Timeline for Deriving the conditional distributions of a multivariate normal distribution
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2023 at 0:29 | history | edited | kjetil b halvorsen♦ | CC BY-SA 4.0 |
added 159 characters in body
|
| Mar 11, 2023 at 23:44 | comment | added | kjetil b halvorsen♦ | The only inverse used here is $\Sigma_{22}^{-1}$, so the proof is valid in the case that $\Sigma$ is singular, but its second block $\Sigma_{22}$ is not. That is worth pointing out! | |
| Feb 14, 2023 at 1:39 | comment | added | schroederadrian | I know this is an old thread but from going from line 2 to 3, why is $E[z|x_2] = E[z]$? I'm under the impression that $E[z|x_2] = E[x_1+Ax_2|x_2] = \mu_1 + AE[x_2|x_2] = \mu_1 + Ax_2$ which is not equal the unconditional expectation $\mu_1 + A\mu_2$? | |
| May 26, 2019 at 13:30 | review | Suggested edits | |||
| May 26, 2019 at 16:22 | |||||
| Jan 3, 2019 at 4:22 | comment | added | Mathews24 | Found these set of notes which contain the above derivation along with associated proofs. | |
| Oct 25, 2018 at 22:01 | comment | added | Mathews24 | This is a beautiful explanation. Is there any way to reference or reproduce/cite it? | |
| May 5, 2018 at 16:03 | comment | added | Ken T | @jakeoung I also don't quite understand that statement. I understand in this way: If $cov(z, x_2)=0$, then $cov(C_1^{-1} z, x_2) = C_1^{-1} cov( z, x_2)=0$. So the value of $C_1$ is somehow an arbitrary scale. So we set $C_1=I$ for simplicity. | |
| Jan 14, 2018 at 14:40 | comment | added | probabilityislogic | @jakeoung - it is not proving that $C_1=I$, it is setting it to this value, so that we get an expression that contains the variables we want to know about. | |
| Jan 13, 2018 at 22:47 | comment | added | jakeoung | @probabilityislogic, How can $p(z|x_2)=p(z)$ leads to $C_1=I$? | |
| Jul 6, 2016 at 10:09 | comment | added | Quirik | @Marco What is the reason of defining Z? How can Z and X2 be independent? | |
| Sep 4, 2015 at 10:09 | review | Suggested edits | |||
| Sep 4, 2015 at 12:32 | |||||
| S Jun 18, 2015 at 8:18 | history | suggested | Naetmul | CC BY-SA 3.0 |
Typo and alignment
|
| Jun 18, 2015 at 7:49 | review | Suggested edits | |||
| S Jun 18, 2015 at 8:18 | |||||
| Mar 20, 2013 at 17:00 | comment | added | mathlete | A textbook reference for the proof you gave would be nice, indeed. Cheers. | |
| Mar 20, 2013 at 13:46 | comment | added | Macro | @Marius, thank you for the close read and for catching that typo. You're right that I didn't prove that the conditional distributions are indeed normal, rather I explicitly appealed to what I thought was a commonly known theorem. In any case, I took the OP's main question to be "Would anyone provide me a derivation steps of deriving $\overline{\boldsymbol\mu}$ and $\overline{\Sigma}$ ?", which is why I only focused on deriving the mean and covariance. When I have time, I may consider adding the piece you mentioned (or, at least, linking to a textbook page). Cheers! | |
| Mar 20, 2013 at 13:44 | history | edited | Macro | CC BY-SA 3.0 |
fixed typo suggested in the edits.
|
| Mar 20, 2013 at 9:17 | comment | added | mathlete | @Macro: Great proof. There is a small typo: A var(x_1) A' should be A var(x_2) A'. Also, I guess you need to justify why the conditional distribution is indeed normal. You showed the form of the mean vector and covariance matrix of this conditional distribution, but not that it is indeed normal (and not, say, multivariate t with that mean and covariance matrix). That should easily follow from the fact that linear combinations are normal again, but I guess you should add a comment on that. | |
| Jul 2, 2012 at 20:06 | comment | added | Macro | @probabilityislogic, I'd actually never thought about the process that resulted in choosing this linear combination but your comment makes it clear that it arises naturally, considering the constraints we want to satisfy. +1! | |
| Jul 2, 2012 at 16:05 | comment | added | probabilityislogic | The current ordering is based on the approach of proposing a linear combination, and seeing if it works. My suggesting goes more towards finding a criterion we want our linear combination to satisfy, and solving for this criterion. This way will work better on other problems. | |
| Jul 2, 2012 at 16:00 | comment | added | probabilityislogic | This is a very good answer (+1), but could be improved in terms of the ordering of the approach. We start with saying we want a linear combination $z=Cx=C_1x_1+C_2x_2$ of the whole vector that is independent/uncorrelated with $x_2$. This is because we can use the fact that $p(z|x_2)=p(z)$ which means $var(z|x_2)=var(z)$ and $E(z|x_2)=E(z)$. These in turn lead to expressions for $var(C_1x_1|x_2)$ and $E(C_1x_1|x_2)$. This means we should take $C_1=I$. Now we require $cov(z,x_2)=\Sigma_{12}+C_2\Sigma_{22}=0$. If $\Sigma_{22}$ is invertible we then have $C_2=-\Sigma_{12}\Sigma_{22}^{-1}$. | |
| Jun 17, 2012 at 15:26 | comment | added | gui11aume | Wow!! +1 for the patience and the care to write all this. | |
| Jun 17, 2012 at 15:15 | vote | accept | Flying pig | ||
| Jun 17, 2012 at 15:02 | comment | added | Macro | @Flyingpig, you're welcome. I believe this is a result of equations $(291),(292)$, combined with an additional property of the variance of the sum of random vectors not written in the Matrix Cookbook - I've added this fact to my answer - thanks for catching that! | |
| Jun 17, 2012 at 15:01 | history | edited | Macro | CC BY-SA 3.0 |
added note not proven in the matrix cookbook.
|
| Jun 17, 2012 at 6:35 | comment | added | Flying pig | Thanks for this brilliant method! There is one matrix algebra does not seem familiar to me, where can I find the formula for opening $var(x_1+Ax_2)$? I haven't found it on the link you sent. | |
| Jun 16, 2012 at 23:59 | history | edited | Macro | CC BY-SA 3.0 |
added 297 characters in body
|
| Jun 16, 2012 at 23:29 | history | answered | Macro | CC BY-SA 3.0 |