I am just getting start with deep reinforcement learning and i am trying to crasp this concept.
I have this deterministic bellman equation
When i implement stochastacity from the MDP then i get 2.6a
My equation is this assumption correct. I saw this implementation 2.6a without a policy sign on the state value function. But to me this does not make sense due to i am using the probability of which different next steps i could end up in. Which is the same as saying policy, i think. and if yes 2.6a is correct, can i then assume that the rest (2.6b and 2.6c) because then i would like to write the action state function like this:
The reason why i am doing it like this is because i would like to explain myself from a deterministic point of view to a non-deterministic point of view.
I hope someone out there can help on this one!
Best regards SΓΈren Koch
No, the value function V(s_t) does not depend on the policy. You see in the equation that it is defined in terms of an action a_t that maximizes a quantity, so it is not defined in terms of actions as selected by any policy.
In the nondeterministic / stochastic case, you will have that sum over probabilities multiplied by state-values, but this is still independent from any policy. The sum only sums over different possible future states, but every multiplication involves exactly the same (policy-independent) action a_t. The only reason why you have these probabilities is because in the nondeterministic case a specific action in a specific state can lead to one of multiple different possible states. This is not due to policies, but due to stochasticity in the environment itself.
There does also exist such a thing as a value function for policies, and when talking about that a symbol for the policy should be included. But this is typically not what is meant by just "Value function", and also does not match the equation you have shown us. A policy-dependent function would replace the max_{a_t} with a sum over all actions a, and inside the sum the probability pi(s_t, a) of the policy pi selecting action a in state s_t.
Yes, your assumption is completely right. In the Reinforcement Learning field, a value function is the return obtained by starting for a particular state and following a policy Ο . So yes, strictly speaking, it should be accompained by the policy sign Ο .
The Bellman equation basically represents value functions recursively. However, it should be noticed that there are two kinds of Bellman equations:
Bellman optimality equation, which characterizes optimal value functions. In this case, the value function it is implicitly associated with the optimal policy. This equation has the non linear maxoperator and is the one you has posted. The (optimal) policy dependcy is sometimes represented with an asterisk as follows:
Maybe some short texts or papers omit this dependency assuming it is obvious, but I think any RL text book should initially include it. See, for example, Sutton & Barto or Busoniu et al. books.
Bellman equation, which characterizes a value function, in this case associated with any policy Ο:
In your case, your equation 2.6 is based on the Bellman equation, therefore it should remove the max operator and include the sum over all actions and possible next states. From Sutton & Barto (sorry by the notation change wrt your question, but I think it's understable):
Related
I have a confusion about the way the LSTM networks work when forecasting with an horizon that is not finite but I'm rather searching for a prediction in whatever time in future. In physical terms I would call it the evolution of the system.
Suppose I have a time series $y(t)$ (output) I want to forecast, and some external inputs $u_1(t), u_2(t),\cdots u_N(t)$ on which the series $y(t)$ depends.
It's common to use the lagged value of the output $y(t)$ as input for the network, such that I schematically have something like (let's consider for simplicity just lag 1 for the output and no lag for the external input):
[y(t-1), u_1(t), u_2(t),\cdots u_N(t)] \to y(t)
In this way of thinking the network, when one wants to do recursive forecast it is forced to use the predicted value at the previous step as input for the next step. In this way we have an effect of propagation of error that makes the long term forecast badly behaving.
Now, my confusion is, I'm thinking as a RNN as a kind of an (simple version) implementation of a state space model where I have the inputs, my output and one or more state variable responsible for the memory of the system. These variables are hidden and not observed.
So now the question, if there is this kind of variable taking already into account previous states of the system why would I need to use the lagged output value as input of my network/model ?
Getting rid of this does my long term forecast would be better, since I'm not expecting anymore the propagation of the error of the forecasted output. (I guess there will be anyway an error in the internal state propagating)
Thanks !
Please see DeepAR - a LSTM forecaster more than one step into the future.
The main contributions of the paper are twofold: (1) we propose an RNN
architecture for probabilistic forecasting, incorporating a negative
Binomial likelihood for count data as well as special treatment for
the case when the magnitudes of the time series vary widely; (2) we
demonstrate empirically on several real-world data sets that this
model produces accurate probabilistic forecasts across a range of
input characteristics, thus showing that modern deep learning-based
approaches can effective address the probabilistic forecasting
problem, which is in contrast to common belief in the field and the
mixed results
In this paper, they forecast multiple steps into the future, to negate exactly what you state here which is the error propagation.
Skipping several steps allows to get more accurate predictions, further into the future.
One more thing done in this paper is predicting percentiles, and interpolating, rather than predicting the value directly. This adds stability, and an error assessment.
Disclaimer - I read an older version of this paper.
I'm a beginner in RL and want to know what is the advantage of having a state value function as well as an action-value function in RL algorithms, for example, Markov Design Process. What is the use of having both of them in prediction and control problems?
I think you mean state-value function and state-action-value function.
Quoting this answer by James MacGlashan:
To explain, lets first add a point of clarity. Value functions
(either V or Q) are always conditional on some policy π. To emphasize
this fact, we often write them as ππ(π ) and ππ(π ,π). In the
case when weβre talking about the value functions conditional on the
optimal policy πβ, we often use the shorthand πβ(π ) and πβ(π ,π).
Sometimes in literature we leave off the π or * and just refer to V
and Q, because itβs implicit in the context, but ultimately, every
value function is always with respect to some policy.
Bearing that in mind, the definition of these functions should clarify
the distinction for you.
ππ(π ) expresses the expected value of following policy π forever
when the agent starts following it from state π .
ππ(π ,π) expresses the expected value of first taking action π
from state π and then following policy π forever.
The main difference then, is the Q-value lets you play a hypothetical
of potentially taking a different action in the first time step than
what the policy might prescribe and then following the policy from the
state the agent winds up in.
For example, suppose in state π Iβm one step away from a terminating
goal state and I get -1 reward for every transition until I reach the
goal. Suppose my policy is the optimal policy so that it always tells
to me walk toward the goal. In this case, ππ(π )=β1 because Iβm just
one step away. However, if I consider the Q-value for an action π
that walks 1 step away from the goal, then ππ(π ,π)=β3 because
first I walk 1 step away (-1), and then I follow the policy which will
now take me two steps to get to the goal: one step to get back to
where I was (-1), and one step to get to the goal (-1), for a total of
-3 reward.
I'm trying to learn policy gradient methods for reinforcement learning but I stuck at the score function part.
While searching for maximum or minimum points in a function, we take the derivative and set it to zero, then look for the points that holds this equation.
In policy gradient methods, we do it by taking the gradient of the expectation of trajectories and we get:
Objective function image
Here I could not get how this gradient of log policy shifts the distribution (through its parameters ΞΈ) to increase the scores of its samples mathematically? Don't we look for something that make this objective function's gradient zero as I explained above?
What you want to maximize is
J(theta) = int( p(tau;theta)*R(tau) )
The integral is over tau (the trajectory) and p(tau;theta) is its probability (i.e., of seeing the sequence state, action, next state, next action, ...), which depends on both the dynamics of the environment and the policy (parameterized by theta). Formally
p(tau;theta) = p(s_0)*pi(a_0|s_0;theta)*P(s_1|s_0,a_0)*pi(a_1|s_1;theta)*P(s_2|s_1,a_1)*...
where P(s'|s,a) is the transition probability given by the dynamics.
Since we cannot control the dynamics, only the policy, we optimize w.r.t. its parameters, and we do it by gradient ascent, meaning that we take the direction given by the gradient. The equation in your image comes from the log-trick df(x)/dx = f(x)*d(logf(x))/dx.
In our case f(x) is p(tau;theta) and we get your equation. Then since we have access only to a finite amount of data (our samples) we approximate the integral with an expectation.
Step after step, you will (ideally) reach a point where the gradient is 0, meaning that you reached a (local) optimum.
You can find a more detailed explanation here.
EDIT
Informally, you can think of learning the policy which increases the probability of seeing high return R(tau). Usually, R(tau) is the cumulative sum of the rewards. For each state-action pair (s,a) you therefore maximize the sum of the rewards you get from executing a in state s and following pi afterwards. Check this great summary for more details (Fig 1).
Learner might be in training stage, where it update Q-table for bunch of epoch.
In this stage, Q-table would be updated with gamma(discount rate), learning rate(alpha), and action would be chosen by random action rate.
After some epoch, when reward is getting stable, let me call this "training is done". Then do I have to ignore these parameters(gamma, learning rate, etc) after that?
I mean, in training stage, I got an action from Q-table like this:
if rand_float < rar:
action = rand.randint(0, num_actions - 1)
else:
action = np.argmax(Q[s_prime_as_index])
But after training stage, Do I have to remove rar, which means I have to get an action from Q-table like this?
action = np.argmax(self.Q[s_prime])
Once the value function has converged (values stop changing), you no longer need to run Q-value updates. This means gamma and alpha are no longer relevant, because they only effect updates.
The epsilon parameter is part of the exploration policy (e-greedy) and helps ensure that the agent visits all states infinitely many times in the limit. This is an important factor in ensuring that the agent's value function eventually converges to the correct value. Once we've deemed the value function converged however, there's no need to continue randomly taking actions that our value function doesn't believe to be best; we believe that the value function is optimal, so we extract the optimal policy by greedily choosing what it says is the best action in every state. We can just set epsilon to 0.
Although the answer provided by #Nick Walker is correct, here it's some additional information.
What you are talking about is closely related with the concept technically known as "exploration-exploitation trade-off". From Sutton & Barto book:
The agent has to exploit what it already knows in order to obtain
reward, but it also has to explore in order to make better action
selections in the future. The dilemma is that neither exploration nor
exploitation can be pursued exclusively without failing at the task.
The agent must try a variety of actions and progressively favor those
that appear to be best.
One way to implement the exploration-exploitation trade-off is using epsilon-greedy exploration, that is what you are using in your code sample. So, at the end, once the agent has converged to the optimal policy, the agent must select only those that exploite the current knowledge, i.e., you can forget the rand_float < rar part. Ideally you should decrease the epsilon parameters (rar in your case) with the number of episodes (or steps).
On the other hand, regarding the learning rate, it worths noting that theoretically this parameter should follow the Robbins-Monro conditions:
This means that the learning rate should decrease asymptotically. So, again, once the algorithm has converged you can (or better, you should) safely ignore the learning rate parameter.
In practice, sometimes you can simply maintain a fixed epsilon and alpha parameters until your algorithm converges and then put them as 0 (i.e., ignore them).
I've been studying up on reinforcement learning, but the thing I don't understand is how a Q value is ever calculated. If you use the Bellman equation Q(s,a) = r + Ξ³*max(Q(s',a')), would't it just go on forever? Because Q(s',a') would need the Q value of one timestep further, and that would just continue on and on. How does it end?
In Reinforcement Learning you normally try to find a policy (the best action to take in a specific state), and the learning process ends when the policy does not change anymore or the value function (representing the expected reward) has converged.
You seem to confuse Q-learning and Value Iteration using the Bellman equation. Q-learning is a model-free technique where you use obtained reward to update Q:
Here the direct reward rt+1 is the reward obtained after having done action at in state st. Ξ± is the learning rate that should be between 0 and 1, if it is 0 no learning is done, if it is 1 only the newest reward is taken into account.
Value iteration with the Bellman equation:
Where a model Pa(s,s') is required, also defined as P(s'|s,a), which is the probability of going from state s to s' using action a. To check if the value function is converged, normally the value function Vt+1 is compared to Vt for all states and if it is smaller than a small value (Ξ΅) the policy is said to be converged:
See also:
Difference between Q-learning and Value Iteration
How do I know when a Q-learning algorithm converges?
Sutton et al.: RL