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).
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 have a question about my own project for testing reinforcement learning technique. First let me explain you the purpose. I have an agent which can take 4 actions during 8 steps. At the end of this eight steps, the agent can be in 5 possible victory states. The goal is to find the minimum cost. To access of this 5 victories (with different cost value: 50, 50, 0, 40, 60), the agent don't take the same path (like a graph). The blue states are the fail states (sorry for quality) and the episode is stopped.
enter image description here
The real good path is: DCCBBAD
Now my question, I don't understand why in SARSA & Q-Learning (mainly in Q learning), the agent find a path but not the optimal one after 100 000 iterations (always: DACBBAD/DACBBCD). Sometime when I compute again, the agent falls in the good path (DCCBBAD). So I would like to understand why sometime the agent find it and why sometime not. And there is a way to look at in order to stabilize my agent?
Thank you a lot,
Tanguy
TD;DR;
Set your epsilon so that you explore a bunch for a large number of episodes. E.g. Linearly decaying from 1.0 to 0.1.
Set your learning rate to a small constant value, such as 0.1.
Don't stop your algorithm based on number of episodes but on changes to the action-value function.
More detailed version:
Q-learning is only garranteed to converge under the following conditions:
You must visit all state and action pairs infinitely ofter.
The sum of all the learning rates for all timesteps must be infinite, so
The sum of the square of all the learning rates for all timesteps must be finite, that is
To hit 1, just make sure your epsilon is not decaying to a low value too early. Make it decay very very slowly and perhaps never all the way to 0. You can try , too.
To hit 2 and 3, you must ensure you take care of 1, so that you collect infinite learning rates, but also pick your learning rate so that its square is finite. That basically means =< 1. If your environment is deterministic you should try 1. Deterministic environment here that means when taking an action a in a state s you transition to state s' for all states and actions in your environment. If your environment is stochastic, you can try a low number, such as 0.05-0.3.
Maybe checkout https://youtu.be/wZyJ66_u4TI?t=2790 for more info.
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):
I'm trying to implement an Inertial Navigation System using an Indirect Kalman Filter. I've found many publications and thesis on this topic, but not too much code as example. For my implementation I'm using the Master Thesis available at the following link:
https://fenix.tecnico.ulisboa.pt/downloadFile/395137332405/dissertacao.pdf
As reported at page 47, the measured values from inertial sensors equal the true values plus a series of other terms (bias, scale factors, ...).
For my question, let's consider only bias.
So:
Wmeas = Wtrue + BiasW (Gyro meas)
Ameas = Atrue + BiasA. (Accelerometer meas)
Therefore,
when I propagate the Mechanization equations (equations 3-29, 3-37 and 3-41)
I should use the "true" values, or better:
Wmeas - BiasW
Ameas - BiasA
where BiasW and BiasA are the last available estimation of the bias. Right?
Concerning the update phase of the EKF,
if the measurement equation is
dzV = VelGPS_est - VelGPS_meas
the H matrix should have an identity matrix in corrispondence of the velocity error state variables dx(VEL) and 0 elsewhere. Right?
Said that I'm not sure how I have to propagate the state variable after update phase.
The propagation of the state variable should be (in my opinion):
POSk|k = POSk|k-1 + dx(POS);
VELk|k = VELk|k-1 + dx(VEL);
...
But this didn't work. Therefore I've tried:
POSk|k = POSk|k-1 - dx(POS);
VELk|k = VELk|k-1 - dx(VEL);
that didn't work too... I tried both solutions, even if in my opinion the "+" should be used. But since both don't work (I have some other error elsewhere)
I would ask you if you have any suggestions.
You can see a snippet of code at the following link: http://pastebin.com/aGhKh2ck.
Thanks.
The difficulty you're running into is the difference between the theory and the practice. Taking your code from the snippet instead of the symbolic version in the question:
% Apply corrections
Pned = Pned + dx(1:3);
Vned = Vned + dx(4:6);
In theory when you use the Indirect form you are freely integrating the IMU (that process called the Mechanization in that paper) and occasionally running the IKF to update its correction. In theory the unchecked double integration of the accelerometer produces large (or for cheap MEMS IMUs, enormous) error values in Pned and Vned. That, in turn, causes the IKF to produce correspondingly large values of dx(1:6) as time evolves and the unchecked IMU integration runs farther and farther away from the truth. In theory you then sample your position at any time as Pned +/- dx(1:3) (the sign isn't important -- you can set that up either way). The important part here is that you are not modifying Pned from the IKF because both are running independent from each other and you add them together when you need the answer.
In practice you do not want to take the difference between two enourmous double values because you will lose precision (because many of the bits of the significand were needed to represent the enormous part instead of the precision you want). You have grasped that in practice you want to recursively update Pned on each update. However, when you diverge from the theory this way, you have to take the corresponding (and somewhat unobvious) step of zeroing out your correction value from the IKF state vector. In other words, after you do Pned = Pned + dx(1:3) you have "used" the correction, and you need to balance the equation with dx(1:3) = dx(1:3) - dx(1:3) (simplified: dx(1:3) = 0) so that you don't inadvertently integrate the correction over time.
Why does this work? Why doesn't it mess up the rest of the filter? As it turns out, the KF process covariance P does not actually depend on the state x. It depends on the update function and the process noise Q and so on. So the filter doesn't care what the data is. (Now that's a simplification, because often Q and R include rotation terms, and R might vary based on other state variables, etc, but in those cases you are actually using state from outside the filter (the cumulative position and orientation) not the raw correction values, which have no meaning by themselves).