User talk:Dorschleber

Answer
Well, I did all this effort to write an answer, and then the question disappeared. For posterity, here is the answer:

The notation, also in the article de:Normalverteilung, is infelicitous and potentially confusing. Take the definition
 * Die Verteilungsfunktion der Normalverteilung ist durch
 * $$F(x) = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{1}{2} \left(\frac{t-\mu}{\sigma}\right)^2} \mathrm dt$$
 * gegeben.

The problem is that there is no such thing as the normal distribution. As long as one is considering one specific normal distribution, that is not an issue, but if (as in the discussion) there are several normally distributed random variables in play with different distributions, confusion may ensue. The issue can be avoided by using
 * Die Verteilungsfunktion der Normalverteilung mit Erwartungswert $$\mu$$ und Varianz $$\sigma^2$$ ist durch
 * $$F_{\mu{,}\sigma}(x) = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{1}{2} \left(\frac{t-\mu}{\sigma}\right)^2} \mathrm dt$$
 * gegeben.

Then $$F_{\mu{,}\sigma}(x)=\Phi\left(\frac{x-\mu}{\sigma}\right).$$ Now we can check that
 * $$F_{\mu{,}\sigma}(\sigma t{+}\mu)=\Phi\left(\frac{(\sigma t{+}\mu)-\mu}{\sigma}\right)=\Phi(t).$$

To get the result from the discussion in some form
 * $$F_{\alpha{,}\beta}(\sigma t{+}\mu)=\Phi\left(\frac{t-\mu}{\sigma}\right),$$

one needs to use $$\alpha=\mu(\sigma+1),\beta=\sigma^2,$$ which looks as implausible as it gets and then some. (Consider that the distribution can be of a random length; in Germany $$\sigma$$ could be 23.1 (cm) while it is 9.1 (inch) in the US, for the same random variable. What is then the meaning of $$\sigma+1$$?) --Lambiam 22:19, 30 June 2023 (UTC)


 * @Lambiam: Thank you so much for taking all that trouble! I shall certainly link this in the ongoing discussion. You were a huge help! Dorschleber (talk) 22:42, 30 June 2023 (UTC)