Rylan Schaeffer

Kernel Papers

3 March 2021

Hyperbolic Discounting and Learning over Multiple Horizons

by Fedus, Gelada, Bengio, Bellemare and Larochelle (Arxiv)


Geometric discounting is ubiquitous in reinforcement learning e.g. Q-learning:

\[Q(s_t, a_t) = r_t + \gamma \max_a Q(s_{t+1}, a)\]

This work shows how geometric discounting arises from a single belief as to the environment’s hazard rate, and how different beliefs over the hazard rate lead to different discounting schemes (e.g. hyperbolic).

Correct Credit

This work was rejected from ICLR; one reviewer criticized the work for adding little beyond preceding work, specifically Zeb Kurth-Nelson and A David Redish’s “Temporal-difference reinforcement learning with distributed representations” (2009) and Peter D Sozou’s “On hyperbolic discounting and uncertain hazard rates” (1998). I don’t have time to review who contributed what, but know that these authors may not deserve credit for the following ideas.

Geometric Discounting from Constant Hazard Rate

We first need two functions: the survival function, \(s(t)\), which gives the probability the agent is alive at time \(t\), and the hazard rate

\[h(t) = - \partial_t \log s(t) = -\frac{1}{s(t)} \partial_t s(t)\]

The authors reason that one should discount a future reward by the probability of being alive to obtain it:

\[V(r_t) = s(t) r_t\]

The authors then show that geometric discounting arises under a constant hazard rate. Specifically, suppose \(h(t) = \lambda > 0\) is constant. Then, solving for \(s(t)\)

\[\begin{align} \lambda &= - \partial_t \log s(t)\\ -\int_0^T \lambda dt &= \log s(t)\\ s(t) &= e^{-\lambda t}\\ &= (e^{-\lambda})^t\\ &= \gamma^t \end{align}\]

where \(\gamma := e^{-\lambda} = e^{-h(t)}\). The intuition is that \(\gamma\) is the per-time-step probability of the episode continuing. For instance, suppose the hazard rate \(h(t) \rightarrow \infty\) i.e. the probability of surviving is basically 0; then, \(\gamma \rightarrow 0\), a very myopic view which makes sense given one’s expectation they won’t survive.

Alternative Discounting from Non-Constant Hazard Rate

If a constant hazard rate gives rise to geometric discounting, what discounting does a non-constant hazard rate give rise to? Suppose we have a prior \(p(\lambda)\) on the hazard rate i.e.

\[s(t) = \int_{\lambda=0}^{\infty} p(\lambda) e^{-\lambda t} d\lambda\]

If \(p(\lambda)\) is exponential with rate \(1/k\), then the survival function is hyperbolic, meaning rewards should be discounted hyperbolically.

\[\begin{align} s(t) &= \int_{\lambda=0}^{\infty} p(\lambda) e^{-\lambda t} d\lambda\\ &= \int_{\lambda=0}^{\infty} \frac{1}{k} e^{-\lambda/k} e^{-\lambda t} d\lambda\\ &= \frac{1}{k} \frac{-1}{1/k + t} e^{\lambda (1/k + t)} \Big\lvert_{\lambda=0}^{\infty}\\ &= -\frac{1}{1 + kt} [0 - 1]\\ &= \frac{1}{1 + kt} \end{align}\]

Other priors over the hazard rate give different discounting schemes. For instance, suppose \(p(\lambda) = \frac{1}{k}\) i.e. uniform over \([0, k]\). Then

\[s(t) = \frac{1}{kt}(1 - e^{-kt})\] tags: reinforcement-learning - discounting - hazard-rate - survival-function