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?


closed as unclear what you're asking by darij grinberg, Eevee Trainer, user98602, Lee David Chung Lin, ancientmathematician Feb 20 at 7:43

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    $\begingroup$ From the sound of your post, "$\approx 1$" could be what you want. But no idea if it's what you mean. $\endgroup$ – darij grinberg Feb 20 at 0:44
  • $\begingroup$ It's not usually used in the contexts you want, but there is the phrase "statistically insignificant" that you may want to look up. Maybe the non-math phrase "the difference is insignificant" suits your purposes for this vague notion? $\endgroup$ – Mark S. Feb 20 at 12:57

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.

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    $\begingroup$ Is this assuming indistinguishability is transitive? $\endgroup$ – Riley Feb 20 at 0:50
  • $\begingroup$ I don't think the inductive proof really works here. You could consider binning the numbers from 0-100. You could easily have a bin including 76 and 74, but not 72. $\endgroup$ – Tyberius Feb 20 at 0:53
  • $\begingroup$ Yes, if indistinguishability isn't transitive, then what is. But thanks, i guess i can't rely on that. Anyway, the Question posted doesn't make much sense. How would you describe such a notion of immeasureability without using some constant distance function? $\endgroup$ – Luke Feb 20 at 0:56
  • $\begingroup$ The bin system would also work, you're right. Still, what would be the sense of such a thing? $\endgroup$ – Luke Feb 20 at 0:59

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.

  • $\begingroup$ For example, if $(p,q) = (0.0001,0.01)$ then the disparity is about $0.7$, whereas if $(p,q) = (0.16,0.18)$ then the disparity is about $0.04$. You can hence choose a threshold for disparity above which you consider $p,q$ to be sufficiently disparate. $\endgroup$ – user21820 Feb 20 at 4:47

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