Considering an i.i.d. sample from a linear model $y_i=\alpha x_i+u_i$ (both $y$ and $x$ are centered with respect to their means) errors are homoscedastic and are distributed as: $$u\sim\frac{1}{\sqrt{2\sigma^2}}e^{\frac{\sqrt{2}{|u_i|}}{\sigma}}$$ How do I calculate the Fisher Information for $\hat\alpha_{ML}$ (maximum likelihood estimator for $\alpha$)?

  • $\begingroup$ I guess that $\hat\alpha_{ML}$ has to be related somehow with the median of $x$ and $y$, but I'm not sure how precisely. $\endgroup$ – GabMac Oct 1 '17 at 9:39
  • $\begingroup$ Trying to compute the second der of the likelihood fnct, I apparently arrived to the conclusion that the Fisher Information has to be null: $$ln(L(y|x,\alpha,\sigma))=-n\ln(\sqrt{2\sigma^2})-\frac{\sqrt{2}}{\sigma}\sum_{i=1}^{n}|y_i-\alpha x_i|\\ \frac{\partial \ln L}{\partial \alpha}=-\frac{\sqrt{2}}{\sigma}\sum_{i=1}^{n}\frac{(y_i-\alpha x_i)(-x_i)}{|y_i-\alpha x_i|}\\ \frac{\partial^2L}{\partial \alpha^2}=0$$ Obviously $L$ is not differentiable when $\alpha=\frac{y_i}{x_i}$, so I'm not sure that this is the right way to proceed to compute the Fischer Information for $\hat\alpha_{ML}$. $\endgroup$ – GabMac Oct 1 '17 at 9:39

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