Timeline for How does $Z = \sqrt{\left( \frac{X_1 - 0}{\sigma} \right)^2 + \left( \frac{X_2 - 0}{\sigma} \right)^2}$ imply that $Z$ has a Rayleigh distribution?

12 events

when toggle format	what		by	license	comment
Dec 3, 2023 at 17:25	history	edited	Hunaphu	CC BY-SA 4.0	deleted 49 characters in body
May 11, 2021 at 11:29	history	edited	Hunaphu	CC BY-SA 4.0	deleted 65 characters in body
Apr 16, 2021 at 10:01	history	edited	Hunaphu	CC BY-SA 4.0	edited body
Apr 16, 2021 at 9:48	history	edited	Hunaphu	CC BY-SA 4.0	edited body
Apr 16, 2021 at 9:42	history	edited	Hunaphu	CC BY-SA 4.0	added 105 characters in body
Apr 14, 2021 at 11:59	history	edited	Hunaphu	CC BY-SA 4.0	added 143 characters in body
Apr 14, 2021 at 11:58	vote	accept	The Pointer
Apr 14, 2021 at 11:57	comment	added	The Pointer		Ahh, ok, that makes sense, since $\int \int_A \ dA = \int_{\theta = \alpha}^{\theta = \beta} \int_{r = g_1 (\theta)}^{r = g_2(\theta)} \ dr d\theta$ for polar coordinates. Thanks for the clarifying edit.
Apr 14, 2021 at 11:52	history	edited	Hunaphu	CC BY-SA 4.0	added 143 characters in body
Apr 14, 2021 at 11:51	comment	added	The Pointer		Thanks for the answer. Can you please explain the reasoning for $\int_0^z$? My polar coordinates are a bit rusty.
Apr 14, 2021 at 11:50	history	edited	Hunaphu	CC BY-SA 4.0	added 143 characters in body
Apr 14, 2021 at 11:43	history	answered	Hunaphu	CC BY-SA 4.0