# Tag Info

## Hot answers tagged stochastic-policy

8

Is the optimal policy always stochastic (that is, a map from states to a probability distribution over actions) if the environment is also stochastic? No. An optimal policy is generally deterministic unless: Important state information is missing (a POMDP). For example, in a map where the agent is not allowed to know its exact location or remember ...

8

A deterministic policy is a function of the form $\pi_{\mathbb{d}}: S \rightarrow A$, that is, a function from the set of states of the environment, $S$, to the set of actions, $A$. The subscript $_{\mathbb{d}}$ only indicates that this is a ${\mathbb{d}}$eterministic policy. For example, in a grid world, the set of states of the environment, $S$, is ...

5

I would say no. For example, consider the multi-armed bandit problem. So, you have $n$ arms which all have a probability of giving you a reward (1 point, for example), $p_i$, $i$ being between 1 and $n$. This is a simple stochastic environment: this is a one state environment, but it is still an environment. But obviously the optimal policy is to choose ...

5

Did AlphaGo and AlphaGo [Zero] play 100 repetitions of the same sequence of boards, or were there 100 different games? There were 100 different games. You can view some example games between AlphaGo [Lee] and AlphaGo Zero here. They are clearly all different. This statement in the question shows a misunderstanding: My understanding of AlphaGo and AlphaGo [...

4

First, the derivative is usually taken with respect to a variable (input) of the function. Hence the notation $\frac{df}{dx}$ for some function $f(x)$. If you look at your equation more carefully $$\nabla log P(\tau^{i};\theta) = \Sigma_{t=0}\nabla_{\theta}log\pi(a_{t}|s_t, \theta).$$ You will see that the gradient is taken with respect to $\theta$, which ...

3

The game of TIC-TAC-TOE can be modelled as a non-deterministic Markov decision process (MDP) if, and only if: The opponent is considered part of the environment. This is a reasonable approach when the goal is to solve playing against a specific opponent. The opponent is using a stochastic policy. Stochastic policies are a generalisation that include ...

2

My understanding of your question is, you have 2 designs: A deterministic policy that outputs 2 scalar for x and y respectively. A value function that outputs the probability of each pixel in the 2D grid. If you choose the max of softmax on (2.), you'll get the same deterministic policy as (1.), assuming there are some tie-breaking designs. So I don't ...

2

Is it possible for value-based methods to learn stochastic policies? Yes, but only in a limited sense, due to the ways it is possible to generate stochastic policies from a value function. For instance, the simplest exploratory policy used by SARSA and Monte Carlo Control, $\epsilon$-greedy, is stochastic. SARSA natually learns the optimal $\epsilon$-...

2

No it is not possible to use Q-learning to build a deliberately stochastic policy, as the learning algorithm is designed around choosing solely the maximising value at each step, and this assumption carries forward to the action value update step $Q_{k+1}(S_t,A_t) = Q_k(S_t,A_t) + \alpha(R_{t+1} +\gamma\text{max}_{a'}Q_k(S_{t+1},a') - Q_k(S_t,A_t))$ - i.e. ...

2

The value function is defined as $v_\pi(s) = \mathbb{E}_\pi[G_t | S_t = s]$ where $G_t$ are the (discounted) returns from time step $t$. The expectation is taken with respect to the policy $\pi$ and the transition dynamics of the MDP. Now, as you pointed out the optimal value function is defined as $v_*(s) = \max_\pi v_\pi(s)\; ; \;\forall s \in \mathcal{S}$....

1

Is the policy (based in the neural network) a stochastic policy? even if the action space is discrete? Yes. A discrete action space does not require a deterministic policy - it is possible to assign arbitrary probabilities to each action in each state provided each probability is in the range $[0,1]$ and the sum across all allowed actions is $1$. The two ...

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