Given ranndom sample $X_1,\ldots,X_n$ from Laplace distribution $f(x)=\frac{1}{2\theta}e^{-\frac{|x-\mu|}{\theta}}$ where $\mu$ is known and knowing the fact that $E|X-\mu|=\theta$. Find maximum likelihood estimator $\hat\theta$.

My attempt was: $$ l = \frac{1}{\left(2\theta\right)^n}e^{-\sum\frac{|x-\mu|}{\theta}} \implies \ln l \implies \frac{\partial \ln l}{\partial\theta}=\frac{-n}{2\theta}-\sum \frac{|x-\mu|}{\theta}=0.$$ With this i was trying to obtain $\hat\theta$ but i do not know how should i do that.

Thus how to find this estimator?


Firstly you have the wrong derivative - It should be $$ \frac{\partial }{\partial \theta}\ln l = -\frac{n}{\theta} +\frac{\sum |x_i-\mu|}{\theta^2} $$ this should be zero so we have $$ \theta = \frac{\sum |x_i-\mu|}{n} $$ This is the maximum likelihood - I am guessing you are trying to prove it is unbiased?

  • $\begingroup$ I am trying to find probability convergence $\hat \theta \rightarrow_{n\rightarrow \infty}$ $\endgroup$ – Mrowkacala Apr 20 '17 at 18:16

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