In RL as probabilistic inference, why do we take a probability to be $\exp(r(s_t, a_t))$? 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.
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.