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1 Advanced Topics in Machine Learning, GI13, 2010/11 Advanced Topics in Machine Learning, GI13, 2010/11 Answer any THREE questions. Each question is worth 20 marks. Use separate answer books Answer any THREE questions. Each question is worth 20 marks. Use separate answer books for PART A and PART B. Gatsby PhD students only: answer either TWO questions from for PART A and PART B. Gatsby PhD students only: answer either TWO questions from PART A and ONE question from PART B; or ONE question from PART and TWO questions PART A and ONE question from PART B; or ONE question from PART A and TWO questions from PART B. from PART B. Marks Marks for for each each part part of of each each question question are are indicated indicated in in square square brackets brackets Calculators Calculators are arenot NOTpermitted PartA: A: Kernel Methods PartB: B: Reinforcement Learning Considerthe the following Markov Decision Process (MDP) with discount factor γ = 0.5. Upper case lettersa, A, B, B, C represent states; arcs represent state transitions; lower case letters ab, ab, ba, ba, bc, bc, ca, ca, cb cb represent actions; signed integers represent rewards; and fractions represent transition probabilities. +2 ba A ab -8 bc B 3/4 1/4 +4 ca C cb Define the thestate-value function V π (s) for a discounted MDP Write down V π (s) the= Bellman E expectation equation for state-value functions π [r t+1 + γr t s t = s] Write down the Bellman expectation equation for state-value functions GI13 1 TURN OVER GI13 1 TURN OVER
2 V π (s) = E π [r t+1 + γv π (s t+1 ) s t = s] or ( ) V π (s) = π(s, a) Rs a + γ Pss a V π (s ) a A s S V π = R π + γp π V π or Consider the uniform random policy π 1 (s,a) that takes all actions from state s with equal probability. Starting with an initial value function of V 1 (A) =V 1 (B) =V 1 (C) = 2, apply one synchronous iteration of iterative policy evaluation (i.e. one backup for each state) to compute a new value function V 2 (s) V 2 (A) = V 1 (B) = 7 V 2 (B) = 0.5( V 1 (A)) + 0.5( V 1 (C)) = 1 V 2 (C) = 0.5( V 1 (B)) + 0.5( (1/4V 1 (A) + 3/4V 1 (C))) = 7 Apply one iteration of greedy policy improvement to compute a new, deterministic policy π 2 (s) π 2 (A) = ab Q 2 (B,ba) = V 2 (A) = 1.5, Q 2 (B,bc) = V 2 (C) = 1.5 = π 2 (B) = bc Q 2 (C,ca) = (1/4V 2 (A) + 3/4V 2 (B)) = 5.75, Q 2 (C,cb) = V 2 (B) = 8.5 = π 2 (C) = cb Consider a deterministic policy π(s). Prove that if a new policy π is greedy with respect to V π then it must be better than or equal to π, i.e. V π (s) V π (s) for all s; and that if V π (s) = V π (s) for all s then π must be an optimal policy. [5 marks] Greedy policy improvement is given by π (s) = argmax Q π (s,a). This is an improvement over one step because for any state s, Q π (s,π (s)) = max a A Qπ (s,a) GI13 2 CONTINUED a A
3 Q π (s,π(s)) =V π (s). It therefore improves the value function, V π (s) Q π (s,π (s)) = E π [r t+1 +γv π (s t+1 ) s t = s] E π [r t+1 +γq π (s t+1,π (s t+1 )) s t = s]... E π [r t+1 + γr t s t = s] = V π (s). If improvements stop, i.e. Q π (s,π (s)) = max a A Qπ (s,a) = Q π (s,π(s)) = V π (s) then V π has satisfied the Bellman optimality equation, so π and must be an optimal policy. Define the optimal state-value function V (s) for an MDP V (s) = max V π (s) π Write down the Bellman optimality equation for state-value functions V (s) = max a V (s) = max π V = max a Rs a + γ Pss a V (s ) or s S E π [r t+1 + γv (s t+1 ) s t = s] or R a + γp a V Starting with an initial value function of V 1 (A) = V 1 (B) = V 1 (C) = 2, apply one synchronous iteration of value iteration (i.e. one backup for each state) to compute a new value function V 2 (s). V 2 (A) = 7,V 2 (B) = 3,V 2 (C) = 9 Is your new value function V 2 (s) optimal? Justify your answer. Applying one more iteration, V 3 (A) = 6.5 V 2 (A) hence V 2 is not a fixed point of the Bellman optimality equation. [Total 20 marks] GI13 3 TURN OVER
4 2. Consider an undiscounted Markov Reward Process with two states A and B. The transition matrix and reward function are unknown, but you have observed two sample episodes: A + 3 A + 2 B 4 A + 4 B 3 terminate B 2 A + 3 B 3 terminate In the above episodes, sample state transitions and sample rewards are shown at each step, e.g. A + 3 A indicates a transition from state A to state A, with a reward of +3. Using first-visit Monte-Carlo evaluation, estimate the state-value function V (A),V (B) V (A) = 1/2(2 + 0) = 1 V (B) = 1/2( 3 + 2) = 5/2 Using every-visit Monte-Carlo evaluation, estimate the state-value function V (A),V (B) V (A) = 1/4( ) = 1/2 V (B) = 1/4( ) = 11/4 Draw a diagram of the Markov Reward Process that best explains these two episodes (i.e. the model that maximises the likelihood of the data - although it is not necessary to prove this fact). Show rewards and transition probabilities on your diagram. [4 marks] 1/4 r=3 A B term 3/4 r=-3 1/2 1/2 GI13 4 CONTINUED
5 Define the Bellman equation for a Markov reward process V (s) = E[r t+1 + γv (s t+1 ) s t = s] or V (s) = R s + γ P ss V π (s ) or s S V = R + γpv Solve the Bellman equation to give the true state-value function V (A),V (B). Hint: solve the Bellman equations directly, rather than iteratively. V (A) = 3 + 1/4V (A) + 3/4V (B) V (B) = 3 + 1/2V (A) V (A) = 2 V (B) = 2 [4 marks] What value function would batch TD(0) find, i.e. if TD(0) was applied repeatedly to these two episodes? The solution to the above MDP, V (A) = 2 V (B) = 2 What value function would batch TD(1) find, using accumulating eligibility traces? The same as every-visit Monte-Carlo V (A) = 1/2 V (B) = 11/4 GI13 5 TURN OVER
6 What value function would LSTD(0) find? The same as batch TD(0) V (A) = 2 V (B) = 2 [Total 20 marks] GI13 6 CONTINUED
7 3. A rat is involved in an experiment. It experiences one episode. At the first step it hears a bell. At the second step it sees a light. At the third step it both hears a bell and sees a light. It then receives some food, worth +1 reward, and the episode terminates on the fourth step. All other rewards were zero. The experiment is undiscounted. Represent the rat s state s by a vector of two binary features, bell(s) {0,1} and light(s) {0,1}. Write down the sequence of feature vectors corresponding to this episode. 1, 0, Approximate the state-value function by a linear combination of these features with two parameters: b bell(s) + l light(s). If b = 2 and l = 2 then write down the sequence of approximate values corresponding to this episode. 2, 2,0 and also 0 for the terminal state Define the λ-return v λ t v (n) t = r t+1 + γr t γ n 1 r t+n + γ n V (s t+n ) v λ t = (1 λ) n=1 λ n 1 v (n) t Write down the sequence of λ-returns v λ t corresponding to this episode, for λ = 0.5 and b = 2,l = 2 v λ 1 v λ 2 v λ 3 = 0.5( ) = 3/4 = 0.5( ) = 1/2 = 0.5(2 1) = 1 GI13 7 TURN OVER
8 Using the forward-view TD(λ) algorithm and your linear function approximator, what are the sequence of updates to weight b? What is the total update to weight b? Use λ = 0.5,γ = 1,α = 0.5 and start with b = 2,l = 2 b 1 = α(v λ 1 V (s 1))bell(s 1 ) = 0.5( 3/4 2)1 = 11/8 b 2 = α(v λ 2 V (s 2))bell(s 2 ) = 0.5(1/2 2)0 = 0 b 3 = α(v λ 3 V (s 3))bell(s 3 ) = 0.5(1 0)1 = 1/2 b = ( 11/8 + 1/2) = 7/8 Define the TD(λ) accumulating eligibility trace e t when using linear value function approximation e t = γλe t 1 + φ(s) Write down the sequence of eligibility traces e t corresponding to the bell, using λ = 0.5,γ = 1 1, 1/2, 5/4 Using the backward-view TD(λ) algorithm and your linear function approximator, what are the sequence of updates to weight b? (Use offline updates, i.e. do not actually change your weights, just accumulate your updates). What is the total update to weight b? Use λ = 0.5,γ = 1,α = 0.5 and start with b = 2,l = 2 b 1 = αδ 1 e 1 = 0.5( )1 = 2 b 2 = αδ 2 e 2 = 0.5( )1/2 = 1/2 b 3 = αδ 3 e 3 = 0.5( )5/4 = 5/8 b = ( 2 + 1/2 + 5/8) = 7/8 [Total 20 marks]
9 END OF PAPER GI13 8 GI13 9 TURN OVER
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