Apologies for asking an object level question, but I probably have Covid and I'm in the UK which is about to experience a nasty heatwave. Do we have a Covid survival guide somewhere?
(EDIT: I lived lol)
Is there a way to alter the structure of a futarchy to make it follow a decision theory other than EDT?
I got a good answer here: https://stats.stackexchange.com/q/579642/5751
Or is that still too closely tied to the explore-exploit paradigm?
Right. The setup for my problem is the same as the 'bernoulli bandit', but I only care about the information and not the reward. All I see on that page is about exploration-exploitation.
What's the term for statistical problems that are like exploration-exploitation, but without the exploitation? I tried searching for 'exploration' but that wasn't it.
In particular, suppose I have a bunch of machines which each succeed or fail independently with a probability that is fixed separately for each machine. And suppose I can pick machines to sample to see if they succeed or fail. How do I sample them if I want to become 99% certain that I've found the best machine, while using the fewest number of samples?
The difference with exploration-exploitation is that this is just a trial period, and I don't care about how many successes I get during this testing. So I want something like Thompson sampling, but for my purposes Thompson sampling oversamples the machine it currently thinks is best because it values getting successes rather than ruling out the second-best options.
The problem is that this measures their amount of knowledge about the questions as well as their calibration.
My model would be as follows. For a fixed source of questions, each person has a distribution describing how much they know about the questions. It describes how likely it is that a given question is one they should say p on. Each person also has a calibration function f, such that when they should say p they instead say f(p). Then by assigning priors over the spaces of these distributions and calibration functions, and applying Bayes' rule we get a posterior describing what we know about that persons calibration function.
Then assign a score to each calibration function which is the expected log score lost by a person using that calibration function instead of an ideal one, assuming that the questions were uniformly distributed in difficulty for them. Then their final calibration score is just the expected value of that score given our distribution of calibration functions for them.
I was expecting a post about tetraethyllead.
I gave some more examples here: https://www.reddit.com/r/math/comments/9gyh3a/if_you_know_the_factors_of_an_integer_x_can_you/e68u5x4/
You just have to carefully do the algebra to get an inductive argument. The fact that the last digit is 5 is used directly.
Suppose n is a number that ends in 5, and such that the last N digits stay the same when you square it. We want to prove that the last N+1 digits stay the same when you square n^2.
We can write n = m*10^N + p, where p has N digits, and so n^2 = m^2*10^(2N) + 2mp*10^N + p^2. Note that since 2p ends in 0, the term 2mp*10^N actually divides by 10^(N+1). Then since the two larger terms divide by 10^N+1, n^2 agrees with p^2 on its last N+1 digits, and so p^2 agrees with p on at least its last N digits. So we may write p^2 = q*10^N + p. Hence n^2 = m^2*10^(2N) + 2mp*10^N + q*10^N + p.
Squaring this yields (n^2)^2 = (terms dividing by 10^(N+1)) + 2qp*10^N + p^2. Again, 2p ends in 0, so 2qp*10^N also divides by 10^(N+1). So the last N+1 digits of this agree with p^2, which we previously established also agrees with n^2. QED
A similar argument shows that the number generated in this way is the only 10-adic number that ends in 5 and squares to itself. So we also have that one minus this number is the only 10-adic ending in 6 that squares to itself. You can also prove that 0 and 1 are the only numbers ending in 0 and 1 that square to themselves. The other digits can't square to themselves. So x^2 = x has precisely four solutions.
If we start with 5 and start squaring we get 5, 25, 625, 390625, 152587890625.... Note how some of the end digits are staying the same each time. If we continue this process we get a number ...8212890625 which is a solution of x^2 = x. We get another solution by subtracting this from 1 to get ...1888119476.