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Find the maximum likelihood estimator of the unknown parameter $\theta$ if $X_1, X_2,..., X_n $ is a sample from a distribution whose density function is $$ f_x(x) = (1/2)e^{-{|x-\theta|}}, -\infty < x < \infty $$ Hint: It might be easier to consider separately the cases where n is odd and n is even.

So far, I used the log-likelihood function to get to: $$ L(X_1...X_n;\theta) = n\ln(1/2)-\sum_{j=1}^n |x_j- \theta| $$ Now, I need to take the derivative with respect to $\theta$, but the summation of the absolute value is throwing me off and I don't know how to approach this. I feel like I can handle the problem once I get through this part, so any help in terms of approaching this would be greatly appreciated.

Thanks!

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2 Answers 2

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$\sum_{j=1}^{n}\left|\theta-x_j\right|$ is minimized by the median of the $x_j$’s. So you should consider the cases $n$ is odd or even.

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The derivative is: $$ \frac{d}{d\theta}|x-\theta|=\frac{x-\theta}{|x-\theta|}=\frac{x-\theta}{\sqrt{(x-\theta)^2}} $$ which should help you finish the problem.

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  • $\begingroup$ Rather than the last equality, I would have ended with saying this is the sign of $x-\theta$. In particular you are going to want it to be zero when $x=\theta$, at least for the $n$ odd case, even if that is not strictly the derivative. $\endgroup$
    – Henry
    Nov 25, 2013 at 8:26

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