Timeline for Subdivide Z into $X_1,X_2$, s.t. $Z=X_1+X_2$ and ($X_1,X_2$) obey bivariate normal
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Nov 21, 2019 at 22:18 | vote | accept | LucasMation | ||
Nov 21, 2019 at 17:31 | comment | added | LucasMation | @DilipSarwate, OP here. Thanks for the insightful explanation. Indeed I was expecting the conditional distribution to be univariate, as X2 is now a transformation of X1 | |
Nov 21, 2019 at 17:05 | history | edited | Dilip Sarwate | CC BY-SA 4.0 |
added 604 characters in body
|
Nov 21, 2019 at 17:00 | history | edited | Dilip Sarwate | CC BY-SA 4.0 |
added 604 characters in body
|
Nov 21, 2019 at 16:36 | comment | added | whuber♦ | Thank you. I don't think it's an effective answer, though, because it reduces the problem to finding the conditional distribution of $X_1,$ which you don't specify. BTW, many people would allow that a degenerate bivariate Normal distribution is still bivariate Normal. Maintaining that distinction would unnecessarily complicate the statements of many basic results about Normal distributions. (E.g., that linear transformations of multivariate Normal variables are multivariate Normal.) | |
Nov 21, 2019 at 16:31 | comment | added | Dilip Sarwate | @whuber Please see revised answer. | |
Nov 21, 2019 at 16:31 | history | edited | Dilip Sarwate | CC BY-SA 4.0 |
added 408 characters in body
|
Nov 21, 2019 at 16:27 | comment | added | whuber♦ | This seems to answer a different question than the one in the title, which asks for some kind of "conditional" distribution. | |
Nov 21, 2019 at 16:25 | history | answered | Dilip Sarwate | CC BY-SA 4.0 |