Suppose $X_1, X_2, ..., X_n$ is a random sample from a population with pdf

$$f(x|\theta) = \dfrac{1}{2\theta}e^{-|x|/\theta},x\in \mathbb{R}$$

The pivotal quantity is $\frac{2}{\theta}\sum_{i=1}^n |X_i|$. How to derive it? How to find pivotal in general?

Update: Is $\frac{2}{\theta}\sum_{i=1}^n |X_i|$ a chi-square distribution?

P.S. this problem is from our previous comprehensive exam. We have no answer for it.

  • $\begingroup$ Hint: $X_i / \theta$ is distributed Laplace (another name for this distribution) with scale equal to one. $\endgroup$
    – dsaxton
    Commented Aug 8, 2015 at 17:35
  • $\begingroup$ Either way I can give some hints: You're trying to estimate $\theta$, which is a scale parameter in the density. Consider, specifically, what happens to $|X|$ when $\theta$ increases or decreases (indeed, consider the distribution of $|X|$ explicitly if that helps). Then consider the definition of a pivotal quantity. $\endgroup$
    – Glen_b
    Commented Aug 9, 2015 at 1:37
  • 1
    $\begingroup$ Could you please show how you arrive at it being chi-square? $\endgroup$
    – Glen_b
    Commented Aug 10, 2015 at 9:35
  • 1
    $\begingroup$ $\frac{1}{2} e^{-|x-0|} \sim Laplace(0,1)$, then $\frac{1}{2\theta} e^{-|x-0|/\theta}\sim Laplace(0, 1/\theta)$. According to the link, $\frac{2}{\theta}\sum_{i=1}^n |X_i|\sim \chi (2n)$. $\endgroup$
    – Gejun
    Commented Aug 10, 2015 at 18:56
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    $\begingroup$ That's right--because $|X_i|$ is Exponential, sums of iid Exponentials are Gamma, and the $\chi^2$ distribution is a special form of the Gamma. BTW, one way to answer questions about finding pivotal quantities is explained and illustrated (with a closely related distribution) in the answer at stats.stackexchange.com/a/10849/919. $\endgroup$
    – whuber
    Commented Aug 10, 2015 at 19:40

1 Answer 1


An outline of one way to approach it.

Step 1: show that $|X_i|/\theta$~Exponential(1)

Step 2: Hence explain how $\frac{2}{\theta}|X_i|\sim\chi^2_2$

Step 3: Hence give the distribution of $\frac{2}{\theta}\sum_i|X_i|$


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