denoted u Ď v, if there exists a word w such that v " u ¨w.We denote by u Ă v the fact that u is a strict prefix of v (i.e.u Ď v and u " v).
For some word u and some integer k ě 0, we denote by u k the word obtained by concatenating k copies of u (with the convention that u 0 " ε).
In this paper we consider full binary node-labelled trees.An A-labelled tree t is a mapping from t0, 1u ˚to A. In this context, an element u P t0, 1u ˚is called a node, and the node u ¨0 (resp.u ¨1) is the left son (resp.right son) of u.The node ε is called the root.We shall refer to |u| as the depth of u.The letter tpuq is called the label of u in t.
A branch is an infinite word π P t0, 1u ω .We write Br " t0, 1u ω for the set of all branches.A node u belongs to a branch π if u is a prefix of π.For an A-labelled tree t and a branch π " π 0 π 1 ¨¨¨we define the label of π as the ω-word tpπq " tpεqtpπ 0 qtpπ 0 π 1 qtpπ 0 π 1 π 2 q ¨¨¨.The cone going through a node u is the set Given a tree t and a node u, the subtree of t rooted at u denoted trus is the tree defined by truspvq " tpu ¨vq.A tree t is said to be regular if it contains only finitely many different subtrees, i.e. the set ttrus | u P t0, 1u ˚u is finite.
We assume that the reader is familiar with basic notions of measure theory and from probability theory and we use [Bauer 2001;1996] as references for all known results related to this field.Let F Br be the σ-algebra generated by the set of cones (i.e. the smallest set of subsets of t0, 1u ω containing the cones and closed under countable union and complementation).Let µ be the unique probability measure on F Br such that for all u P t0, 1u ˚, µpConepuqq " 2 ´|u| .The existence and uniqueness of µ are guaranteed by Carathéodory's extension theorem [Bauer 2001, Theorem 5.6, p. 24].For all 0 ă p ă 1, a probability measure µ p is similarly defined by taking µ p pConepuqq " p |u|0 p1 ´pq |u|1 where |u| 0 and |u| 1 respectively designate the number of occurrences of 0 and 1 in u.In particular, the measure µ corresponds to µ 1{2 .

Tree Automata and Regular Tree Languages
A tree automaton A is a tuple xA, Q, q ini , ∆, Accy where A is the input alphabet, Q is the finite set of states, q ini P Q is the initial state, ∆ Ď Q ˆA ˆpQ ˆQq is the transition relation and Acc Ď Q ω is the acceptance condition.In the following, we use the notation q a Ñ pq 0 , q 1 q as a shorthand for pq, a, pq 0 , q 1 qq P ∆.An automaton is deterministic if q a Ñ pq 0 , q 1 q and q a Ñ pq 1 0 , q 1 1 q implies q 0 " q 1 0 and q 1 1 " q 1 1 .An automaton is complete if, for all q P Q and a P A there is at least one pair pq 0 , q 1 q P Q 2 such that q a Ñ pq 0 , q 1 q.Given an A-labelled tree t, a run of A over t is a Q-labelled tree ρ such that the root is labelled by the initial state, i.e. ρpεq " q ini ; for all nodes u, pρpuq, tpuq, ρpu 0q, ρpu ¨1qq P ∆.
A branch π P t0, 1u ω is accepting in the run ρ if ρpπq P Acc.A run ρ is accepting if all its branches are accepting.Finally, a tree t is accepted if there exists an accepting run of A over t.The set of all trees accepted by A is denoted LpAq.
We consider the following classical acceptance conditions: -A reachability condition is given by a subset F Ď Q of final states by letting ReachpF q " Q ˚F Q ω , i.e. a branch is accepting if it contains a final state.
-A Büchi condition is given by a subset F Ď Q of final states by letting BuchipF q " pQ ˚F q ω , i.e. a branch is accepting if it contains infinitely many final states.-A co-Büchi condition is given by a subset F Ď Q of forbidden states by letting coBuchipF q " Q ˚pQzF q ω , i.e. a branch is accepting it contains finitely many forbidden states.
-A parity condition is given by a colouring mapping Col : Q Ñ N by letting P arity " tq 0 q 1 q 2 ¨¨¨| lim infpColpq i qq i is evenu, i.e. a branch is accepting if the smallest colour appearing infinitely often is even.
All these conditions are examples of ω-regular acceptance conditions, i.e.Acc is regular set of ω-words [Perrin and Pin 2004].
Remark 1.The parity condition is expressive enough to capture the general case of an arbitrary ω-regular condition.Indeed it is well known that Acc is accepted by a deterministic parity word automaton.By taking the synchronised product of this automaton with the tree automaton, we obtain a parity tree automaton accepting the same language (see e.g.[Perrin and Pin 2004]).
When it is clear from the context, we may replace, in the description of A, Acc by F (for a reachability, Büchi or co-Büchi condition) or Col (for a parity condition), and we shall refer to the automaton as a reachability (resp.Büchi, co-Büchi, parity) tree automaton.A set L of trees is a regular language if there exists a parity tree automaton A such that L " LpAq.The class of regular tree languages is robust, as illustrated by the following statement.
Theorem 2. [Rabin 1969], [Emerson et al. 1993] The class of regular tree languages is a Boolean algebra.
A regular tree language is non-empty if and only if it contains a regular tree.Testing the emptiness of a regular tree language (defined by a given parity automaton) is in NP X coNP.

Perfect Information Setting.
A probability distribution over a countable set X is a mapping d : X Ñ r0, 1s such that ř xPX dpxq " 1.In the sequel we denote by DpXq the set of probability distributions over X.In this paper, all probabilities will be rational numbers, which will be described in binary when dealing with encoding.
An arena is a tuple G " xS, s ini , Σ, ζy where S is a countable set of states, s ini is an initial state, Σ is a finite set of actions and ζ : S ˆΣ Ñ DpSq is the transition (total) function.
A play in such an arena proceeds as follows.It starts in state s ini and Éloïse picks an action σ, and a successor state is chosen according to the probability distribution ζps ini , σq.Then Éloïse chooses a new action and the state is updated and so on forever.Hence a play is an infinite sequence s 0 s 1 s 2 ¨¨¨P S ω such that s 0 " s ini and for every i ě 0, there exists a σ P Σ with ζps i , σqps i`1 q ą 0. In the sequel we refer to a prefix of a play as a partial play and we denote by Plays the set of all plays.
A (pure) strategy4 for Éloïse is a function ϕ : S ˚Ñ Σ assigning to every partial play an action.Of special interest are those strategies that do not require memory: a strategy ϕ is memoryless if ϕpλ ¨sq " ϕpλ 1 ¨sq for all partial play λ, λ 1 and all states s (i.e.ϕ only depends on the current state).A play λ " s 0 s 1 s 2 ¨¨¨is consistent with a strategy ϕ if ζps i , ϕpv 0 ¨¨¨v i qqps i`1 q ą 0, for all i ě 0. Now, for any partial play λ, the cylinder for λ is the set Cylpλq " λS ω XPlays.Let F P be the σ-algebra generated by the set of cylinders.Then, pPlays, F P q is a measurable space.
A strategy ϕ induces a probability space over pPlays, F P q as follows: one defines a measure µ ϕ on cylinders and then uniquely extends it to a probability measure on F P using the Carathéodory's unique extension theorem.For this, we first define inductively µ ϕ on cylinders: -as all plays start from s ini , we let µ ϕ pCylps ini qq " 1; -for any partial play λ ending in some state s, we let µ ϕ pCylpλ ¨s1 qq " µ ϕ pCylpλqq ζps, ϕpλqqps 1 q.
We also denote by µ ϕ the unique extension of µ ϕ to a probability measure on F. Then pPlays, F P , µ ϕ q is a probability space.
An objective is a measurable set O Ď Plays: a play is winning if it belongs to O. A Markov decision process (MDP, aka one-and-half-player game) is a pair G " pG, Oq where G is an arena and O is an objective.In the sequel we should focus on ω-regular objectives (which are easily seen to be measurable), whose definitions are the same as for the acceptance condition on tree automata (the only differences is that we may have an infinite set of states and that we restrict ourselves to the set Plays).
When the set of actions Σ is reduced to one element, the MDP pG, Oq is called a Markov chain and we omit the unique action from all the definitions.The set Plays is called the set of traces of the Markov chain and is denoted Traces.We write µ G the probability measure associated with the unique strategy.We say that the Markov chain almost-surely fulfils its objective if µ G pOq " 1.
MDPs over finite arenas enjoy many good properties.
Theorem 3. [Courcoubetis and Yannakakis 1990;Chatterjee et al. 2004] Let G be an MDP over a finite arena with a parity objective.Then, one can decide in polynomial time whether Éloïse almost-surely ( resp.positively) wins.Moreover, Éloïse always has an optimal memoryless strategy.
2.3.2Imperfect Information Setting.Now we consider the case where Éloïse has imperfect information about the current state.For this, we consider an equivalence relation " over S. We let rss " be the equivalence class of s for " and S{ " be the set of equivalence classes of " over S.
The intuitive meaning of " is that two states s 1 " s 2 cannot be distinguished by Éloïse.We easily extend " to partial plays: s 0 s 1 ¨¨¨s n " s 1 0 s 1 1 ¨¨¨s 1 n if and only if s i " s 1 i for all i " 0, ¨¨¨, n.As two equivalent plays λ 1 " λ 2 cannot be distinguished by Éloïse she should therefore behave the same in both of them.
Hence, we should only consider so-called observation-based strategies.An observation-based (pure) strategy is a function ϕ : pS{ " q ˚Ñ Σ, i.e., to choose her next action, Éloïse considers the sequence of observations she got so far5 .In particular, an observation-based strategy ϕ is such that ϕpλq " ϕpλ 1 q whenever λ " λ 1 .In this context, a memoryless strategy is a function from S{ " Ñ DpΣq, i.e. it only depend on the current equivalence class.
A partial observation Markov decision process (POMDP, aka one-andhalf-player imperfect information game) is a triple pG, ", Oq where G is an arena, " is an equivalence relation over states and O is an objective.We say that Éloïse almost-surely wins (resp.positively wins) G if she has an almost-surely (resp.positively) winning observation-based strategy.Finally the value of G is defined as ValpGq " sup ϕ µ ϕ pOq where ϕ ranges over observation-based strategies; optimality is defined as previously.
The following decidability results are known for POMDP: Theorem 4. [Baier et al. 2008] In a POMDP with a Büchi ( resp.co-Büchi) objective, deciding whether Éloïse almost-surely ( resp.positively) wins is ExpTimecomplete.Moreover if Éloïse has an almost-surely ( resp.positively) winning strategy, she has an almost-surely ( resp.positively) winning strategy with finite memory.
Remark 5.The results in Theorem 3 and 4 do not depend on the encoding of probability distributions, as the only relevant information is which probabilities are non zero.

Definitions
In the classical definition, a run of a tree automaton A is accepting if all its branches satisfy the acceptance condition.In this article, we introduce a more relaxed notion of acceptation: a run is qualitatively accepting if almost every (in the sense of the measure µ) branch in it is accepting.More formally, consider a tree automaton A with an ω-regular acceptance condition Acc.A run ρ of A is qualitatively accepting if the set AccBrpρq " tπ P t0, 1u ω | ρpπq P Accu has measure 1, i.e. µpAccBrpρqq " 1.Note that, thanks to Proposition 6 below, the set AccBrpρq is indeed measurable.A tree t is qualitatively accepted by A if there exists a qualitatively accepting run of A over t and the set of all trees qualitatively accepted by A is denoted L Qual pAq.Finally, a qualitative tree language is a set L of trees such that there exists a parity automaton A such that L Qual pAq " L. Proposition 6.Let A be a tree automaton equipped with an ω-regular acceptance condition, and let ρ be a run of A. The set AccBrpρq is measurable.
Proof.Below is a simple ad-hoc proof of Proposition 6.Nevertheless, one should notice that Corollary 37 (see Section 4.1.2)directly implies Proposition 6.
The reachability case (with final states F ) is obvious as the set AccBrpρq is the (countable) union of all cones Cpuq with ρpuq P F .
We focus now on the co-Büchi case (with forbidden states F ). Recall that a branch is accepting if and only if there is a node after which there are only states in QzF .
We claim that Therefore the set AccBrpρq is measurable.The parity conditions follows from the previous case as one can express them as a Boolean combination of co-Büchi conditions.The general case of an arbitrary ωregular condition is obtained as follows.First, one considers a deterministic parity word automaton recognising Acc, and then takes the synchronised product of this automaton with the tree automaton.This leads to a parity tree automaton accepting the same language and whose runs are in bijection with the ones of the original automaton.Moreover as this bijection preserves the set of accepting branches one can conclude from the parity case.
Example 7. Let L a be the language of ta, bu-labelled trees whose set of branches containing at least one a has measure 1.This language is recognised by the following reachability deterministic automaton A " xta, bu, tq ini , q f u, q ini , ∆, tq f uy where: If one considers A as a Büchi automaton, the accepted language consists of those trees whose set of branches containing infinitely many a has measure 1.
Example 8. Let L 1 be the language of trees t such that in almost every branch, there is a node u labelled by a such that the subtree trus has only a on its leftmost branch.This language is recognised by the non-deterministic reachability automaton A " xA, Q, q w , ∆, tq acc uy with A " ta, bu, Q " tq w , q l , q acc , q rej u, and ∆ contains the following transitions: q w Ñ pq w , q w q, q w a Ñ pq l , q acc q, q l a Ñ pq l , q acc q, q l b Ñ pq rej , q rej q, q acc Ñ pq acc , q acc q, q rej Ñ pq rej , q rej q (here ˚is a shorthand for an arbitrary letter).Intuitively, the automaton can wait in state q w as long as it wants.
Using the second transition, the automaton can guess that the node u (labeled by a) has a leftmost branch containing only a.This assumption is checked by sending on the leftmost branch the state q l and the accepting state q acc on all other branches.As long as the nodes are labelled by a state q l is propagated to the left son.If all nodes on the leftmost branch starting at u are labelled by a, this branch will be rejecting, but this does not affect the measure as there are only countably many such branches).If a node v labelled by b is encountered in state q l the non-accepting state q rej is propagated on all branches.This last scenario cannot occur in an accepting run as these cones of rejecting branches have a strictly positive measure.Hence the automaton is penalised for wrong guesses.
For the same reasons as for regular tree languages (cf.Remark 1), the parity condition is expressive enough to capture any ω-regular conditions: for any automaton A with an ω-regular acceptance condition, there exists a parity automaton B such that L Qual pAq " L Qual pBq.
Thanks to the following proposition, we can only focus on complete automata.
Proposition 9.For any tree automaton A with an ω-regular acceptance condition, there exists a complete tree automaton B with the same acceptance condition and such that L Qual pAq " L Qual pBq.
Proof.Let A " xA, Q, q ini , ∆, Accy be a possibly incomplete automaton.Define a complete automaton B " xA, Q 1 " Q Z tq rej u, q ini , ∆ 1 , Accy, where the set ∆ 1 is ∆ augmented with tq a Ñ pq rej , q rej q | E q a Ñ pq 0 , q 1 q P ∆u Y tq rej a Ñ pq rej , q rej q | a P Au; i.e. we add a transition to the sink state q rej whenever a transition is missing.
Note that as the acceptance condition is unchanged, a branch going through q rej is rejecting.In particular, as q rej is a sink state, a run that contains q rej as a rejecting cone, hence is rejecting (as the measure of the rejecting cone is strictly positive).Therefore, the only accepting runs of B are actually runs that only uses transitions in A, hence are runs of A. Therefore L Qual pAq " L Qual pBq.

Unsurprisingly determinism is a restriction.
Proposition 10.There is a qualitative tree language that cannot be qualitatively accepted by any deterministic automaton.
Proof.Let L a be the qualitative language (see Example 7) of ta, bu-labelled trees whose set of branches containing at least one a has measure 1.Consider now the language L 1 a " tt | tr0s P L a or tr1s P L a u.Clearly L 1 a is qualitative (it suffices to guess one subtree and check that it belongs to L a while accepting without any further consideration the other subtree).By contradiction assume that there is a deterministic automaton A that qualitatively accepts L 1 a .Indeed, consider the two trees t a,b and t b,a defined as follows: t a,b pεq " t b,a pεq " b, the left subtree of the root of t a,b (resp.t b,a ) has only a's (resp.b's) and the right subtree of the root of t a,b (resp.t b,a ) has only b's (resp.a's).Both t a,b and t b,a belongs to L 1 a .The run of A 1 on t a,b start with the same transition at the root, and therefore one can combine them and get a run for the tree t b whose nodes are all labeled by b, which leads a contradiction as t b R L 1 a .

On the Choice of Measure µ
The choice of the measure µ though natural is arbitrary.Considering the measure µ p for some 0 ă p ă 1 2 would not affect the results obtained in this article (provided that definitions of the games are modified accordingly).However note that changing the measure does change the accepted language for a given automaton.
Proposition 11.Let 0 ă p ă q ă 1 be two reals.Let A be the (deterministic and complete) automaton of Example 7.Then, there is a tree t such that µ p pAccpA, tqq " 0 and µ q pAccpA, tqq " 1, where AccpA, tq denotes the set of accepting branches for the unique run of A over t.
Proof.First recall that A accepts those trees whose set of branches containing infinitely many a has measure 1.
We define the ta, bu-labeled tree t by letting for all u P t0, 1u ˚tpuq " a if and only if µ p pConepuqq ď µ q pConepuqq.
Let ε be the maximum among p, q, 1 ´p, 1 ´q.
Following the proof of Kakutani's theorem [Kakutani 1948] as presented in [Muchnik et al. 1998] and [Bienvenu 2008], we consider the measure µ r for r " p `q 2 .
We first establish that for all w P t0, 1u Let K n be the set of words w P t0, 1u n such that µ p pConepwqq ď µ q pConepwqq and let C n be the disjoint union of all Conepwq for all w P K n .
For all w P K n we have µ r pConepwqq ě µ p pConepwqqα n{2 , and for all w R K n we have µ r pConepwqq ě µ q pConepwqqα n{2 .
As the sequence of sets pD k q kě0 is decreasing for the inclusion, we have thanks to [Bauer 2001, Theorem 3.2, p. 10] that: " µ p pAccpA, tqq " lim kÑ8 µ p pD k q " 0 µ q pAccpA, tqq " lim kÑ8 µ q pD k q " 1 A more general definition is to associate with any letter a in the alphabet a pair pp 0 a , p 1 a q P r0, 1s 2 with p 0 a `p1 a " 1 and then to define the measure of a cone in a tree t by letting µpConepu 1 ¨¨¨u n qq " p u1 tpεq p u2 tpu1q . . .p un tpu1¨¨¨un´1q .Intuitively, the node label determines the respective weights of the left and right sons in the definition of the measure.In particular the measure µ p is the one obtained by letting pp 0 x , p 1 x q " pp, 1 ´pq for all letters x in the alphabet.Again, with such a measure the results obtained in this article (provided that definitions of the games are modified accordingly) remains correct.
Remark 12.Following Proposition 11, a natural question is whether the class of qualitative tree languages is the same for each distribution µ p with 0 ă p ă 1.And the same for the above variant with probability values p 0 a , p 1 a for all letters in the alphabet.From the statement of Proposition 11 we conjecture that the answer to these questions is no but we leave it open.

Pumping Lemma
Let t be a tree and u P t0, 1u ˚be a node.A pair ∆ " pt, uq is called a pointed tree.With a pointed tree ∆ 1 " pt 1 , u 1 q and a tree t 2 , we associate a new tree, ∆ 1 ¨t2 , by plugging t 2 in t 1 instead of the subtree rooted at u 1 .Formally, ∆ 1 ¨t2 puq " t 1 puq if u 1 is not a prefix of u and ∆ 1 ¨t2 puq " t 2 pu 1 q if u " u 1 u 1 for some u 1 P t0, 1u ˚.We can also define the product of two pointed trees ∆ 1 " pt 1 , u 1 q and ∆ 2 " pt 2 , u 2 q by letting ∆ 1 ¨∆2 " p∆ 1 ¨t2 , u 1 ¨u2 q.Finally, with a pointed tree ∆ " pt, uq, we associate a tree ∆ ω by taking an ω-iteration of the product: ∆ ω pvq " tpv 1 q where v 1 is the shortest word s.t.v " u k v 1 for some k ě 0.
Qualitative tree languages enjoy a pumping lemma (see Figure 13 for an illustration), which contrasts with regular tree languages.Intuitively, pumping does not change acceptance for the qualitative semantics as it may only introduce a set of rejecting branches of measure zero (while for the classical semantics introducing a rejecting branch would make the run rejecting).
Lemma 13.Let A be an n-states parity automaton, t be a tree in L Qual pAq and u be a node of depth greater that n.Then there exists three pointed trees ∆ 1 , ∆ 2 and ∆ 3 such that t " ∆ 1 ¨∆2 ¨∆3 ¨trus and ∆ 1 ¨∆ω 2 P L Qual pAq.
Proof.Let ρ be an accepting run of A over t.Since u has depth greater than the number of states, there are two nodes u 1 and u 2 such that u 1 Ď u 2 Ď u and ρpu 1 q " ρpu 2 q " q for some state q.We now decompose t as t " ∆ 1 ¨∆2 ¨∆3 ¨trus with ∆ 1 " pt, u 1 q, ∆ 2 " ptru 1 s, u 2 q and ∆ 3 " ptru 2 s, uq.We decompose ρ in a similar way, by letting ρ " Θ 1 ¨Θ2 ¨Θ3 ¨ρrus with Θ 1 " pρ, u 1 q, Θ 2 " pρru 1 s, u 2 q and Θ 3 " pρru 2 s, uq.Note that Θ 1 ¨Θω 2 is a run of A over ∆ 1 ¨∆ω 2 .A branch in Θ 1 ¨Θω 2 can be rejecting for three reasons: it is a rejecting branch in ρzρru 1 s; or it is a branch in some of the copies of ρru 1 szρru 2 s; or it is the newly created branch u 1 v ω where v is the word such that u 2 " u 1 ¨v.As ρ is accepting, the measure of each of these three sets of rejecting branches is 0 (for the second set, it is a finite union of sets of measure 0).Therefore Θ 1 ¨Θω 2 is accepting, hence ∆ 1 ¨∆ω 2 P L Qual pAq.

Closure Properties
We now investigate the closure properties of qualitative tree languages under Boolean operations.
Proposition 14. Qualitative tree languages are closed under union and intersection.
Proof.The union is for free thanks to non-determinism of the automata model.For the intersection of two languages, L Qual pA 1 q and L Qual pA 2 q, we consider the usual product automaton whose states set is the Cartesian product of the states set of A 1 and of A 2 , and that simulates A 1 (resp.A 2 ) on its first (resp.second) component.Hence runs of the product automaton are in bijection with pairs of runs of A 1 and A 2 .The (ω-regular) acceptance condition requires that the acceptance condition of A 1 (resp.A 2 ) is fulfilled on the first (resp.second) component.Hence the set of accepting branches in a run of the product automaton is the intersection of the set of accepting branches in the corresponding runs of A 1 and A 2 .Hence, it has measure 1 if and only if both sets have measure 1, i.e. a run is accepting if and only if both corresponding runs in A 1 and A 2 are accepting.
Unsurprisingly, qualitative tree languages are not closed under complement.This is a simple consequence of the pumping lemma.
Proposition 15.Qualitative tree languages are not closed under complement.
Proof.Let L a be the set of ta, bu-labelled trees whose set of branches containing an a has measure 1.Clearly, L a is a qualitative tree language.However, we will show that its complement L a does not satisfy the pumping lemma, hence is not qualitative.By contradiction assume that L a " L Qual pAq for some n-state automaton A. Now define a tree t by letting u " 0 n and tpvq " b if v P trus and tpvq " a otherwise.Hence in t a tree is labelled by a b if and only if it belongs to the subtree rooted at u, and therefore the branches in Conepuq contains finitely many a, implying that t P L a .Now, if one uses Lemma 13 for t and u, the tree ∆ 1 ¨∆ω 2 (as in the statement of Lemma 13) is the tree whose nodes are all labelled by a, i.e. ∆ 1 ¨∆ω 2 R L a , leading a contradiction.

Emptiness Problem
It is well known that tree automata (as acceptors of regular languages) and twoplayer (perfect information) game are closely related [Gurevich and Harrington 1982;Grädel et al. 2002].In particular, the emptiness problem for regular tree languages and the problem of deciding the winner in a parity game on a finite graph are polynomially equivalent.From the proof of this result also follows that a regular tree language is non-empty if and only if it contains a regular tree.We show that a similar connection exists between tree automata as acceptors of qualitative tree languages and MDP.For this, fix a parity tree automaton A " xA, Q, q ini , ∆, Coly and a tree t.Consider the arena, depicted in Figure 2, G A,t " xS, s ini , Σ, ζy where S " Q ˆt0, 1u ˚Y tKu, s ini " pq ini , εq, Σ " ∆ and ζ is defined as follows.First we let d K be the distribution defined by d K psq " 1 if s " K and d K psq " 0 otherwise, and, for all q 0 , q 1 P Q and u P t0, 1u ˚, we let d q0,q1,u be the distribution such that d q0,q1,u pq 0 , u0q " d q0,q1,u pq 1 , u1q " 1{2 and d q0,q1,u psq " 0 q, u K q 0 , u0 q 1 , u1 for δ " pq, tpuq, q 0 , q 1 q for δ " pq 1 , a, q 0 , q 1 q with q 1 ‰ q or a ‰ tpuq for all other s P S. Then we let ζppq, uq, pq 1 , a, q 0 , q 1 qq " d K if q ‰ q 1 or a ‰ tpuq, ζppq, uq, pq, tpuq, q 0 , q 1 qq " d q0,q1,u and ζpK, σq " d K for all σ P ∆.Finally, we define a colouring function ρ by letting ρppq, uqq " Colpqq and ρpKq " 1, and we call G A,t " pG A,t , O ρ q the MDP equipped with the parity objective O ρ defined by ρ.
Then, the following holds: Theorem 16.The tree t belongs to L Qual pAq if and only if Éloïse almost-surely wins in G A,t .
Proof.In G A,t , a partial play that does not visit K is a sequence of the form pq 0 , u 0 qpq 1 , u 1 q ¨¨¨pq k , u k q where for all i, u i`1 " u i x i for some x i " 0, 1.Now, with any strategy ϕ of Éloïse we can associate a strategy ϕ 1 of Éloïse that is defined only on those partial plays that are compatible with it (meaning that if in some partial play λ Éloïse does not respect ϕ 1 then ϕ 1 pλq is undefined) and coincide with ϕ when defined; it is easily seen that ϕ and ϕ 1 leads the same plays and define the same probability measure.This means that one can only focus on such strategies for Éloïse.We should also only focus on strategies that never reach K, i.e. it always chooses an action of the form pq, tpuq, q 0 , q 1 q if the current state is pq, uq.Now, one can remark that for any node u there is a unique partial play λ where Éloïse respects ϕ 1 and that ends in a state of the form pq, uq for some q such that ϕ 1 is defined on λ.Therefore, an equivalent way to see ϕ 1 is as a Q-labelled tree, that should additionally verify the local properties imposed by ∆, namely the tree should be a run of A over t.It should also be clear that the set of plays where Éloïse respects ϕ 1 is the set of branches in the run associated with ϕ 1 and that this map preserves the measure (in particular the set of winning plays has the same measure as the set of accepting branches in the run), meaning that a strategy with value ξ is mapped to a run whose set of accepting branches has measure ξ and vice-versa.As this map, from strategies to runs is bijective, this concludes the proof: a tree t belongs to L Qual pAq if and only if there is an accepting run of A over t if and only if there is an almost surely winning strategy for Éloïse in G A,t .
Consider the (finite) arena G A " xS, s ini , Σ, ζy, depicted in Figure 3, where S " Q ˆt0, 1u Y tq ini , Ku, s ini " q ini , Σ " ∆ and ζ is defined as follows.First we let d K be the distribution defined by d K psq " 1 if s " K and d K psq " 0 otherwise, and, for all q 0 , q 1 P Q, we let d q0,q1 be the distribution such that d q0,q1 ppq 0 , 0qq " d q0,q1 ppq 1 , 1qq " 1{2 and d q0,q1 psq " 0 for all other s P S. Then we let ζppq, iq, pq, a, q 0 , q 1 qq " d q0,q1 , ζppq, iq, pq 1 , a, q 0 , q 1 qq " d K if q ‰ q 1 , ζpq ini , pq ini , a, q 0 , q 1 qq " for δ " pq, a, q 0 , q 1 q for δ " pq 1 , a, q 0 , q 1 q with q 1 ‰ q Fig. 3.The arena G A of the emptiness game.d q0,q1 , ζpq ini , pq, a, q 0 , q 1 qq " d K if q ‰ q ini , and ζpK, σq " d K for all σ P ∆.Finally, we define a colouring function ρ by letting ρppq, iqq " Colpqq and ρpKq " 1, and we call G A " pG A , O ρ q the MDP equipped with the parity objective O ρ defined by ρ.Then, the following hold: Theorem 17.The language L Qual pAq is non empty if and only if Éloïse almostsurely wins in G A from q ini .Proof.Following the same lines as for the proof of Theorem 16 it is easily seen that strategies that avoid state K are in bijection with pairs of pt, ρq where t is an A-labeled tree and ρ is a run of A over t.Moreover this map preserves the measure, meaning that a strategy with value ξ is mapped to a pair pt, ρq such that the set of accepting branches in ρ has measure ξ.In particular, Éloïse almost surely wins in G A if and only if there is a tree t that is qualitatively accepted by A.
Corollary 18.Let A be a parity tree automaton.Then one can decide whether L Qual pAq " H in polynomial time.Moreover, if L Qual pAq " H, it contains a regular tree, and such a tree can be constructed in polynomial time.
Proof.Emptiness in polynomial time follows from theorems 17 and 3. Now, if Éloïse has an almost surely strategy, then she has a memoryless one (Theorem 3).Then the tree and the run associated with this strategy are regular, the run is qualitatively accepting.As one can compute (when exists) an almost surely winning positional strategy, one can also compute (when exists) in polynomial time a regular tree in L Qual pAq.
Remark 19.Motivated by a decision problem for qualitative tree language we designed a polynomial reduction of the emptiness problem to the problem of deciding almost-surely winning in a finite MDP.
As already mentioned, a similar connection exists between tree automata (as acceptors of regular tree languages) and two-player (perfect information) games (see e.g.[Gurevich and Harrington 1982;Grädel et al. 2002]).Indeed, the emptiness problem for regular tree languages and the problem of deciding the winner in a parity game on a finite graph are polynomially equivalent.
Hence, one may ask whether conversely the problem of deciding almost-surely winning in a finite MDP can be polynomially reduced to the emptiness problem for qualitative tree languages.It is indeed possible and the proof is very similar to the one from a two-player game to a regular tree language.We briefly sketch the proof below.
First note that one can, up to coding, restrict its attention to finite MDP with only two actions and such that from any state and any action there are always two possible successors that can both be reached with same probability 1{2.Then one designs a deterministic tree automaton whose states are identified with the ones of the MDP, whose input alphabet is identified with the set of actions of the MDP, and whose transition function mimics the one of the MDP.Then one concludes by noting that there is a bijection between trees and strategies and that this bijection is such that the measure of the accepting branches in the (unique) run of the automaton on a tree equals the value of the corresponding strategy in the MDP.

Regular Tree Languages and Qualitative Tree Languages are Incomparable
In this section, we prove that regular tree languages and qualitative tree languages are incomparable.
Proposition 20.There is a regular tree language that is not qualitative.
Proof.Consider the regular tree language L of those ta, bu-labelled trees that contains at least one node labeled by b.By contradiction, assume that L " L Qual pAq for some n-state automaton A and let t P L be the tree defined by tpuq " b if u " 0 n and tpuq " a otherwise.Apply Lemma 1 to t and u: the tree ∆ 1 ¨∆ω 2 (using the notations of Lemma 13) is the tree whose nodes are all labelled by a, thus ∆ 1 ¨∆ω 2 R L, leading to a contradiction.Theorem 21.There is a qualitative tree language that is not regular.
Proof.Let L a be the language of trees whose set of branches containing at least one a has measure 1.This language is qualitative as noticed in Example 7. In the sequel, we prove that L a is not regular.
We first prove that, for any regular tree t, if there is no cone in t whose branches only contain the letter b, then t P L a .Let t be a regular tree, we can assume w.l.o.g. that if there is a node labelled by a then all its descendants are labeled by a. Then the property "there is no cone in t whose branches only contain b" is the same as "every subtree contains a subtree made only of a".Let X 1 , . . ., X n be the n different subtrees of t, and for all i, let µ i be the measure of the set of branches containing a in X i (we call it the value of X i ).We can assume w.l.o.g. that @i µ 1 ď µ i .If X i1 and X i2 are the two sons of X 1 , we know that µ 1 " µi 1 `µi 2 2 . Since µ 1 ď µ i for i " i 1 , i 2 , µ i1 " µ i2 " µ 1 .Hence we can prove by induction that for all X i of minimal value, all the subtrees of X i have minimal value too.Since there is a subtree of a (of value 1) in X 1 , µ 1 " 1 hence for all i, µ i " 1, hence the value of t is 1, hence t P L a .
We assume by contradiction that L a is regular.The closure properties of regular tree languages implies that the following language L is also regular: t ˇˇt R L a ^"there is no cone in t whose branches only contain b" ( Using our previous characterisation of regular trees in L a it follows that L does not contain any regular tree, hence L is empty (Theorem 2).Then, to raise a contradiction, we build a (non-regular) tree t 0 P L.
For every node u P t0, 1u ˚, we let n u be the integer whose binary representation is 1 ¨u.We define a tree t 0 as follows: let v P t0, 1u ˚, if there exists some u such that v " u.0 nu`1´|u| then t 0 pvq " a, otherwise t 0 pvq " b (see Figure 4).We now establish that t 0 R L a .First note that the set of branches in t 0 that contains at least one a is obtained by taking the union of those cones Cpuq such that t 0 puq " a. Then remark that, for every level , there is one and only one node u of depth labelled by a (except for " 0, 1 where there are no such u).Thus we can bound the measure µ of the set of branches in t 0 that contains at least one a: This proves that t 0 R L a .Moreover, it follows from the definition, that for every node u, there is a branch (the leftmost one) in the cone Cpuq that contains an a, hence t 0 P L, which contradicts the fact that L is empty.

The Value of a Tree May not be Reached
So far we defined qualitative acceptance of a tree by the existence of a run whose set of accepting branches has measure 1.We can refine this notion by defining the value of a tree as follows.For a tree automaton A, and a tree t we let In particular L Qual pAq is the set of trees t whose value is 1 and is reached for some run (i.e. the sup is a max).The following result proves that the value may not be reached by some run.
Theorem 22.There is a reachability automaton A and a tree t such that Val A ptq " 1 but t R L Qual pAq.
Proof.Let A " xA, Q, q w , ∆, tq acc uy with A " ta, bu, Q " tq w , q l , q acc , q rej u, and ∆ contains the following transitions: q w Ñ pq w , q w q, q w Ñ pq l , q acc q, q l a Ñ pq acc , q l q, q l b Ñ pq rej , q rej q (here ˚is a shorthand for an arbitrary letter).Intuitively, the automaton can wait in state q w as long as it wants.It can at some point use the second transition: this leads to accept (all branches in) the subtree rooted at ACM Transactions on Computational Logic, Vol.V, No. N, Month 20YY.
the left son, as well as those subtrees rooted in the branch going left and then right forever as long as this branch does not contain a node labelled b, in which case the subtree rooted at this node is rejected.
We claim that any run ρ t of A on t is such that µpAccBrpρ t qq ă 1.Indeed, either the only transition used in ρ t is q w Ñ pq w , q w q and therefore all branches are rejecting.Otherwise the transition q w Ñ pq l , q acc q is used at least once.Then pick u a node of minimal depth where this transition is used: the cone Conephpuqq is rejecting and therefore µpAccBrpρ t qq ď 1 ´2|hpuq| ă 1.
Now, for all integer i ě 1, consider the run ρ i t where the transition q w Ñ pq w , q w q is used for all nodes at depth ă i, and the transition q w Ñ pq l , q acc q is used for all nodes at depth i (then there is no more freedom in defining for the rest of the run).Then for all node u at depth i, the branches in Conephpuqq are rejecting and those in ConepuqzConephpuqq are accepting.Therefore µpAccBrpρ t qq " 1 ´2i ˆ2´pi`1`iq " 2 ´pi`1q .Thus, sup ρ i t µpAccBrpρ t qq " 1, implying that Val A ptq " 1.Actually, the proof of Theorem 16 directly leads the following.
Corollary 23.Let A be a parity tree automaton and let t be a tree.Then Val A ptq " ValpG A,t q.

Positive Tree Languages
So far, we favoured the almost-sure acceptance condition (i.e.requiring the measure to be equal to 1) over the positive one (i.e.requiring the measure to be strictly positive).However, the decidability results on MDP stated in Theorem 3 still hold if we replace the almost-sure acceptance by the positive acceptance [Courcoubetis and Yannakakis 1990;Chatterjee et al. 2004].We now discuss the impact when considering positive acceptance instead of almost-sure acceptance, and motivate our choice to focus on almost-sure acceptance.
We say that a run ρ of a tree automaton A is positively accepting if the measure of its set of accepting branches is (strictly) positive, i.e. µpAccBrpρqq ą 0. A tree t is positively accepted if there exists a positively accepting run of A over t, and we denote by L Pos pAq the set of all trees positively accepted by A. Finally, a positive tree language is a language L of trees such that there exists a parity automaton A with L Pos pAq " L.
Note that, contrarily to the qualitative semantics, we can no longer assume that our automata are complete.In particular, the naive idea of adding a sink state does not work, as it would have new runs that may be accepting (going to the sink state yields a rejecting cone, but this may not affect the positivity of the measure of the set of accepting branches).
Example 24.Consider the language L ą0 a of ta, bu-labeled trees that have a nonnegligible set of branches containing infinitely many a's.This language is positively accepted by a deterministic Büchi tree automaton (that goes in a final state whenever an a is read and in a non-final state otherwise) and hence is a positive tree language.
Remark 25.The complement of a language qualitatively accepted by a deterministic tree automaton A is positively accepted by a deterministic tree automaton B. Indeed, it suffices to define B starting from A and dualise its acceptance condition (namely increment the value of the colouring function by 1).
Let us now give two examples of languages that are not positive tree languages.For this we introduce the language L a^a of ta, bu-labelled trees containing an a in both their left and right subtrees that is formally defined as L a^a " tt | Du, v P t0, 1u ˚s.t.tp0uq " tp1vq " au.
Proposition 26.The language L a of ta, bu-labelled trees whose set of branches containing an a has measure 1 and the language L a^a of ta, bu-labelled trees containing an a in both their left and right subtrees are not positive tree languages.
Proof.The proofs for both languages rely on the same family of counterexamples.For every n ě 0, let t n be the tree defined by t n puq " b for |u| ď n and t n puq " a otherwise and let t b be the tree such that t b puq " b for all u P t0, 1u ˚.
We are going to show that if t n`2 is positively accepted by a parity automaton A with n states then A also accepts either the tree t 0 n`2 obtained by replacing the left subtree of t n`2 by t b or the tree t 1 n`2 obtained by replacing the right subtree of t n`2 by t b .
This property will allow us to conclude as for all n ě 0, t n belongs to L a but t 0 n and t 1 n do not, L a is not a positive language.The same arguments work for L a^a .Let ρ be a positively accepting run of A on t n`2 .Without loss of generality, we can assume that the set of accepting branches in the right subtree has a strictly positive measure.We are going to show that A admits a run ρ 1 (not necessarily a positively accepting one) on t b starting from ρp0q.By replacing the left subtree of ρ by ρ 1 , we then obtain a positively accepting run for A on t 1 n`2 .For all i ě 1, consider the set Q i " tρp0uq | u P t0, 1u ˚and 0 ď |u| ď i ´1u of those states appearing in the left subtree of ρ at depth at most i.As the sequence of the Q i is increasing, we have Q n " Q n`1 .For all q P Q n , there exists a least one transition δ q of A of the form pq, b, q 1 , q 2 q with q 1 and q 2 P Q n (recall that in t n`2 the first n `2 levels only contains b's).By using the transition δ q when in state q and starting from state q 0 , we easily construct a run of A on t b starting in state ρp0q.
Proposition 27.The classes of positive and qualitative tree languages are incomparable.
Proof.Let L a be the qualitative language of those ta, bu-labelled trees whose set of branches containing an a has measure 1.We proved in Proposition 15 that its complement L a is not qualitative.But L a is a positive tree languages, as it is the complement of a language qualitatively accepted by a deterministic tree automaton (see Remark 25).
Conversely we have seen in Proposition 26 the qualitative language L a is not positive.

Remark 28.
A natural open question is whether a tree language which is both positively and qualitatively accepted is always regular.

¨21
As it is the case for qualitative languages, the positive languages are incomparable with the regular languages.
Proposition 29.The class of positive tree languages is incomparable with the class of regular languages.
Proof.The language L a^a of ta, bu-labelled trees containing an a in both their left and right subtrees is an example of a regular language that is not positive tree language (cf.Proposition 26).
We are now going to show that the positive tree language L ą0 a of ta, bu-labeled trees that have a non-negligible set of branches containing infinitely many a's (cf.Example 24) is not a regular language.
Toward a contradiction, we assume that L ą0 a is accepted by a parity tree automaton A " xA, Q, q ini , ∆, Coly (with the standard acceptance condition for runs).
For every i ě 1, we let r i be the (finite) tree whose domain consists of all words of length at most i and in which only 0 i is labeled by a and all other nodes are labelled by b.We denote by |r i | " i the height of r i .The probability p ri to visit an a-labelled node in r i when starting from the root is 1 2 i .We let T denote the set tr i | i ě 1u of all such trees.
To every sequence s " pt i q iě0 of trees in T , we associate an infinite ta, bu-labelled tree t s which is intuitively obtained as follows: start with t 1 and glue to every leaf a copy of t 2 , to every leaf glue a copy of t 3 and so on.Formally t s puq " a if u is of the form v0 |ti| with |v| " ř jăi |t j | and t s puq " b otherwise.By Borel-Cantelli's lemma [Bauer 1996, Lemma 11.1 -p. 70], the tree t s belongs to L ą0 a if and only if ř iě0 p ri " `8.We extend the notion of runs for A to finite trees.A run of A on finite tree t : t0, 1u ďn Þ Ñ ta, bu is a finite tree ρ : t0, 1u ďn Þ Ñ Q such that for all u P t0, 1u ăn , pρpuq, tpaq, ρpu0q, ρpu1qq belongs to ∆.We can summarise the run ρ by a finite information called the profile of the run and denoted Profpρq.The profile Profpρq of ρ is defined as: Profpρq " pρpεq, Rq where R " tpm, ρpuqq | u P t0, 1u n and m " inf vĎu Colpvqu.
By extension, the profile of a finite tree t, denoted Profptq, is the set tProfpρq | ρ run of A on tu of all possible profiles of runs of A on t.
This notion induces an equivalence relation on finite trees denoted " A which intuitively equates trees that cannot be distinguished by A. Formally for all finite tree t and t 1 , t " A t 1 if and only if Profptq " Profpt 1 q.
The following claim is straightforward.
Claim 30.Let s " pt i q iě1 and s 1 " pt 1 i q iě1 be two sequences of finite trees in T .If for all i ě 1, t i " A t 1 i then A accepts t s if and only if A accepts t s 1 .Now, note that there are only finitely many different possible profiles hence, there exists an infinite sequence s " pr ij q jě0 of trees with the same profile and such that the sequence pi j q jě0 is strictly increasing.As the pi j q jě0 sequence is strictly increasing it implies that ř jě0 p ri j ă `8, hence the tree t s does not belong to L ą0 a .Now consider the constant sequence s 1 " r i0 , r i0 , . . .: one has ř jě0 p ri 0 " `8 hence t s 1 belongs to L ą0 a .This leads a contradiction with A accepting L ą0 a as A behaves the same on both t s and t s 1 ,i.e.A accepts t s if and only if A accepts t s 1 .
The most notable difference between positive and qualitative tree languages is that positive tree languages are not closed under intersection.
Proposition 31.The class of positive tree languages is closed under union but neither under intersection nor complementation.
Proof.The closure under union is immediate thanks to non-determinism.The non-closure under complementation will be a consequence of the closure under union and the non closure under intersection.However, one can also derive it directly by considering the language L a : we have shown in the proof of Proposition 26 that this language is not positive while its complement is positive.
We now prove the non-closure under intersection.Consider the language L 0 (resp.L 1 ) of those trees t containing a non-negligible set of branches containing an a in the left (resp.right) subtree of the root.Both languages are positively accepted by a deterministic reachability automaton.But we have seen in Proposition 26 that their intersection L a^a is not a positive tree language.
However, positive tree languages enjoy all decidability properties of qualitative languages presented in Section 3.5.
Theorem 32.The tree t belongs to L Pos pAq if and only if Éloïse positively wins in G A,t .
Proof.The proof is identical to that of Theorem 16.
Theorem 33.The language L Pos pAq is non empty if and only if Éloïse positively wins in G A from q ini .
Proof.The proof is identical to that of Theorem 17.
Corollary 34.Let A be a parity tree automaton.Then one can decide whether L Pos pAq " H in polynomial time.Moreover, if L Pos pAq " H, it contains a regular tree, and such a tree can be constructed in polynomial time.
Proof.As the properties expressed in Theorem 3 of MDP with almost-sure winning condition also hold for the positive winning condition, we can use the same proof as for Corollary 18.

BEYOND NON-DETERMINISTIC AUTOMATA: THE PROBABILISTIC SETTING
Following [Rabin 1963] for finite words and [Baier and Größer 2005;Baier et al. 2008;Baier et al. 2012] for infinite words we investigate probabilistic automata on infinite trees.That is the set of transitions of an automaton is replaced by a probability distribution over the set of all transitions which induces a probability measure on the set of runs of the automaton.Now, a tree is accepted if almost every run over the input tree is accepting.For the run, we may use either the classical or the qualitative acceptance criterion.A probabilistic tree automaton A is a tuple xA, Q, q ini , δ, Accy where A is the input alphabet, Q is a finite set of states, q ini P Q is the initial state, Acc Ď Q ω is the acceptance condition and δ is a mapping from Q ˆA ˆQ ˆQ to r0, 1s such that for all q P Q and a P A, ř q0,q1PQ δpq, a, q 0 , q 1 q " 1. Intuitively, the value δpq, a, q 0 , q 1 q is the probability for a transition q a Ñ pq 0 , q 1 q to be used by the automaton when it is in state q and reads the symbol a.
This probability distribution on the transitions induces a probability measure on the set of runs of A. In this setting, a run of A is simply a Q-labeled tree whose root is labeled by the initial state q ini .We denote by RunspAq (or simply Runs if A is clear from the context) the set of all runs of A. We denote by AccRunspAq the set of accepting runs of A and by QualAccRunspAq the set of qualitatively accepting runs of A.
In the sequel, we will show that, for a given tree t the sets AccRunspAq and QualAccRunspAq are measurable (Proposition 38), and this will allows us to define almost-sure acceptance of a tree by a probabilistic automaton.4.1.2Measurability of AccRunspAq and QualAccRunspAq.We first define a σalgebra for RunspAq.A partial run is a partial function λ : t0, 1u ˚Ñ Q with λpεq " q ini and such that Dompλq is finite, non-empty, prefix-closed and proper, i.e. such that for all w P t0, 1u ˚, w0 P Dompλq iff w1 P Dompλq.If λ is a partial run, we denote by Innerpλq the set tw P Dompλq | w0 P Dompλq and w1 P Dompλqu of non-leaf nodes.The cylinder associated to λ, denoted Cylpλq, is the set of runs consistent with λ, i.e.
Cylpλq " tρ P Runs | @w P Dompλq, ρpwq " λpwqu Let F R be the σ-algebra generated by the cylinders.By Carathéodory's extension theorem, there exists a unique probability measure µ t on the measurable space pRuns, F R q such for all partial run λ : t0, 1u ˚Ñ Q, µ t pCyl A pλqq " ź wPInnerpλq δpλpwq, tpwq, λpw0q, λpw1qq Note that both µ t and pRuns, F R q depend on t.Also note that the definition of µ t does not make sense without the properness condition in the definition of a partial run.
Remark 35.A balanced partial run λ is a mapping from t0, 1u n to Q for some n ě 0 with λpεq " q ini .Hence it is a special kind of partial play in the previous sense.It is straightforward to note that the σ-algebra generated by balanced partial play is F R .Moreover one has µ t pCyl A pλqq " ź wPt0,1u ďn´1 δpλpwq, tpwq, λpw0q, λpw1qq Let A be an automaton and define the mapping f A : Runs ˆBr Ñ r0, 1s associating to a pair pρ, πq P Runs ˆBr the value 1 if ρpπq belongs to Acc and 0 otherwise.The following lemma proves that f A is integrable.
Lemma 36.Let A be a probabilistic tree automaton with an ω-regular acceptance condition and let t be a tree.The mapping f A is integrable in the product space pRunspAq, F R , µ t q b pBr, F Br , µq.
Proof.The product space pRuns, F R q b pBr, F Br q is generated by the sets Conepwq ˆCylpλq for all w P t0, 1u ˚and all partial run λ (cf.[Bauer 2001, Theorem 22.1 -p. 132]).Thanks to Remark 35 we can restrict our attention to balanced partial runs.
According to [Bauer 2001, Theorem 9.1 -p. 50], it is enough to show that AccPairspAq " tpρ, πq | ρpπq P Accu is measurable in the product space.
We first establish the result for a Büchi acceptance condition.For that, we let A " xA, Q, q ini , δ, F y be probabilistic tree automaton with a Büchi acceptance condition.
For u P t0, 1u ˚, we introduce the set R u of all pairs pρ, πq P Runs ˆBr such that π admits u as a prefix and ρpvq R F for all u Ĺ v Ď π.The set AccPairspAq is the complement of the countable union Ť uPt0,1u ˚Ru .Hence, it is enough to show R u is measurable for all u P t0, 1u Let us fix a node u P t0, 1u ˚.For all n ą |u|, let C n be the union all sets Cylpλq ˆConepwq where λ : t0, 1u n Ñ Q is a partial run and w P t0, 1u n Ľu is a word such that λpvq R F for all u Ĺ v Ď w.We claim that the set R u is equal to Ş ną|u| C n and hence is measurable.The direct inclusion, R u Ď Ş ną|u| C n , is obvious.For the converse one, let pρ, πq be in Ş ną|u| C n .For all n ą |u|, the node w n P t0, 1u n witnessing that pρ, πq belongs to C n is indeed the unique node of π of depth n.As n can be arbitrary large, it follows that every node v in π of depth ą |u| is such ρpvq R F , which exactly means that pρ, πq P R u .
The case of the parity conditions follows from the previous case as one can express them as a Boolean combination of Büchi conditions.
The general case of an arbitrary ω-regular condition is obtained as follows.First, one considers a deterministic parity word automaton recognising Acc, and then takes the synchronised product of this automaton with the tree automaton.This leads to a parity tree automaton accepting the same language and whose runs are in bijection with the ones of the original automaton.Moreover as this bijection preserves the set of accepting branches, one can conclude from the parity case.
Note that the previous lemma implies that the set of accepting branches in a run of a tree automaton with an ω-regular acceptance condition is measurable in pBr, F Br q.This leads an alternative proof of Proposition 6.
Corollary 37. Let A be a tree automaton equipped with an ω-regular acceptance condition, and let ρ be a run of A over some tree t.Then the set AccBrpρq of accepting branches in t is measurable.
Proof.Let A be a tree automaton with an ω-regular acceptance condition, t be a tree and ρ be a run of A on t.The set AccBrpρq is equal to tλ P Br | ρpλq P Accu.
As by the proof of Lemma 36, AccPairspAq is measurable in the product space, AccBrpρq is also measurable (cf.[Bauer 2001, Lemma 23.1 -p. 135]).
We are now ready to establish the measurability of the sets AccRunspAq and QualAccRunspAq.
Proposition 38.For all probabilistic tree automata A with an ω-regular acceptance condition, the sets AccRunspAq and QualAccRunspAq are measurable.
Proof.We first consider the case of AccRunspAq.As seen in Proposition 6, it is enough to establish the property for the Büchi acceptance condition.
Let xA, Q, q ini , δ, F y be probabilistic tree automaton with a Büchi acceptance condition.
For u P t0, 1u ˚, we introduce the set R u of all runs ρ of A for which there exists an infinite branch π such that u Ď π and for all u Ď v Ĺ π, ρpvq R F .The set of accepting runs AccRunspAq is the complement of the countable union Ť uPt0,1u ˚Ru .Hence it is enough to show R u is measurable for all u P t0, 1u Let us fix a node u P t0, 1u ˚.For all n ą |u|, let C n be the union of all cylinders Cylpλq where λ : t0, 1u n Ñ Q is a (balanced) partial run for which there exists a word w P t0, 1u n Ě u such that λpvq R F for all u Ĺ v Ď w.
We claim that R u is equal to Ş ną|u| C n and is hence measurable.The direct inclusion, R u Ď Ş ną|u| C n , is obvious.For the converse one, let ρ be a run in Ş ną|u| C n .For all n ą |u|, let w n be a word in t0, 1u n witnessing that ρ belongs to C n .Let T be the prefix-closure of the set tw n | n ą |u|u.By Koenig's lemma, T contain an infinite branch π.Clearly u is a prefix of π and for all u Ď v Ĺ π, ρpvq R F as v is a prefix of one of the w n .Hence ρ belongs to R u .
Let us now consider the case of QualAccRunspAq.From Lemma 36 we have that AccPairspAq " tpρ, πq | ρpπq P Accu " f ´1 A pt1uq is measurable in the product space.Hence the numerical function g : Runs Ñ r0, 1s associating to a run ρ the measure of its set of accepting branches (i.e.µpAccPairspAq ρ q where AccPairspAq ρ " tλ P Br | ρpλq P Accu) is measurable (cf.[Bauer 2001, Lemma 23.1 -p. 135]).As QualAccRunspAq is equal to g ´1pt1uq, it is measurable.4.1.3Almost-Surely Accepted Trees.Proposition 38 shows that the following definition is well-founded.
A tree t is (almost-surely) accepted by A with the classical (resp.qualitative) semantics if almost all runs of A on t are accepting (resp.qualitatively accepting), i.e. µ t pAccRunspAqq " 1 (resp.µ t pQualAccRunspAqq " 1).We denote by L "1 pAq (resp.L "1 Qual pAqq the set of trees accepted by A with the classical (resp.qualitative) semantics for runs.More formally, we define L "1 pAq " tt | µ t pAccRunspAqq " 1u and L "1 Qual pAq " tt | µ t pQualAccRunspAqq " 1u.
Remark 39.Our motivation for considering almost-sure acceptation and not positive acceptation is discussed in Section 4.5.
The following easy lemma is used later to prove Proposition 42 and Proposition 43.
Remark 41.The definition of [Baier and Größer 2005] (for ω-words) allows for incomplete automata ( i.e. for all q P Q, the sum ř q 1 PQ δpq, a, q 1 q " 1 or 0).However it is easy to verify that every automaton can be rendered complete without changing the acceptance condition and the measure of acceptation.Remark that removing all states q such that ř q1,q2PQ δpq, a, q 1 , q 2 q " 0 does not change the measure of the set of accepting runs (either in the classical or qualitative sense).By iteratively applying this process, we obtain an equivalent complete automata.
Combining Lemma 36 and Lemma 40, we can show that the almost-sure acceptation of a tree t by an automaton A for the qualitative semantics can be defined by integrating the mapping f A .
Proposition 42.Let A be a probabilistic tree automaton with an ω-regular acceptance condition and let t be a tree.Then we have: Proof.As f A is measurable (Lemma 36), by Tonelli's theorem [Bauer 2001, Theorem 23.6 -p. 138 [Bauer 2001, Theorem 23.6 -p. 138] 4. 1.4 Examples.We conclude this section with examples of languages accepted by probabilistic tree automata.
For an ω-word language L Ď ta, bu ω , we denote by Path "1 pLq the set of trees labeled by ta, bu with almost all their branch labels in L (i.e.µptπ P Br | tpπq P Luq " 1).It is easy to see that, for any ω-regular language L, the tree language Path "1 pLq is a qualitative tree language.
More interestingly, if L is almost-surely accepted by a probabilistic ω-word automaton6 with an ω-regular acceptance condition, we can show that Path "1 pLq is accepted by a probabilistic tree automaton (with the qualitative semantics).
Proposition 43.Given a probabilistic ω-word automaton B with an ω-regular acceptance condition, there exists a probabilistic tree automaton A with the same acceptance condition such that L "1  Qual pAq is equal to Path "1 pL "1 pBqq.Proof.Let B " xA, Q, q ini , δ, Accy be a (complete) probabilistic ω-word automaton with an ω-regular condition.Consider the probabilistic tree automaton A simulating B on all branches.Formally A is equal to xA, Q, q ini , δ 1 , F y where, for all p, q P Q and x P A, δ 1 pp, x, q, qq " δpp, x, qq.
Let t be a tree and let g : Br Þ Ñ r0, 1s be the mapping associating to a branch π the value ş f A p¨, πqdµ t .We have the following claim.
Proof of Claim 44.Let π " a 1 a 2 ¨¨¨P Br and let w " tpπq.It is enough to show that: gpπq " ş f A p¨, πqdµ t is equal to µ w pAccRunspBqq.By definition ş f A p¨, πqdµ t is equal to µ t pAccRuns π pAqq where µ t pAccRuns π pAqq is the set of runs ρ of A such that ρpπq is accepting.
Consider the mapping ψ : RunspAq Ñ RunspBq associating to a run ρ of A the corresponding run ρpεqρpa 1 q ¨¨¨of B. We claim that ψ is measurable and that µ w is the image of µ t under ψ.This claim in particular implies that: To substantiate our claim, it is enough to show that ψ ´1pCylpλqq is a measurable subset of RunspAq such that µ t pψ ´1pCylpλqqq " µ w pCylpλqq.The first part implies that ψ is measurable and the second that the image measure of µ t under ψ, denoted ψpµ t q, coincide with µ w on the cylinders.By Carathéodory's unique extension theorem, this in turn implies that µ w and ψpµ t q coincide.Let λ " q 0 . . .q n be a partial run of B. Consider the partial balanced run η of A defined for i P r0, n ´1s by ηpa 1 ¨¨¨a i´1 0q " ηpa 1 ¨¨¨a i´1 1q " q i .It is clear that ψ ´1pCylpλqq " Cylpηq.Moreover we have: µ t pCylpηqq " ś jPr0,n´1s δ 1 pq j , tpa 1 ¨¨¨a j q, q j`1 , q j`1 q " ś jPr0,n´1s δpq j , tpa 1 ¨¨¨a j q, q j`1 q " µ w pCylpλqq

Proof of Claim 44
We are now ready to conclude: Tonelli's Theorem ô gpπq " 1 almost everywhere Lemma 40 ô µptπ | tpπq P L "1 pBquq " 1 Claim 44 ô t P Path "1 pL "1 pBqq 4.2 Acceptance Game for Qualitative Probabilistic Tree Automata Fix a probabilistic tree automaton A " xA, Q, q ini , δ, Accy and a tree t.We define an infinite Markov chain M A,t " pG A,t , O Acc q such that M A,t almost-surely fulfils its objective iff t belongs to L "1 Qual pAq.Compared with the acceptance game for qualitative tree automata, the transition is no longer chosen by Éloïse: it is now randomly chosen with the probability distribution given by A. Hence, this is why we simply obtain a Markov chain instead of an MDP.
Note that a trace in M A,t can be uniquely represented by an infinite sequence ppp 0 , q 0 0 , q 1 0 q, a 0 qppp 1 , q 0 1 , q 0 1 q, a 1 q . . .labeled by ∆ ˆt0, 1u such that p 0 " q ini and for all i ě 0, p i`1 " q ai i .The objective O Acc is defined as the set of those traces ppp 0 , q 0 0 , q 1 0 q, a 0 qppp 1 , q 0 1 , q 0 1 q, a 1 q . . .such that p 0 p 1 . . .belongs to Acc.Proposition 45.Let A be a probabilistic tree automaton with an ω-regular acceptance condition and let t be a tree.We have t P L "1 Qual pAq iff M A,t almost-surely fulfils its objective.
Proof.Let AccPairspAq " tpρ, πq | ρpπq P Accu Ď Runs ˆBr.By Proposition 42, to establish the desired equivalence, it is sufficient to show that µ t b µpAccPairspAqq " µ M A,t pO Acc q.
We claim that ψ is measurable and that µ M A,t is the image of µ t b µ under ψ.In particular this implies that µ M A,t pO Acc q " µ t b µpAccPairspAqq.
To substantiate our claim, it is enough to show, for any partial trace θ, that ψ ´1pCylpθqq is a measurable subset of Runs ˆBr such that µ t b µpψ ´1pCylpθqqq " µ M A,t pCylpθqq.The first part implies that ψ is measurable and the second that the image measure of µ t b µ under ψ, denoted ψpµ t b µq, coincide with µ M A,t on the cylinders.By Carathéodory's unique extension theorem, this in turn implies that µ M A,t and ψpµ t b µq coincide.

Decidability and Undecidability Results
In this section, we show that the emptiness problem for probabilistic Büchi tree automata is decidable for the qualitative semantics for runs.We reduce this problem to deciding almost-surely winning in a POMDP, and the reduction works for any ω-regular acceptance condition.However, the corresponding decision problem on POMDPs is only decidable for the Büchi condition.Hence we only obtain decidability in the Büchi case.Let A " xA, Q, q ini , δ, Accy be a probabilistic automaton with an ω-regular acceptance condition and let ∆ " Q ˆQ ˆQ.
Define a POMDP G A " pG, ", Oq as follows.The arena G is equal to xS, s ini , A, ζy where S " Qˆt0, 1, Kuˆp∆YtKuq, s ini " pq 0 , K, Kq and ζ is defined as follows.For all a P A and pp, x, tq P S, ζppp, x, tq, aq is the distribution that assigns 1 2 δpp, a, q 0 , q 1 q to pq y , y, pp, q 0 , q 1 qq where y " 0, 1 and 0 to all other state.The objective O is the set of plays for which the sequence of states obtained when projecting on the first component belongs to Acc.The equivalence " is defined by pq, x, tq " pq 1 , x 1 , t 1 q iff x " x 1 .Intuitively in G A , Éloïse describes a branch along a tree and Random builds a piece of run along this branch.As Éloïse does not observe the state in the run constructed by Random, it does not influence her choice for the branch.
Theorem 46.Let A be a probabilistic tree automaton with an ω-regular acceptance condition.The language L "1 Qual pAq is non-empty if and only if Éloïse almostsurely wins in G A .
Proof.From the definitions, we easily remark that trλs " | λ P Playsu is in bijection with " t0, 1u ˚.Hence, strategies are in bijection with A-labeled trees.Once such a strategy ϕ t (seen as a tree t) is fixed, the resulting Markov chain is, up to renaming, G A,t , meaning that the value of ϕ t is the value of G A,t .In particular, Éloïse almost-surely wins in G A iff there is some t such that ValpG A,t q " 1 iff t P L "1 Qual pAq (thanks to Proposition 45).For the Büchi acceptance condition, this leads a decidability result for the emptiness problem.
Corollary 47.Let A be a probabilistic Büchi tree automaton.Deciding emptiness of L "1 Qual pAq is an ExpTime-complete problem.Moreover, if L "1 Qual pAq ‰ H, it contains a regular tree.
Proof.The ExpTime upper-bound follows from the polynomial time reduction to deciding almost-surely winning in a Büchi POMDP.The lower bound follows from Proposition 43: emptiness of probabilistic Büchi ω-word automata with the almost-sure acceptation (which is ExpTime-complete [Baier et al. 2008;Baier et al. 2012]) can be reduced to our problem.Indeed, if B is a probabilistic Büchi ω-word automaton we can construct a probabilistic Büchi tree automaton A of linear size (cf.Proposition 43) such that L "1 Qual pAq is equal to Path "1 pL "1 pBqq and in particular L "1 Qual pAq is empty if and only if L "1 pBq is empty.The proof of Theorem 46 together with the fact that a finite memory optimal strategy always exists in POMDP imply the existence of a regular tree when L "1 Qual pAq ‰ H. On the negative side, we show that the emptiness problem for probabilistic co-Büchi tree automata is undecidable for both the classical and qualitative semantics for runs.These results are obtained by reduction to the undecidability of the emptiness problem for co-Büchi ω-word automata with the almost-sure acceptation [Baier et al. 2008].
Theorem 48.The following problems are undecidable : ACM Transactions on Computational Logic, Vol.V, No. N, Month 20YY.example of a qualitative tree language that no probabilistic automaton (regardless of its semantics) can accept and another example of a regular tree language that no probabilistic automaton (regardless of its semantics) can accept.Hence, this proves incomparability of probabilistic tree languages with both qualitative and regular tree languages.
For the first result, we consider, for all 0 ă λ ă 1, the ω-word language L λ over ta, bu defined by In [Baier et al. 2008], L λ is shown to be almost-surely accepted by a co-Büchi probabilistic automaton (actually they show that L λ is positively accepted by a Büchi automaton hence, as previously remarked in the introduction, is accepted by a co-Büchi probabilistic automaton).Therefore, by Proposition 43, Path "1 pL λ q is a co-Büchi probabilistic qualitative tree language.
Proposition 49.For all 0 ă λ ă 1, Path "1 pL λ q is neither a regular tree language nor a qualitative tree language.
Proof.Let 0 ă λ ă 1.We will show that Path "1 pL λ q does not contain any regular tree.Hence, thanks to Theorem 2 it is not a regular tree language and thanks to Corollary 18 it is not a qualitative tree language.
Toward a contradiction, assume that L λ contains a regular tree t 0 .Consider the Markov chain M whose arena G is given by pS, s ini , ζq.The set of states S is the finite set of subtrees of t.The initial state s ini is t 0 .The transition function ζ is defined for all t P S by ζptqptr0sq " ζptqptr1sq " 1 2 and 0 otherwise.Clearly there exists a measure preserving bijection between the traces of M and the branchs of t 0 .
To define the objective of the M, consider the mapping ψ for S to ta, bu associating to a state t the label of its root tpεq.This mapping can canonically be extended to associate a trace of M an ω-word over ta, bu.The objective O of M is the set of traces whose image under ψ belongs to L λ .
As t 0 belongs to Path "1 pL λ q, M almost-surely fulfils its objective.
We claim that there exists some k ą 0 such that the following event has a strictly positive probability: "The Markov chain from its current state reaches a state t with ψptq " b and in less that k steps another state t 1 with ψpt 1 q " b." Indeed, as t 0 belongs to Path "1 pL λ q, for all t P S there exists u Ď v such that tpuq " tpvq " b.Hence, it suffices to take k to be the maximum over all t of the minimum of |v| ´|u| over all u Ď v such that tpuq " tpvq " b.
By Borel-Cantelli lemma, this event occurs infinitely often along a trace with probability 1.In particular, this implies that with probability 1 for a trace θ with ψpθq " a k1 ba k2 b . . .| k 1 , k 2 , . . .ą 0 there are infinitely many k i 's which are equal to k 1 ď k.Hence ψpθq does not belongs to L λ (as for all n ą 0, ś 8 i"1 p1 ´λki q ď p1 ´λk 1 q n ).This contradicts the fact that M fulfils its objective with probability 1.
Remark 50.Using the correspondence with POMDP introduced in Theorem 46, ACM Transactions on Computational Logic, Vol.V, No. N, Month 20YY.
any co-Büchi automaton accepting Path "1 pL λ q gives rise to an example of a co-Büchi POMDP G A in which Éloïse needs infinite memory to almost-surely win.We now give an example of a qualitative tree language that no probabilistic tree automaton can accept.For this we consider the language L 0_1 a of ta, bu-labelled trees such that either the left subtree or the right subtree of the root belongs to the language L a of example 7 (recall that L a is the language of ta, bu-labelled trees whose set of branches containing at least one a has measure 1).Formally L 0_1 a " tt | tr0s P L a or tr1s P L a u.One easily verifies that L 0_1 a is a qualitative tree language.However, L 0_1 a cannot be recognised by a probabilistic tree automaton.
Proposition 51.The language L 0_1 a cannot be recognised by a probabilistic tree automaton (regardless of its semantics).
Proof.By contradiction assume that there is some probabilistic parity tree automaton A " xta, bu, Q, q ini , δ, Coly such that L 0_1 a " L "1 Qual pAq (the proof going exactly the same for the case where L 0_1 a " L "1 pAq we omit it).Let t a (resp.t b ) be the tree whose nodes are all labeled by a (resp.b), i.e. t a puq " a (resp.t b puq " b) for all node u P t0, 1u ˚.Now let t 0 be the tree defined by t 0 pεq " b, t 0 r0s " t a and t 0 r1s " t b ; and t 1 be the tree defined by t 1 pεq " b, t 1 r0s " t b and t 1 r1s " t a .Obviously, both t 0 and t 1 belong to L 0_1 a " L "1 Qual pAq.For any state q P Q, let A q " xta, bu, Q, q, δ, Coly be the automaton obtained from A by changing its initial state to be q.Then one easily checks that a tree t belongs to L "1 Qual pAq if and only if for all pq 0 , q 1 q P Q 2 such that δpq ini , tpεq, q 0 , q 1 q ą 0 one has tr0s P L "1 Qual pA q0 q and tr1s P L "1 Qual pA q1 q.Now, for any pq 0 , q 1 q P Q 2 such that δpq ini , b, q 0 , q 1 q ą 0 one has that t b P L "1 Qual pA q0 q (because of t 1 P L "1 Qual pAq) and t b P L "1 Qual pA q1 q (because of t 0 P L "1 Qual pAq).Therefore the tree t defined by tpεq " b and tr0s " tr1s " t b belongs to L "1 Qual pAq, which leads a contradiction as t " t b and t b R L 0_1 a " L "1 Qual pAq.We now give an example of a regular tree language that no probabilistic tree automaton can accept.For this, let t a (resp.t b ) be the tree whose nodes are all labeled by a (resp.b), i.e. t a puq " a (resp.t b puq " b) for all node u P t0, 1u ˚.Let t 0 be the tree defined by t 0 pεq " b, t 0 r0s " t a and t 0 r1s " t b ; and t 1 be the tree defined by t 1 pεq " b, t 1 r0s " t b and t 1 r1s " t a .Finally let L 0_1 " tt 0 , t 1 u: it is a regular tree language as it consists of two regular trees.However, L 0_1 cannot be recognised by a probabilistic tree automaton.
Proposition 52.The language L 0_1 cannot be recognised by a probabilistic tree automaton (regardless of its semantics).
Proof.One reasons by contradiction exactly as for Proposition 51 and concludes similarly that if there is a probabilistic automaton accepting L 0_1 then it also accepts tree t b R L 0_1 , which leads a contradiction.

Variants
A natural variant is to replace the almost-sure acceptance condition on the set of runs by the probable one.That is a tree t is probably accepted by A if µ t pAccRunspAqq ą 0.
Similarly to Proposition 43, we obtain a connection with ω-word languages.For an ω-word language L Ď ta, bu ω , we denote by Path ą0 pLq the set of trees labeled by ta, bu with a positive set of branch labels in L (i.e.µptπ P Br | tpπq P Luq ą 0).In the following statement, for a probabilistic word automaton B, we denote by L ą0 pBq the language positively-accepted by B.
Proposition 55.Given a probabilistic ω-word automaton B with an ω-regular acceptance condition, there exists a probabilistic tree automaton A with the same acceptance condition such that L ą0 Pos pAq is equal to Path ą0 pL ą0 pBqq.
Proof.Let B " xA, Q, q ini , δ, Accy be a (complete) probabilistic ω-word automaton with an ω-regular condition.Consider the probabilistic tree automaton A simulating B on all branches.Formally A is equal to xA, Q, q ini , δ 1 , F y where, for all p, q P Q and x P A, δ 1 pp, x, q, qq " δpp, x, qq.
Let t be a tree and let g : Br Þ Ñ r0, 1s be the mapping associating to a branch π the value ş f A p¨, πqdµ t .We have the following claim whose proof is identical to that of Claim 44 in Proposition 43.
Claim 56.For any π P Br, gpπq ą 0 if and only if tpπq belongs to L ą0 pBq.
We are now ready to conclude: Pos pAq ô ş f A dµ t b µ ą 0 Proposition 54 ô ş gdµ " 1 Tonelli's Theorem ô gpπq ą 0 on a non-null set rBauer2001, T heorem 13.2 ´p.71s ô µptπ | tpπq P L ą0 pBquq ą 0 Claim 56 ô t P Path ą0 pL ą0 pBqq We can also transfer the results on the acceptance game (Proposition 45 and Theorem 46).Proposition 57.Let A be a probabilistic tree automaton with an ω-regular acceptance condition and let t be a tree.We have t P L ą0 Pos pAq iff M A,t positively fulfils its objective.
Proof.Let AccPairspAq " tpρ, πq | ρpπq P Accu.By Proposition 54, to establish the desired equivalence, it is sufficient to show that µ t b µpAccPairspAqq " µ M A,t pO Acc q.The latter was done already in the proof of Proposition 45.
Theorem 58.Let A be a probabilistic tree automaton with an ω-regular acceptance condition.The language L ą0 Pos pAq is non-empty if and only if Éloïse positively wins in G A .
Proof.The proof only differs from the one of Theorem 46 in its conclusion.Namely, Éloïse positively wins in G A iff there is some t such that ValpG A,t q ą 0 iff t P L ą0 Pos pAq (thanks to Proposition 57).If we consider a co-Büchi acceptance condition, this leads a decidability result for the emptiness problem (which is a dual version of Corollary 47).
AccBrpρq and let u be a node after which all node labels in b belongs to QzF .Then b P Cpuq and for all v Ě u, ρpvq R F : hence, b belongs to Cpuqz Ť uĎv,ρpvqPF Cpvq.Now, let b R AccBrpρq: b contains infinitely many occurrences of forbidden states.Therefore, for all node u in b there is another node v below u on b that is labelled by a state in F , meaning that b P Ť uĎv,ρpvqPF Cpvq.Equivalently, b R Cpuqz Ť uĎv,ρpvqPF Cpvq.

Fig. 2 .
Fig. 2. The arena G A,t of the acceptance game.
], the mapping g : Runs Ñ r0, 1s associating to a run ρ P Runs the value ş Br f A pρ, ¨qdµ is measurable.t belongs to L "1