Timeline for The expected square of the determinant of a random row stochastic matrix
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 22, 2022 at 6:56 | comment | added | Guillaume Aubrun | No (If for example $g(x)=\sqrt{x}$, then $g(||Z||_2)$ is a function of $Z/g(||Z||_2)$). You also need to use the fact that the polar decomposition induces a bijection between $\mathbf{R}^n \setminus \{0\}$ and $(0,\infty) \times S^{n-1}$; after passing to polar coordinates the density has a product form, hence independence. In the case of exponential variables, the "polar" decomposition induces a bijection between $(0,\infty)^n$ and $(0,\infty) \times \Delta_{n-1}$ where $\Delta_{n-1}$ is the standard simplex. | |
| Jun 22, 2022 at 6:20 | comment | added | Machinato | Does it imply that since the density of $Z$ depends actually on any strictly monotone function $ g(\left \| Z \right \|_2)$, so the pair of random variables $Z/g(\left \| Z \right \|_2)$ and $g(\left \| Z \right \|_2)$ are independent as well? | |
| Jun 22, 2022 at 4:31 | comment | added | Guillaume Aubrun | This is because the density of $(Y_1,\dots,Y_n)$ at a point $(y_1,\dots,y_n)$ depends only on $y_1+\dots+y_n$. It is similar to the following situation, which is maybe more standard: if $Z=(Z_1,\dots,Z_n)$ are i.i.d. $N(0,1)$ random variables, then $Z/||Z||_2$ is independent from $||Z||_2$ because the density of $Z$ depends only on $\|Z\|_2$ (the standard Euclidean norm). | |
| Jun 21, 2022 at 11:59 | comment | added | Machinato | I haven't understood why $(X_1,\ldots ,X_n)$ is independent from $S = Y_1 + \cdots + Y_n$ ? | |
| Feb 6, 2018 at 20:57 | history | bounty ended | Rodrigo de Azevedo | ||
| Feb 2, 2018 at 7:59 | history | edited | Guillaume Aubrun | CC BY-SA 3.0 |
deleted 1 character in body
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| Jan 31, 2018 at 15:27 | history | answered | Guillaume Aubrun | CC BY-SA 3.0 |