nips nips2011 nips2011-245 knowledge-graph by maker-knowledge-mining
Source: pdf
Author: Odalric-ambrym Maillard, Daniil Ryabko, Rémi Munos
Abstract: The problem of selecting the right state-representation in a reinforcement learning problem is considered. Several models (functions mapping past observations to a finite set) of the observations are given, and it is known that for at least one of these models the resulting state dynamics are indeed Markovian. Without knowing neither which of the models is the correct one, nor what are the probabilistic characteristics of the resulting MDP, it is required to obtain as much reward as the optimal policy for the correct model (or for the best of the correct models, if there are several). We propose an algorithm that achieves that, with a regret of order T 2/3 where T is the horizon time. 1
Reference: text
sentIndex sentText sentNum sentScore
1 net Abstract The problem of selecting the right state-representation in a reinforcement learning problem is considered. [sent-6, score-0.164]
2 Without knowing neither which of the models is the correct one, nor what are the probabilistic characteristics of the resulting MDP, it is required to obtain as much reward as the optimal policy for the correct model (or for the best of the correct models, if there are several). [sent-8, score-0.511]
3 We propose an algorithm that achieves that, with a regret of order T 2/3 where T is the horizon time. [sent-9, score-0.48]
4 1 Introduction We consider the problem of selecting the right state-representation in an average-reward reinforcement learning problem. [sent-10, score-0.164]
5 Each state-representation is defined by a model φj (to which corresponds a state space Sφj ) and we assume that the number J of available models is finite and that (at least) one model is a weakly-communicating Markov decision process (MDP). [sent-11, score-0.128]
6 This problem is considered in the general reinforcement learning setting, where an agent interacts with an unknown environment in a single stream of repeated observations, actions and rewards. [sent-13, score-0.181]
7 Our goal is to construct an algorithm that performs almost as well as the algorithm that knows both which model is a MDP (knows the “true” model) and the characteristics of this MDP (the transition probabilities and rewards). [sent-15, score-0.173]
8 Given a potentially large number of feature combinations of this kind, we want to find a policy whose average reward is as good as that of the best policy for the right combination of features. [sent-21, score-0.45]
9 Again, we would like to find a policy that is as good as the optimal policy for the right discretization. [sent-28, score-0.37]
10 In fact, the constructed algorithm never “knows” which model is the right one; it is “only” able to get the same average level of reward as if it knew. [sent-33, score-0.161]
11 Such a problem has been pioneered in the reinforcement learning literature by [7] and then improved in various ways by [4, 11, 12, 6, 3]; UCRL2 achieves a regret of the order DT 1/2 in any weakly-communicating MDP with diameter D, with respect to the best policy for this MDP. [sent-37, score-0.861]
12 The diameter D of a MDP is defined in [6] as the expected minimum time required to reach any state starting from any other state. [sent-38, score-0.188]
13 A similar approach has been considered in [10]; the difference is that in that work the probabilistic characteristics of each model are completely known, but the models are not assumed to be Markovian, and belong to a countably infinite (rather than finite) set. [sent-40, score-0.206]
14 In the usual bandit setting, the rewards are assumed to be i. [sent-44, score-0.304]
15 thus one can estimate the mean value of the arms while switching arbitrarily from one arm to the next (the quality of the estimate only depends on the number of pulls of each arm). [sent-47, score-0.127]
16 However, in our setting, estimating the average-reward of a policy requires playing it many times consecutively. [sent-48, score-0.24]
17 We show that despite the fact that the true Markov model of states is unknown and that nothing is assumed on the wrong representations, it is still possible to derive a finite-time analysis of the regret for this problem. [sent-51, score-0.605]
18 This is stated in Theorem 1; the bound on the regret that we obtain is of order T 2/3 . [sent-52, score-0.486]
19 The intuition is that if the “true” model φ∗ is known, but its probabilistic properties are not, then we still know that there exists an optimal control policy that depends on the observed state sj ∗ ,t only. [sent-53, score-0.318]
20 Therefore, the optimal rate of rewards can be obtained by a clever exploration/exploitation strategy, such as UCRL2 algorithm [6]. [sent-54, score-0.221]
21 Since we do not know in advance which model is a MDP, we need to explore them all, for a sufficiently long time in order to estimate the rate of rewards that one can get using a good policy in that model. [sent-55, score-0.461]
22 2 Notation and definitions We consider a space of observations O, a space of actions A, and a space of rewards R (all assumed def to be Polish). [sent-61, score-0.533]
23 Definition 1 We say that an environment e with a state representation function φ is Markov, or, for short, that φ is a Markov model (of e), if the process (st,φ , at , rt ), t ∈ N is a (weakly communicating) Markov decision process. [sent-69, score-0.227]
24 For that purpose we define the regret of any strategy at time T , like in [6, 3], as T def ∆(T ) = T ρ − rt , t=1 where rt are the rewards received when following the proposed strategy and ρ is the average optimal T 1 value in the best Markov model, i. [sent-74, score-1.171]
25 , ρ = limT T E( t=1 rt (π )) where rt (π ) are the rewards received when following the optimal policy for the best Markov model. [sent-76, score-0.602]
26 Note that this definition makes sense since when the MDP is weakly communicating, the average optimal value of reward does not depend on the initial state. [sent-77, score-0.166]
27 Also, one could replace T ρ∗ with the expected sum of rewards √ obtained in T steps (following the optimal policy) at the price of an additional O( T ) term. [sent-78, score-0.253]
28 In the next subsection, we describe an algorithm that achieves a sub-linear regret of order T 2/3 . [sent-79, score-0.418]
29 Each stage consists in 2 phases: an exploration and an exploitation phase. [sent-83, score-0.77]
30 In the exploration phase, BLB plays the UCRL2 algorithm on each model (φj )1 j J successively, as if each model φj was a Markov model, for a fixed number τi,1,J of rounds. [sent-84, score-0.242]
31 The exploitation part consists in selecting first the model with highest lower bound, according to the empirical rewards obtained in the previous exploration phase. [sent-85, score-0.788]
32 This model is initially selected for the same time as in the exploration phase, and then a test decides to either continue playing this model (if its performance during exploitation is still above the corresponding lower bound, i. [sent-86, score-0.712]
33 if the rewards obtained are still at least as good as if it was playing the best model). [sent-88, score-0.276]
34 Until the exploitation phase (of fixed length τi,2 ) finishes and the next stage starts. [sent-90, score-0.757]
35 3 Parameters: f, δ For each stage i 1 do Set the total length of stage i to be τi := 2i . [sent-91, score-0.649]
36 – Compute the corresponding average empirical reward µi,1 (φj ) received during this exploration phase. [sent-99, score-0.359]
37 While the current length of the exploitation part is less than τi,2 do – Select j = argmax µi,1 (φj ) − 2B(i, φj , δ) (using (3)). [sent-106, score-0.4]
38 j∈J – Run UCRL2 with parameter δi (δ) using φj : update at each time step t the current average empirical reward µi,2,t (φj ) from the beginning of the run. [sent-107, score-0.131]
39 Provided that the length of the current run is larger than τi,1,J , do the test µi,2,t (φj ) µi,1 (φj ) − 2B(i, φj , δ) . [sent-108, score-0.163]
40 def The length of stage i is fixed and defined to be τi = 2i . [sent-115, score-0.66]
41 Thus for a total time horizon T , the number def of stages I(T ) before time T is I(T ) = log2 (T + 1) . [sent-116, score-0.494]
42 Each stage i (of length τi ) is further decomposed into an exploration (length τi,1 ) and an exploitation (length τi,2 ) phases. [sent-117, score-0.847]
43 All the models {φj }j J are played one after another for the same amount of def τ τi,1,J = i,1 . [sent-119, score-0.391]
44 J time Each episode 1 j J consists in running the UCRL2 algorithm using the model of states and transitions induced by the state-representation function φj . [sent-120, score-0.188]
45 Note that UCRL2 does not require the horizon T in advance, but requires a parameter p in order to ensure a near optimal regret bound with probability higher than 1 − p. [sent-121, score-0.646]
46 We define this parameter p to be δi (δ) in stage i, where def δi (δ) = (2i − (J −1 + 1)22i/3 + 4)−1 2−i+1 δ . [sent-122, score-0.583]
47 The average empirical reward received during each episode is written µi,1 (φj ). [sent-123, score-0.306]
48 We use the empirical rewards µi,1 (φj ) received in the previous exploration part of stage i together with a confidence bound in order to select the model to play. [sent-125, score-0.824]
49 Moreover, a model φ is no longer run for a fixed period of time (as in the exploration part of stage i), but for a period τi,2 (φ) that depends on some test; we first initialize J := {1, . [sent-126, score-0.523]
50 , J} and then choose def j = argmax µi,1 (φj ) − 2B(i, φj , δ) , (2) j∈J where we define τ def B(i, φ, δ) = 34f (τi − 1 + τi,1 )|Sφ | A log( δi,1,J ) i (δ) , (3) τi,1,J where δ and the function f are parameters of the BLB algorithm. [sent-129, score-0.624]
51 In parallel we test whether the average empirical reward we receive during this exploitation phase is high enough; at time t, if the length of the current episode is larger than τ1,i,J , we test if µi,2,t (φj ) µi,1 (φj ) − 2B(i, φj , δ). [sent-131, score-0.826]
52 Now, if the test fails, then the model j is discarded (until the end of stage i) i. [sent-133, score-0.347]
53 We repeat those steps until the total time τi,2 of the exploitation phase of stage i is over. [sent-136, score-0.742]
54 4 Remark Note that the model selected for exploitation in (2) is the one that has the best lower bound. [sent-137, score-0.381]
55 We know that if the right model is selected, then with high probability, this model will be kept during the whole exploitation phase. [sent-139, score-0.452]
56 If this is not the right model, then either the policy provides good rewards and we should keep playing it, or it does not, in which case it will not pass the test (4) and will be removed from the set of models that will be exploited in this phase. [sent-140, score-0.561]
57 If there are several such models, let φ be the one with the highest average reward of the optimal policy of the corresponding MDP. [sent-143, score-0.289]
58 Importantly, the algorithm considered here does not know in advance the diameter D of the true model, nor the time horizon T . [sent-148, score-0.34]
59 log(t)) on this diameter, which result in the additional regret term c(f, D) and the additional factor f (T ); knowing D would enable to remove both of them, but this is a strong assumption. [sent-151, score-0.418]
60 Choosing f (t) := log2 t + 1 gives a bound which is of order T 2/3 in T but is exponential in D; taking f (t) := tε we get a bound of order T 2/3+ε in T but of polynomial order 1/ε in D. [sent-152, score-0.136]
61 The “wrong” models are used during exploitation stages only as long as they are giving rewards that are higher than the rewards that could be obtained in the “true” model. [sent-155, score-0.882]
62 All the models are explored sufficiently long so as to be able to estimate the optimal reward level in the true model, and to learn its policy. [sent-156, score-0.21]
63 As long as the rewards are high we do not care where the model we are using is correct or not. [sent-163, score-0.256]
64 While passing from a finite to a countably infinite set appears rather straightforward, getting rid of the assumption that this set contains the true model seems more difficult. [sent-167, score-0.128]
65 What one would want to obtain in this setting is sub-linear regret with respect to the performance of the optimal policy in the best model; this, however, seems difficult without additional assumptions on the probabilistic characteristics of the models. [sent-168, score-0.689]
66 The reader familiar with adversarial bandit literature will notice that our bound of order T 2/3 is worse than T 1/2 that usually appears in this context (see, for example [2]). [sent-173, score-0.189]
67 The reason is that our notion of regret is different: in adversarial bandit literature, the regret is measured with respect to the best choice of the arm for the given fixed history. [sent-174, score-0.997]
68 In contrast, we measure the regret with respect to the best policy (for knows the correct model and its parameters) that, in general, would obtain completely different (from what our algorithm would get) rewards and observations right from the beginning. [sent-175, score-0.937]
69 As previously mentioned, a possibly large additive constant c(f, D) appears in the regret since we do not known a bound on the diameter of the MDP in the “true” model, and use log T instead. [sent-177, score-0.67]
70 Finding a way to properly address this problem by estimating online the diameter of the MDP is an interesting open question. [sent-178, score-0.151]
71 First, we notice that, as reported in [6], when we compute an optimistic model based on the empirical rewards and transitions of the true model, the span of the corresponding optimistic value function sp(V + ) is always smaller than the diameter D. [sent-180, score-0.681]
72 This span increases as we get more rewards and transitions samples, which gives a natural empirical lower bound on D. [sent-181, score-0.376]
73 In [3], the authors derive a regret bound that scales with the span of the true value function sp(V ), which is also less than D, and can be significantly smaller in some cases. [sent-183, score-0.595]
74 However, since we do not have the property that sp(V + ) sp(V ), we need to introduce an explicit penalization in order to control the span of the computed optimistic models, and this requires assuming we know an upper bound B on sp(V ) in order to guarantee a final regret bound scaling with B. [sent-184, score-0.737]
75 Then we decompose the regret into the sum of the regret in each stage i. [sent-189, score-1.107]
76 After analyzing the contribution to the regret in stage i, we then gather all stages and tune the length of each stage and episode in order to get the final regret bound. [sent-190, score-1.698]
77 What is interesting is that this diameter does not need to be known by the algorithm. [sent-193, score-0.151]
78 Also by carefully looking at the proof of UCRL, it can be shown that the following bound is also valid with probability higher than 1 − δ : 1 τ t1 +τ −1 rt − ρ 34D|Sφ | t=t1 τ A log( δ ) . [sent-194, score-0.226]
79 τ We now define the following quantity, for every model φ, episode length τ and δ ∈ (0, 1) τ A log( δ ) def BD (τ, φ, δ ) = 34D|Sφ | . [sent-195, score-0.545]
80 2 Regret of stage i In this section we analyze the regret of the stage i, which we denote ∆i . [sent-197, score-0.96]
81 Note that since each stage i I is of length τi = 2i except the last one I that may stop before, we have I(T ) ∆(T ) = ∆i , (8) i=1 where I(T ) = log2 (T +1) . [sent-198, score-0.373]
82 We further decompose ∆i = ∆1,i +∆i,2 into the regret corresponding to the exploration stage ∆1,i and the regret corresponding to the exploitation stage ∆i,2 . [sent-199, score-1.877]
83 τi,1 is the total length of the exploration stage i and τi,2 is the total length of the exploitation stage i. [sent-200, score-1.255]
84 def τ For each model φ, we write τi,1,J = i,1 the number of consecutive steps during which the UCRL2 J algorithm is run with model φ in the exploration stage i, and τi,2 (φ) the number of consecutive steps during which the UCRL2 algorithm is run with model φ in the exploitation stage i. [sent-201, score-1.678]
85 Let us now introduce the two following sets of models, defined after the end of the exploration stage, i. [sent-203, score-0.176]
86 def Gi = {φ ∈ Φ ; µi,1 (φ) − 2B(i, φ, δ) ≥ µi,1 (φ ) − 2B(i, φ , δ)}\{φ∗ } , def Bi = {φ ∈ Φ ; µi,1 (φ) − 2B(i, φ, δ) < µi,1 (φ ) − 2B(i, φ , δ)} . [sent-206, score-0.624]
87 1 Regret in the exploration phase Since in the exploration stage i each model φ is run for τi,1,J many steps, the regret for each model φ = φ is bounded by τi,1,J ρ . [sent-210, score-1.272]
88 Now the regret for the true model is τi,1,J (ρ − µ1 (φ )), thus the total contribution to the regret in the exploration stage i is upper-bounded by ∆i,1 τi,1,J (ρ − µ1 (φ )) + (J − 1)τi,1,J ρ . [sent-211, score-1.433]
89 2 Regret in the exploitation phase By definition, all models in Gi ∪ {φ } are selected before any model in Bi is selected. [sent-214, score-0.492]
90 Let us consider some φ ∈ Gi and an event Ωi under which the exploitation phase does not reset. [sent-216, score-0.492]
91 We deduce that the total contribution to the regret of all the models φ ∈ Gi in the exploitation stages on the event Ωi is bounded by max{τi,1,J ρ , (τi,2 (φ) − 1)(ρ − µi,1 (φ ) + 2B(i, φ , δ)) + ρ } . [sent-218, score-1.108]
92 Now we show that, with high probability, the true model φ passes all the tests (equation (4)) until the end of the episode i, and thus equivalently, with high probability no model φ ∈ Bi is selected, so that τi,2 (φ) = 0. [sent-222, score-0.302]
93 φ∈Bi For the true model, after τ (φ , t) τi,1,J , there remains at most (τi,2 −τi,1,J +1) possible timesteps where we do the test for the true model φ . [sent-223, score-0.19]
94 Now provided that f (ti ) D, then BD (τi,1,J , φ , δi (δ)) B(i, φ , δ) , thus the true model passes all tests until the end of the exploitation part of stage i on an event Ω3,i of probability higher than def 1 − (τi,2 − τi,1,J + 2)δi (δ). [sent-226, score-1.185]
95 The total length of stage i is by definition τi = τi,1 + τi,2 = τi,1,J J + τi,2 , def def 2/3 τ 2/3 2/3 where τi = 2i . [sent-237, score-1.002]
96 A log def We now define δi (δ) such that δi (δ) = (2i − (J −1 + 1)22i/3 + 4)−1 2−i+1 δ . [sent-240, score-0.345]
97 We conclude by using the fact that since I(T ) 1 − δ, the following bound on the regret holds ∆(T ) A log cf (T )S AJ log(Jδ)−1 log2 (T ) 1/2 log2 (T + 1), then with probability higher than T 2/3 + c DS A log(δ −1 ) log2 (T )T 1/2 + c(f, D) . [sent-243, score-0.624]
98 def for some constant c, c , and where c(f, D) = i∈I0 2i ρ . [sent-244, score-0.312]
99 REGAl: a regularization based algorithm for reinforcement learning in weakly communicating mdps. [sent-258, score-0.211]
100 Optimistic linear programming gives logarithmic regret for irreducible mdps. [sent-292, score-0.418]
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