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Apr 27, 2020 at 13:14 comment added DglPr .Thank you for clarification.
Apr 22, 2020 at 10:20 comment added Oriol B I don't know what part of the last line you are referring to, so I'll walk through. I found the MLE by simply maximising the $\ell$ with respect to $\sigma$. Equivariance states that if $g:\Theta \to \Theta'$ is a bijection and $\hat{\theta}$ is the MLE of $\theta$, then $g(\hat{\theta})$ is the MLE of $\theta$. Finally I say that your problem is easier by noting that $Z:=X-\mu$ has a distribution $Z\sim Exp\left(\frac{1}{\sigma}\right)$ and the MLE of $\sigma$ (or $\lambda$ in the usual notation) is well-known.
Apr 20, 2020 at 20:01 comment added DglPr I could not understand you last line. Could you please elaborate it?
Apr 20, 2020 at 19:42 history answered Oriol B CC BY-SA 4.0