Reinforcement Learning

REINFORCEMENT LEARNING

AGENDA Online learning Reinforcement learning

Model-free vs. model-based Passive vs. active learning Exploration-exploitation tradeoff

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INCREMENTAL (“ONLINE”) FUNCTION LEARNING Data is streaming into learner

x1,y1, …, xn,yn yi = f(xi) Observes xn+1 and must make prediction for

next time step yn+1 “Batch” approach:

Store all data at step n Use your learner of choice on all data up to time

n, predict for time n+1 Can we do this using less memory?

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EXAMPLE: MEAN ESTIMATION yi = q + error term (no x’s) Current estimate qn = 1/n Si=1…n yi

qn+1 = 1/(n+1) Si=1…n+1 yi = 1/(n+1) (yn+1 + Si=1…n yi) = 1/(n+1) (yn+1 + n qn) = qn + 1/(n+1) (yn+1 - qn)

q5

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EXAMPLE: MEAN ESTIMATION yi = q + error term (no x’s) Current estimate qt = 1/n Si=1…n yi


q5

y6

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EXAMPLE: MEAN ESTIMATION yi = q + error term (no x’s) Current estimate qt = 1/n Si=1…n yi


q5 q6 = 5/6 q5 + 1/6 y6

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EXAMPLE: MEAN ESTIMATION qn+1 = qn + 1/(n+1) (yn+1 - qn) Only need to store n, qn

q5 q6 = 5/6 q6 + 1/6 y6

LEARNING RATES In fact, qn+1 = qn + an (yn+1 - qn) converges to

the mean for any an such that: an 0 as n San San

2 C < O(1/n) does the trick If an is close to 1, then the estimate shifts

strongly to recent data; close to 0, and the old estimate is preserved

REINFORCEMENT LEARNING RL problem: given only

observations of actions, states, and rewards, learn a (near) optimal policy

No prior knowledge of transition or reward models

We consider: fully-observable, episodic environment, finite state space, uncertainty in action (MDP)

WHAT TO LEARN?

Policy p Action-utilityfunction Q(s,a)

Less online deliberation

More online deliberationUtility

function UModel of R and TLearn:

Online: p(s) arg maxa Q(s,a)

Model free Model based

Simpler execution Faster learning?

arg maxaSs P(s’|s,a)U(s’)

Solve MDP

Method: Learning from demonstration

Q-learning, SARSA

Direct utility estimation, TD-learning

Adaptive dynamic programming

FIRST STEPS: PASSIVE RL Observe execution trials of an agent that

acts according to some unobserved policy p Problem: estimate the utility function Up

[Recall Up(s) = E[St gt R(St)] where St is the random variable denoting the distribution of states at time t]

DIRECT UTILITY ESTIMATION

1. Observe trials t(i)=(s0(i),a1

(i),s1(i),r1

(i),…,ati(i),sti

(i),rti(i)) for i=1,…,n

2. For each state sS:3. Find all trials t(i) that pass through s4. Compute subsequent utility Ut(i)(s)=St=k to ti gt-k rt

(i)

5. Set Up(s) to the average observed utility

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0.390.610.660.71

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0.870.81 0.92

ONLINE IMPLEMENTATION

1. Store counts N[s] and estimated utilities Up(s)2. After a trial t, for each state s in the trial:

3. Set N[s] N[s]+14. Adjust utility Up(s) Up(s)+a(N[s])(Ut(s)-Up(s))

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0.390.610.660.71

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0.870.81 0.92

• Simply supervised learning on trials• Slow learning, because Bellman equation is

not used to pass knowledge between adjacent states

TEMPORAL DIFFERENCE LEARNING

1. Store counts N[s] and estimated utilities Up(s)2. For each observed transition (s,r,a,s’):

3. Set N[s] N[s]+14. Adjust utility Up(s) Up(s)+a(N[s])(r+gUp(s’)-Up(s))

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-0.02-0.02 0.48 With learning rate a=0.5




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0.21-0.04 0.72 With learning rate a=0.5




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• For any s, distribution of s’ approaches P(s’|s,p(s))

• Uses relationships between adjacent states to adjust utilities toward equilibrium

• Faster learning than direct estimation

With learning rate a=0.5

“OFFLINE” INTERPRETATION OF TD LEARNING

1. Observe trials t(i)=(s0(i),a1

(i),s1(i),r1

(i),…,ati(i),sti

(i),rti(i)) for i=1,…,n

2. For each state sS:3. Find all trials t(i) that pass through s4. Extract local history at (s,r(i),a(i),s’(i)) for each trial5. Set up constraint Up(s) = r(i) + gUp(s’(i))

6. Solve all constraints in least squares fashion using stochastic gradient descent

[Recall linear system in policy iteration: u = r+Tpu]

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1. Store counts N[s],N[s,a],N[s,a,s’],estimated rewards R(s), and transition model P(s’|s,a)

2. For each observed transition (s,r,a,s’):3. Set N[s] N[s]+1, N[s,a] N[s,a]+1, N[s,a,s’]

N[s,a,s’]+14. Adjust reward R(s) R(s)+a(N[s])(r-R(s))5. Set P(s’|s,a) = N[s,a,s’]/N[s,a]6.Solve policy evaluation using P, R, p

ADAPTIVE DYNAMIC PROGRAMMING

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• Faster learning than TD, because Bellman equation is exploited across all states

• Modified policy evaluation algorithms make updates faster than solving linear system (O(n3))

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R(s)

P(s’|s,a)

ACTIVE RL Rather than assume a policy is given, can we

use the learned utilities to pick good actions? At each state s, the agent must learn

outcomes for all actions, not just the action p(s)

GREEDY RL Maintain current estimates Up(s) Idea: At state s, take action a that maximizesSs’ P(s’|s,a) Up(s’)

Very seldom works well! Why?

EXPLORATION VS. EXPLOITATION Greedy strategy purely exploits its current

knowledge The quality of this knowledge improves only for

those states that the agent observes often A good learner must perform exploration in

order to improve its knowledge about states that are not often observed But pure exploration is useless (and costly) if it is

never exploited

RESTAURANT PROBLEM

OPTIMISTIC EXPLORATION STRATEGY Behave initially as if there were wonderful

rewards scattered all over the place Define a modified optimistic Bellman update

U+(s) R(s)+g maxa f( Ss P(s’|s,a)U+(s’) , N[s,a])

Truncated exploration function:

f(u,n) = R+ if n < Ne u otherwise

[Here the agent will try each action in each state at least Ne times.]

COMPLEXITY Truncated: at least Ne·|S|·|A| steps are

needed in order to explore every action in every state Some costly explorations might not be

necessary, or the reward from far-off explorations may be highly discounted

Convergence to optimal policy guaranteed only if each action is tried in each state an infinite number of times!

This works with ADP… But how to perform action selection in TD? Must also learn the transition model P(s’|s,a)

Q-VALUES Learning U is not enough for action selection

because a transition model is needed Solution: learn Q-values: Q(s,a) is the utility

of choosing action a in state s Shift Bellman equation

U(s) = maxa Q(s,a) Q(s,a) = R(s) + g Ss P(s’|s,a) maxa’ Q(s’,a’)

So far, everything is the same… but what about the learning rule?

Q-LEARNING UPDATE Recall TD:

Update: U(s) U(s)+a(N[s])(r+gU(s’)-U(s)) Select action: a arg maxa f( Ss P(s’|s,a)U(s’) , N[s,a])

Q-Learning: Update: Q(s,a) Q(s,a)+a(N[s,a])(r+g maxa’Q(s’,a’)-

Q(s,a)) Select action: a arg maxa f( Q(s,a) , N[s,a])

Key difference: average over P(s’|s,a) is “baked in” to the Q function

Q-learning is therefore a model-free active learner

MORE ISSUES IN RL Model-free vs. model-based

Model-based techniques are typically better at incorporating prior knowledge

Generalization Value function approximation Policy search methods

LARGE SCALE APPLICATIONS Game playing

TD-Gammon: neural network representation of Q-functions, trained via self-play

Robot control

RECAP Online learning: low memory overhead TD-learning: learn U through incremental

updates. Cheap, somewhat slow learning. ADP: learn P and R, derive U through policy

evaluation. Fast learning but computationally expensive.

Q-learning: learn state-action function Q(s,a), allows model-free action selection

Action selection requires trading off exploration vs. exploitation Infinite exploration needed to guarantee that the

optimal policy is found!

INCREMENTAL LEAST SQUARES Recall Least Squares estimate

q = (ATA)-1 AT b Where A is matrix of x(i)’s, b is vector of y(i)’s

(laid out in rows)

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A = x(1)

x(2)

x(N)

…

b = y(1)

y(2)

y(N)

…NxM Nx1

DELTA RULE FOR LINEAR LEAST SQUARES Delta rule (Widrow-Hoff rule): stochastic

gradient descentq(t+1) = q(t)+a x (y-q(t)Tx)

O(n) time and space

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INCREMENTAL LEAST SQUARES Let A(t), b(t) be A matrix, b vector up to time t

q(t) = (A(t)TA(t))-1 A(t)T b(t)

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A(t+1) =

x(t+1)

b(t+1) =

y(t+1)

(T+1)xM

(t+1)x1

b(t)A(t)


q(t+1) = (A(t+1)TA(t+1))-1 A(t+1)T b(t+1)

A(t+1)T b(t+1) =A(t)T b(t) + y(t+1)x(t+1)

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A(t+1) =

x(t+1)

b(t+1) =

y(t+1)

(T+1)xM

(t+1)x1

b(t)A(t)


q(t+1) = (A(t+1)TA(t+1))-1 A(t+1)T b(t+1)

A(t+1)T b(t+1) =A(t)T b(t) + y(t+1)x(t+1)

A(t+1)TA(t+1) = A(t)TA(t) + x(t+1)x(t+1)T

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A(t+1) =

x(t+1)

b(t+1) =

y(t+1)

(T+1)xM

(t+1)x1

b(t)A(t)


q(t+1) = (A(t+1)TA(t+1))-1 A(t+1)T b(t+1)

A(t+1)T b(t+1) =A(t)T b(t) + y(t+1)x(t+1)


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A(t+1) =

x(t+1)

b(t+1) =

y(t+1)

(T+1)xM

(t+1)x1

b(t)A(t)


q(t+1) = (A(t+1)TA(t+1))-1 A(t+1)T b(t+1)

A(t+1)T b(t+1) =A(t)T b(t) + y(t+1)x(t+1)


Sherman-Morrison Update (Y + xxT)-1 = Y-1 - Y-1

xxT Y-1 / (1 – xT Y-1 x)

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INCREMENTAL LEAST SQUARES Putting it all together Store

p(t) = A(t)Tb(t)

Q(t) = (A(t)TA(t))-1

Updatep(t+1) = p(t) + y x

Q(t+1) = Q(t) - Q(t)

xxT Q(t) / (1 – xT Q(t) x)q(t+1) = Q(t+1)p(t+1)

O(M2) time and space instead of O(M3+MN) time and O(MN) space for OLS

True least squares estimator for any t, (delta rule works only for large t) 44

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Reinforcement Learning