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I have a linear model $Y \sim N(X\beta,\sigma^2*I)$

The mean is parameterized into the form

$E[Y]=Z\gamma$

Where $\gamma=B\beta$ and $Z=XB^{-1}$

I have to show that the maximum likelihood estimator of $\gamma$ is given by:

$\hat{\gamma}=B\hat{\beta}$

I have tried to calculate the mle by differentiating the log likelihood function of the normal distribution in relation to gamma, I set the equation equal to zero and then solve for gamma. I then substitute the expressions for Z and $\gamma$ into the equation, but from this point I am stuck.

Does someone know if this is the right way to solve this problem, or do I just have to check if the estimator is biased in some way?

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I guess this is self study so I will try to give some hints rather than a full answer.

Under the reparametrization you have the following normal model

$$ Y = Z\gamma + \sigma^2I_n \sim N(Z\gamma, \sigma^2I_n) $$

You probably know by now that the ML estimate $\hat{\gamma}$ is equal to the OLS estimate $(Z^TZ)^{-1}Z^TY$

Now substitue back $Z = XB^{-1}$ and try to simplify this expression.

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  • $\begingroup$ 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? $\endgroup$
    – Christoffer GT
    Commented Jan 29, 2018 at 20:18
  • $\begingroup$ 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 $\endgroup$
    – Łukasz Grad
    Commented Jan 30, 2018 at 8:31
  • $\begingroup$ 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 . $\endgroup$
    – Christoffer GT
    Commented Jan 30, 2018 at 11:16
  • $\begingroup$ 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? $\endgroup$
    – Christoffer GT
    Commented Jan 30, 2018 at 13:09

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