Timeline for Generate marginally dependent (with predetermined covariance) but conditionally independent data from a Mixture of Gaussians
Current License: CC BY-SA 4.0
11 events
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| Sep 28, 2023 at 11:44 | history | edited | Yves | CC BY-SA 4.0 |
Add generalisation
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| Sep 27, 2023 at 17:45 | vote | accept | Sergio | ||
| Sep 27, 2023 at 15:43 | comment | added | Yves | Yes! You can also start from non diagonal matrices $\boldsymbol{\Sigma}_0$ and $\boldsymbol{\Sigma}_1$ corresponding to a negative correlation and get a marginal $\boldsymbol{\Sigma}$ with positive correlation illustrating Simspon's paradox. | |
| Sep 27, 2023 at 15:19 | comment | added | Sergio | Ohh thank you. This is super interesting. Does that imply that, if I am only interested in choosing $\sigma_{0,1}$, then the choice of $\Sigma,\Sigma_0$ and $\Sigma_1$, with $\Sigma_0=\Sigma_1$ that guarantee the rank-one condition are the following: $\Sigma=\begin{pmatrix} 2\sigma_{0,1} & \sigma_{0,1}\\ \sigma_{0,1} & 2\sigma_{0,1}\end{pmatrix}$ and $\Sigma_0=\Sigma_1=\begin{pmatrix} \sigma_{0,1} & 0\\ 0 & \sigma_{0,1}\end{pmatrix}$? Then we would have $\Delta=\begin{pmatrix}\sigma_{0,1} &\sigma_{0,1}\\ \sigma_{0,1}&\sigma_{0,1}\end{pmatrix}$, so that $\delta_0=\delta_1=\sqrt{\sigma_{0,1}}$. | |
| Sep 27, 2023 at 13:49 | comment | added | Yves | For the solution to exist, $\boldsymbol{\Delta}$ must be of rank one which is actually the case only when $\sigma_{0,1} =1$. If you want to escape this rank-one condition you need to use a mixture of $3$ normal distributions. The eigendecomposition of $\boldsymbol{\Delta}$ provides the group means. Written as $\boldsymbol{\Delta} = \mathbf{P}\mathbf{D}\mathbf{P}^\top$ were $\mathbf{P}$ is orthogonal gives $\boldsymbol{\Delta}$ as the sum of the $d_i \mathbf{p}_i\mathbf{p}_i^\top$ | |
| Sep 27, 2023 at 13:29 | comment | added | Sergio | Given your formulas, then we would have that $\delta\delta^\top = \Delta = \begin{pmatrix} 2 & 2\sigma_{0,1}\\ 2\sigma_{0,1} & 2\end{pmatrix}$ so that, $\delta_0=\delta_1=\sqrt{2}$ but $\delta_0\delta_1=2\sigma_{0,1}$ which is not true in the general case where $\sigma_{0,1}\neq 1$. Could you please point to my misunderstanding, if there is any? | |
| Sep 27, 2023 at 13:24 | comment | added | Sergio | Thank you for your answer. I have been digesting it for a couple of days and tried to do a simple example by myself and failed. Let me show you. Suppose $\mu=[0,0], \Sigma=\begin{pmatrix}3 & 2\sigma_{0,1}\\ 2\sigma_{0,1} & 3\end{pmatrix}$ and $\Sigma_0=\Sigma_1=I$. Then according to your answer, we should be able to derive the mean vectors that produce $\Sigma$. If $\delta = [\delta_0, \delta_1]$, we have $\delta\delta^\top = \begin{pmatrix} \delta_0^2 & \delta_0\delta_1\\ \delta_0\delta_1 & \delta_1^2\end{pmatrix}$. | |
| Sep 25, 2023 at 6:00 | history | edited | Yves | CC BY-SA 4.0 |
edited body
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| Sep 24, 2023 at 15:03 | history | edited | Yves | CC BY-SA 4.0 |
added 2 characters in body
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| Sep 24, 2023 at 14:21 | history | edited | Yves | CC BY-SA 4.0 |
typo
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| Sep 24, 2023 at 14:12 | history | answered | Yves | CC BY-SA 4.0 |