Timeline for Gaussian-to-gaussian transformations
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2022 at 14:58 | comment | added | whuber♦ | @Kaira I don't think you need independence; but that is indeed the case: uncorrelated multivariate Normal variables are always independent. | |
| Dec 27, 2022 at 23:58 | comment | added | Kaira | Is it fine to say that since each of $f(X)=(Y_1,\cdots , Y_n)$ is uncorrelated in the case of standard normal, they are independent(since $f(X)$ is multivariate normal), so they cannot depend on more than one $X_i$? If that's the case, I think I can reduce it to the 1d case. | |
| Dec 27, 2022 at 23:38 | comment | added | whuber♦ | @Kaira Or maybe I am mistakenly trivializing it ;-). Note that you must assume the distribution is nondegenerate, for otherwise the proposition fails. Under that assumption, this proposition is equivalent to asserting the only diffeomorphism that preserves the standard multivariate Normal distribution is the identity. | |
| Dec 27, 2022 at 23:34 | comment | added | Kaira | I agree, but I am failing to prove that. Maybe I am missing something trivial? | |
| Dec 27, 2022 at 23:28 | comment | added | whuber♦ | @Kaira I might be wrong, but wouldn't it be straightforward to show that the Jacobian of $f$ must be constant in that case? | |
| Dec 27, 2022 at 23:04 | comment | added | Kaira | I was referring to the Proposition. Specifically I was trying to prove that for diffeomorphism $f:\mathbb {R}^n\to \mathbb{R}^n$ and a multivariate Gaussian $X$, if $f(X)$ is multivariate Gaussian, then $f$ is affine. I will look into Sklar’s theorem. | |
| Dec 27, 2022 at 22:25 | comment | added | whuber♦ | @Kaira It's unclear which "result" you might be referring to. See Sklar's Theorem for one natural generalization related to these ideas. | |
| Dec 27, 2022 at 21:40 | comment | added | Kaira | Is the result still true in the multivariate case? I was trying to prove it in the case $f$ is a diffeomorphism, but it got me nowhere. | |
| Jun 11, 2020 at 14:32 | history | edited | CommunityBot |
Commonmark migration
|
|
| Feb 26, 2020 at 1:40 | comment | added | MSIS | @Whuber: Is it the case that affine transformations are the only ones that preserve normality? | |
| Feb 19, 2019 at 22:01 | vote | accept | avid | ||
| Feb 18, 2019 at 12:54 | comment | added | whuber♦ | @Artem Thank you for pointing out those implications. The continuity is used to conclude $f$ is a surjection. Although it's not a necessary assumption here, I first introduced it in my considerations when pondering extensions to non-continuous random variables, and left it in. Obviously, though, if $f$ is not continuous, then $f(X)$ cannot be a continuous variable. | |
| Feb 18, 2019 at 1:52 | comment | added | Artem Mavrin | Also, I think your conditions may be relaxed to just “$f$ is strictly increasing” since strictly increasing functions on $\mathbb{R}$ are automatically Borel-measurable (as are continuous functions, if that assumption is needed) | |
| Feb 18, 2019 at 1:50 | comment | added | Artem Mavrin | Great answer! Can you explain where the assumption that $f$ is continuous is used in your Proposition? | |
| Feb 18, 2019 at 0:05 | history | answered | whuber♦ | CC BY-SA 4.0 |