# How many sunrises are worth observing?

The one-sun version of Laplace's sunrise problem provides a Bayesian argument that, if on all $$n$$ mornings in recorded history the Sun has risen, its probability of doing so tomorrow is $$\frac{n+1}{n+2}$$. In fact, if the Sun rises on a proportion $$p$$ of mornings, the Bayesian prior $$p\sim U(0,\,1)$$ becomes a Bayesian posterior $$p\sim \operatorname{B}(k+1,\,n-k+1)$$ if $$k$$ of $$n$$ mornings had sunrises. So if we consider an event we can seriously imagine only happens sometimes (but I'll talk of mornings sunrises for convenience), $$n$$ data change our estimate of $$p$$'s distribution from a uniform distribution to a Beta distribution.

In theory, we get more confident with every new observation. Returning to the simple case $$k=0$$, the probability $$\frac{n+1}{n+2}=\mathbb{E}p$$ gets slightly larger as $$n$$ increases. However, some kinds of sunrises take real effort to monitor, e.g. "electrons will always spiral left in this magnetic field". There comes a point where it's not worth checking again. In theory, one could defend this with the solution of an optimisation problem. For example, if we discount the increase $$\frac{n+1}{n+2}-\frac{1}{2}$$ in $$\mathbb{E}p$$ with a $$\frac{1}{\sqrt{n}}$$ factor to represent the cost of reducing empirical standard errors, we're left with the question of which $$n$$ maximises $$\frac{1}{\sqrt{n}}(\frac{n+1}{n+2}-\frac{1}{2})$$. The answer's $$2$$, by the way, but I'm sure this isn't the right or standard way to do this.

Can we quantify in terms of some trade-off the optimum amount of data to gather? In other words, do what I just did above, but... probably some more sensible choice of the utility function.

Also, I really struggled with the choice of tags for this question, so I'd appreciate better ones if there are any.

• @RobinRyder Thank you for that tag suggestion; I think it works. – J.G. Oct 18 '18 at 21:05
• Two comments: 1) a uniform distribution is a Beta(1,1) so the statement "from a uniform distribution to a beta distribution" isn't accurate and 2) yes, you can determine the optimum amount of data to gather by minimizing the expected loss for an appropriate loss function. With the information given, I cannot help you determine an appropriate loss function. What you will need is information about how much it costs to obtain an observation compared with the cost on the accuracy of p. It seems like you need information on the cost of new observations. – jaradniemi Oct 18 '18 at 21:56
• @jaradniemi I know uniform distributions are special cases of Beta; priors and posteriors are often in the same family but with different parameters. But before the Bayesian analysis, it's not obvious which family should generalise the trivial prior. As for 2), it sounds like the utility function should be of the form $f(n)-cn$ with $f$ increasing and $c>0$, but I think I need as much help with choosing $f$ as choosing $c$. – J.G. Oct 18 '18 at 22:09
• @J.G. There's not, and cannot be, a general answer to "what utility function should I use." $f(n) - c n$ makes sense assuming constant cost (and that you always observe "yes" – in general it should be $f(k, n)$ – but $f$ is going to look very different if it's "my nuclear reactor is operating within safety tolerances" than if it's "my toast is just the right amount of crispy." – Dougal Oct 19 '18 at 2:37