Timeline for How to show that this integral of the normal distribution is finite?
Current License: CC BY-SA 4.0
10 events
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| Jan 12, 2022 at 16:30 | comment | added | Sextus Empiricus | It is right that you were not using unrelated expressions, but using the rule of l'Hôpital relates to the limit and not to asymptotic behaviour. Or maybe you can use the rule like that for the asymptotic behaviour to infinity? | |
| Jan 12, 2022 at 16:10 | history | edited | Roger V. | CC BY-SA 4.0 |
expanded the answer
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| Jan 12, 2022 at 16:06 | comment | added | Roger V. | @SextusEmpiricus I have modified the answer to reflect our discussions. I holt to the first approach, because I find it more satisfying for personal reasons (my background is in physics). But, of course, from mathematical view point your remarks are correct. | |
| Jan 12, 2022 at 16:05 | history | edited | Roger V. | CC BY-SA 4.0 |
expanded the answer
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| Jan 12, 2022 at 15:50 | comment | added | Roger V. | @SextusEmpiricus I am talking here about the asymptotic form of the integrand. The limit notation might be somewhat misleading here - I am not equating any unrelated expressions (as in your last comment), I was simply too lazy to write explicitly the coefficient. The approach suggested in your first comment works as well, but it would look less intuitive to me. | |
| Jan 12, 2022 at 15:01 | comment | added | Sextus Empiricus | Counterexample $\lim_{x \to \infty} 1/x = \lim_{x \to \infty} 1/x^2 = 0$ | |
| Jan 12, 2022 at 15:00 | comment | added | Sextus Empiricus | I find it a nice short proof, but can we really say that if $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = 0$ and if the integral of g(x) converges, then also the integral of f(x) converges? | |
| Jan 12, 2022 at 14:54 | comment | added | Roger V. | @SextusEmpiricus or one could show that $xe^{-\frac{x^2}{2\sigma^2}}$ decays faster than $1/x$ - either by the second application of the L'Hôpital's rule or by noting that the integral of this quantity is known to converge - which is what I mean by my last remark. Perhaps, I should add this comment to my answer. | |
| Jan 12, 2022 at 14:44 | comment | added | Sextus Empiricus | This shows that the term is decaying to $0$ but not that it isn't decaying slower than $1/x$. I guess that you can still use the same technique though, by showing $$\lim_{x\rightarrow -\infty} \frac{x\phi^2(x)}{\Phi(x)}=0$$ | |
| Jan 12, 2022 at 8:59 | history | answered | Roger V. | CC BY-SA 4.0 |