Fisher information for Laplace Distribution

Question

Assume you have $X_1,\ldots,X_n$ iid with the below pdf and let $x_i$ be the observations of the random variable $X_i$.

I am doing some revision on fisher information functions and I stumbled upon a problem asking to derive the expected information for a Laplace distribution with pdf given by $$f(x;θ)=\frac{1}{2θ}\exp(-\frac{|x|}{θ}) $$

I derived the log likelihood function as $$l(θ)=-n\log(θ)-\frac{\Sigma|x_i|}{θ}-n\log2 $$

$$l'(θ)=\frac{-n}{θ}+\frac{\Sigma|x_i|}{θ^{2}} $$

and $$l''(θ)=\frac{n}{θ^{2}}-2\frac{\Sigma|x_i|}{θ^{3}}$$ and since $E|x_i|=0$ I get that the fisher information is $$-\frac{n}{θ^{2}}$$ which is obviously wrong since it cannot be negative. Any tips on what I have done wrong would be appreciated! Thanks!

EDIT: $E|x_i|=0$ this expectation was wrong

Answer 1

If $$f(x;θ)=\frac{1}{2θ}exp(-\frac{|x|}{θ})$$

then $$l(\theta):=\log f(x;θ) = -\log2 - \log\theta - \frac{|x|}{\theta}$$ $$\frac{\partial l(\theta)}{\partial \theta} = -\frac{1}{\theta} + \frac{|x|}{\theta^2}$$ $$\frac{\partial^2 l(\theta)}{\partial \theta^2} = \frac{1}{\theta^2} - 2\frac{|x|}{\theta^3}$$ then for each measurement the expected information is, $$I_\theta = E_{X|\theta}[-\frac{\partial^2 l(\theta)}{\partial \theta^2}] = E_{X|\theta}[2\frac{|x|}{\theta^3}-\frac{1}{\theta^2}] \\ = \frac{2}{\theta^3} \int\limits_{-\infty}^\infty f(x,\theta) |x|~dx - \frac{1}{\theta^2} \\=\frac{2}{\theta^3} \int\limits_{-\infty }^\infty \frac{1}{2θ}exp(-\frac{|x|}{θ}) |x|~dx - \frac{1}{\theta^2}\\ = \frac{1}{\theta^4}\int\limits_{-\infty}^\infty exp(-\frac{|x|}{θ}) |x|~dx- \frac{1}{\theta^2}\\ = \frac{2}{\theta^4} \int\limits_0^\infty exp(-\frac{x}{θ}) x~dx - \frac{1}{\theta^2} \\ (integrating ~ by ~ parts)= \frac{2}{\theta^4} \theta^2 - \frac{1}{\theta^2} \\ = \frac{1}{\theta^2}$$

Then for independent measurements the expected information simply adds and so because they are iid, from measurments of $X_1,...,X_n$ the expected information in then,

$$ nI_\theta = \frac{n}{\theta^2} $$

Answer 2

If $f(x, \theta) = \frac{1}{2 \theta} \exp (- \frac{|x|}{\theta})$ then $$\partial_\theta \log f(x, \theta) = 2 \exp \Bigl(\frac{|x|}\theta\Bigr) \cdot \theta \cdot \Bigl(-\frac{\exp (- \frac{|x|}{\theta})}{2\theta^2} + \frac{\exp (- \frac{|x|}{\theta})|x|}{2 \theta^3}\Bigr) = \frac{|x|-\theta}{\theta^2}$$

and thus $$f(x, \theta) (\partial_\theta \log f(x, \theta))^2 = \exp\Bigl(-\frac{|x|}\theta\Bigr) \cdot \frac{(|x|-\theta)^2}{2 \theta^5}.$$

The integral in definition of Fisher information is easy to calculate, as

$$\int \exp\Bigl(-\frac{x}\theta\Bigr) \cdot \frac{(x-\theta)^2}{2 \theta^5} \,\textrm{d}x = \exp\Bigl(-\frac{x}\theta\Bigr) \cdot \frac{\theta^2+x^2}{2 \theta^4},$$

notice the missing absolute value here.