Timeline for Rationale behind defining distribution function with strict inequality
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2022 at 2:32 | comment | added | User1865345 | In fact, there is also the same phenomenon happening over defining Stieltjes premeasure on Borel field. While many define it using Stieltjes function which should be right continuous, authors like Schilling used left continuous condition. | |
| Nov 23, 2022 at 2:29 | comment | added | User1865345 | I think your footnote might be the "explicit" reason. At least it explains a lot although what Kolmogorov was actually thinking, I can't say. As for $[a, b) $ check equation $(2) $ in p. $23.$ | |
| Nov 22, 2022 at 23:25 | comment | added | Sextus Empiricus | Ok, so maybe the reasoning might still not be clear/explicit, but at least the goalposts have been moved. Given that one wishes to define the distribution function as $F^{(x)}(a) = P^{(x)}(-\infty,a)$ it is a simple consequence that we get $F^{(x)}(a) = P \{ x<a \}$ (I see this as the explicit reason for using $<$ instead of $\leq$). But indeed, it still remains a question why one wishes to use $(-\infty,a)$ | |
| Nov 22, 2022 at 23:22 | comment | added | Sextus Empiricus | He didn't choose [𝑎,𝑏) over (𝑎,𝑏], he chose (a,b). | |
| Nov 22, 2022 at 23:18 | comment | added | Sextus Empiricus | @User1865345 It is not just a 'choice' on the right side endpoint bracket. $$(-\infty,a) \text{ versus } (-\infty,a]$$ it is also the left side where the bracket was chosen to be non-inclusive. That left side has nothing to do with the sign $<$ or $\leq$. The reasons may have to do with $-\infty$ and $\infty$ not being part of the sample space. The value $F^{(x)}(-\infty) = P^{(x)}(-\infty,-\infty) = 0$, but what about $F^{(x)}(-\infty) = P^{(x)}[-\infty,-\infty)$ or $F^{(x)}(-\infty) = P^{(x)}[-\infty,-\infty]$ ? | |
| Nov 22, 2022 at 23:01 | comment | added | User1865345 | That is what I am wondering as to why he chose $[a, b) $ over $(a, b].$ Was there any precedent? Or was it simply just a choice? That is not explicit. | |
| Nov 22, 2022 at 19:48 | comment | added | Sextus Empiricus | @User1865345 the reasoning here is explicit. He was defining the distribution in terms of sets $F^{(x)}(a) = P^{(x)}(-\infty,a)$ and the expression $F^{(x)}(a) = P \{ x<a \}$ is derived from that. The use of $<$ instead of $\leq$ stems from the use of $(-\infty,a)$ instead of $[-\infty,a]$. | |
| Nov 22, 2022 at 7:25 | comment | added | User1865345 | That is an interesting take. Hm. Unfortunately it would be hard to trace any explicit reason as to why he stuck to this for $\leq$ wasn't uncommon back then either. | |
| Nov 22, 2022 at 7:20 | history | answered | Sextus Empiricus | CC BY-SA 4.0 |