All of 52ceccf20f20130d0f8c2716521d24de's Comments + Replies

methods based on rounding probabilities are hot flaming garbage

I think this depends a lot on what you're interested in, i.e. what scoring rules you use. Someone who runs the same analysis with Brier instead of log-scores might disagree.

More generally, I'm not convinced it makes sense to think of "precision" as a constant, let alone a universal one, since it depends on

• the scoring rule in question: Imagine a set of forecasts that's awfully calibrated on values <1% and >99%, but perfectly calibrated on values between 1% and 99%. With the log-score, this

Just to confirm: Writing $p_t$, the probability of $A$ at time $t$, as $p_t=E[1_A\mid F_t]$ (here $F_t$ is the sigma-algebra at time $t$), we see that $p_t$ must be a martingale via the tower rule.

The log-odds $x_t=\log \frac{p_t}{1-p_t}$ are not martingales unless $p_t\equiv \text{const}$ because Itô gives us
$\begin{array}{cc}d{x}_t& \stackrel{\text{def}}{=}d\log \frac{p_t}{1-p_t}\\ & \stackrel{\text{Itô}}{=}\underset{\text{martingale part}}{\underset{}{\frac{1}{p_t(1-p_t)}d{p}_t}}+\frac{1}{2}\underset{\text{drift part}}{\underset{}{(\frac{1}{(1-p_t{)}^2}-\frac{1}{p_t^2})d[p{]}_t}}.\end{array}$

So unless $p_t$ is continuous and of bounded variation (⇒ $d[p{]}_t=0$, but this also implies... (read more)

(Why) are you not happy with Velenik's answer or "a probabilistic theory $(\Omega ,F,P)$ tells us that if we look at an event $A\in F$ and perform the same experiment $N\to \infty$ times, then the fraction of experiments where $A$ happened approaches $P\left(A\right)$ in a LLN-like manner"? Is there something special about physical phenomena as opposed to observables?

>$[0,1]$ can be written as the union of a meager set and a set of null measure. This result forces us to make a choice as to which class of sets we will neglect, or otherwise we will end up neglecting the whole space&n... (read more)