Is there a symbol for immeasurability?

Question

I don't know if "immeasurability" is the term I'm after. Whenever I search for "immeasurable" I get references to infinity, which is not what I'm looking for.

If I say I'm 74% certain of something or 76% certain of something, although there is a mathematical difference, in practical terms the two statements are identical, i.e. I wouldn't make any decisions differently based on the two statements. However, choosing between 50% certainty and 90% certainty, I would potentially make different decisions.

$$76\% - 74\% = immeasurable$$

$$90\% - 50\% = measurable$$

$$1\% - 0.01\% = measurable$$

Or possibly:

$$76\% \div 74\% = immeasurable$$

$$90\% \div 50\% = measurable$$

$$1\% \div 0.01\% = measurable$$

although I might consider there to be a difference between 80% and 90%, but not 16% and 18%.

As an example, if the weather bureau says there's a 74% chance of rain, or a 76% chance of rain, it won't affect my decision about whether to take an umbrella or not.

The percentage only makes sense for events that haven't passed yet. Tomorrow might have a 75% chance of rain, but yesterday it either rained or it didn't, so it can't be expressed as any percentage except 0% or 100%.

Is there a mathematical way of describing this?

Answer 1

No, because mathematically there is a big difference. If you say you wouldn't distinguish 76% and 74%. Then you also wouldn't distinguish between 74$ and 72%. If you let this go on, inductively, you wouldn't distinguish between 100% and 0%. The term measureable/imeasureable comes from measure theory, where probability is a sub theme of.

Answer 2

This strictly isn't about mathematics, but here's a possible idea. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

For likelihoods $p,q ∈ [0,1]$, define the disparity between $p$ and $q$ to be $\lfrac{|p-q|}{\sqrt{pq}}·\sqrt{(r)(1-r)}$ where $r = \lfrac{p+q}2$.

The idea is that if the average $r$ of $p,q$ is in the middle of $[0,1]$, you want the answer to be roughly the difference $|p-q|$ divided by $\sqrt{pq}$, so that you get roughly the relative proportional difference but give more weight if the ratio $p:q$ is far from $1$. But if $p,q$ are both close to $0$ or both close to $1$, you want something less. One way is to introduce the weight $\sqrt{(r)(1-r)}$ that is $1$ when $r = \lfrac12$ but is $0$ when $r = 0$ or $r = 1$. In fact, this weighting function is a semi-circular arc.