1
$\begingroup$

In section 2 the paper Reinforcement Learning and Control as Probabilistic Inference: Tutorial and Review the author is discussing formulating the RL problem as a probabilistic graphical model. They introduce a binary optimality variable $\mathcal{O}_t$ which denotes whether time step $t$ was optimal (1 if so, 0 otherwise). They then define the probability that this random variable equals 1 to be

$$\mathbb{P}(\mathcal{O}_t = 1 | s_t, a_t) = \exp(r(s_t, a_t)) \; .$$

My question is why do they do this? In the paper they make no assumptions about the value of the rewards (e.g. bounding it to be non-positive) so in theory the rewards can take any value and thus the RHS can be larger than 1. This is obviously invalid for a probability. It would make sense if there was some normalising constant, or if the author said that the probability is proportional to this, but they don't.

I have searched online and nobody seems to have asked this question which makes me feel like I am missing something quite obvious so I would appreciate if somebody could clear this up for me please.

$\endgroup$
2
  • $\begingroup$ Some papers that I've come across assumed that the reward is normalized to be in a certain range. Not sure if the author is implicitly assuming that in that context or not. Alternatively, he may be using $\operatorname{exp}$ to denote any exponential function with codomain $[0, 1]$, for instance, the sigmoid, but that seems a bit uncommon, given that $\operatorname{exp}$ is typically used to denote $e$. My last guess is that the author was careless. I didn't really read the paper and these are more guesses. You could try to e-mail him. $\endgroup$ – nbro Dec 19 '20 at 18:56
  • 1
    $\begingroup$ @nbro I’d noticed it in another paper I’d read briefly but I can’t remember it now (it wasn’t really relevant for my research) so I can’t remember if they made the assumptions on the range of the rewards. I think I’ll try and email the author and maybe answer this myself. $\endgroup$ – David Ireland Dec 19 '20 at 19:45
0
$\begingroup$

After doing some further reading, it turns out that negative rewards are an assumption for this distribution to hold. However, the author notes that as long as you don't receive a reward of infinity for any action then it is possible to re-scale your rewards by subtracting the maximum value of your potential rewards so that they are always negative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.