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In one of the easy cases of inverse reinforcement learning, we try to derive the unknown reward function assuming that an optimal policy is known and MDP is completely known.

Let $P_{a_1}$ be the probability transition matrix from one state to another, following the optimal policy $\pi(s) = a_1$. Assuming $n$ states, then this matrix is $n \times n$. On the other hand, $P_a$ is the transition matrix of all other policies that are not $a_1$. Still $n \times n$.

If $I$ is the identity matrix and $\gamma$ is the reward discount factor (scalar), what is the dimension of $R$ in the follow rule? (from Algorithms for inverse reinforcement learning, by Ng and Russell, 2000)

$$(P_{a_1} - P_a)(I - \gamma P_{a_1})^{-1}R \geq 0 $$

Is $R$ a vector? or $n \times n$ matrix? And what would this mean?

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  • $\begingroup$ is this presumed to be a linear function in a linear system? $\endgroup$ – EngrStudent Nov 30 '15 at 13:29
  • $\begingroup$ I actually am not sure how to answer your question. What do you mean by linear system? $\endgroup$ – cgo Nov 30 '15 at 13:31
  • $\begingroup$ Is $P_{a}$ constant? $\endgroup$ – EngrStudent Dec 1 '15 at 4:28
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As the paper of Ng and Russell (2000) indicates in section 2.1, the reinforcement function $R$, takes as input a state, and as output has the reward, a real number. Therefore $R$ should be a vector of $n$ items. The result of equation (4) of the paper:

$$(P_{a_1} - P_a)(I - \gamma P_{a_1})^{-1}R$$

therefore also is a vector of $n$ items.

Note that the reward function can also be defined with the parameters state ($s$), next state ($s'$), and action ($a$): $R_{ss'}^a$ as done by Sutto and Barto (1998, section 3.6).

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  • $\begingroup$ Hi, thank you for your reply. I am not sure if you are familiar with the Q-learning example of a robot in a house with different rooms. The aim of the robot is to exit the house, and until then, the reward it gets is 0 for every move to a wrong room. Some rooms are connected say room 0 with room 4; room 4 with room 3; etc. In this case, the reward is $n \times n$. The column is the current state the robot is in, the row is the next possible state. I am asking this because for me, it would make sense that $R$ is $n \times n$, but I could not understand what it could mean when it is a vector. $\endgroup$ – cgo Nov 30 '15 at 15:31
  • $\begingroup$ When it is a vector you simply define the reward in function of its current state only: $R(s)$. The algorithm you use to calculate the value function will then probably need some iterations more to arrive to a same kind of policy as when you predefine the reward in function of the current and next state: $R(s,s')$. $\endgroup$ – agold Nov 30 '15 at 15:39
  • $\begingroup$ Alright, let me think about what you said. Thank you for your help. $\endgroup$ – cgo Nov 30 '15 at 15:43

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