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I am given that the order statistics $X_{(1)}, ..., X_{(n)}$ are iid from a distribution with pdf

$$f_X(x;\theta) = \frac{1}{2}e^{-|x-\theta|}, x\in \Bbb R, \theta\in \Bbb R$$

I approached with a likelihood function

$$L(\theta) = 2^{-n}\exp\left[-\sum_{i=1}^n |X_i-\theta|\right]$$

and the log likelihood

$$l(\theta) = -n \ln2-\sum_{i=1}^n |X_i-\theta|$$

I am not comfortable taking the derivative from here since the range of $X_i$ and $\theta$ does not let me guarantee that it will be $1$ or $-1$ and it also bothers me that the pdf is not the distribution of the order statistics that I am given.

My ultimate goals is to do a likelihood ratio test where

$$H_0: \theta = \theta_0 \quad vs \quad H_1: \theta \ne \theta_0$$

so that is why I am trying to find the MLE.

My notes suggest that the median is the MLE but I have no idea why that would be true.

I appreciate your input.

hyg17
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  • Related questions: https://math.stackexchange.com/questions/1678740/mle-of-double-exponential?noredirect=1&lq=1, https://math.stackexchange.com/questions/240496/finding-the-maximum-likelihood-estimator, https://math.stackexchange.com/questions/113270/the-median-minimizes-the-sum-of-absolute-deviations-the-l-1-norm?noredirect=1&lq=1. – StubbornAtom Jul 13 '19 at 16:02

1 Answers1

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Differentiating wrt $\theta$ gives: $$\frac{\partial l}{\partial \theta}=-\sum_{i=0}^n\operatorname{sgn}\left(\theta-X_i\right)$$

This is zero iff the number of $X_i$'s that are larger than $\theta$ is equal to the number of $X_i$'s that are less than $\theta$. That is, when $\theta$ is the median of the $X_i$'s.

Stefan Lafon
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  • Ohhh and if the number of $X_i$s that are less than $\theta$ exceeds the median the value just becomes smaller and smaller. That makes sense! Thank you! – hyg17 Jul 11 '19 at 17:21
  • But $\ell$ is not differentiable at any vector $X$ with has any entry equal to $\theta$. – ViktorStein Apr 27 '22 at 16:51