Timeline for How to check if I got the right MLE
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30, 2018 at 13:09 | vote | accept | Christoffer GT | ||
| Jan 30, 2018 at 13:09 | comment | added | Christoffer GT | I think I got it now!! ((XB^-1)^T.XB^-1)^-1.(XB^-1)^T.Y=γ => ((X^-1B)^T.X^-1B).(XB^-1)^T.Y => X^-1B.Y=γ => Y=XB^-1γ which is equal to: Y=Zγ => Y=XB^-1Bβ => Y=X β, Sorry for the mess, but does it look right? | |
| Jan 30, 2018 at 11:16 | comment | added | Christoffer GT | Thanks again. I think I end up with something usefull now, but I'm still unsure wheter X is invertible or how commutative my matrices are?: ((XB^-1)^T.XB^-1)^-1.(XB^-1)^T.Y=γ => ((X^-1B)^T.X^-1B).(XB^-1)^T.Y => I.B^T.B(XB^-1)^T.Y => IXB.Y=γ => IY=XB^-1γ => Y=XB^-1γ => Y=Zγ . | |
| Jan 30, 2018 at 8:31 | comment | added | Łukasz Grad | Don't multiply by $Z^TZ$ - keep in mind that $(AB)^{-1} = B^{-1}A^{-1}$ and try to get rid of some inverses since $AA^{-1} = I$ of course | |
| Jan 29, 2018 at 20:18 | comment | added | Christoffer GT | Thanks alot for your answer! This is where I previously went before going to a hault. I end up with this expression: (XB)^T.Y=γ(XB^-1)^T.XB^-1 . (Hope you can read this). But from this point I become very unsure, because my matrix algebra is a bit rusty. I assume I somehow how to end up with the linear model that you wrote. But I'm not quite sure how to reach that point? | |
| Jan 29, 2018 at 18:23 | history | answered | Łukasz Grad | CC BY-SA 3.0 |