I have been looking at this problem and attempted calculating the Fisher information. I got zero, so I become unsure and wanted to my answer checked.
$X_1,X_2, …, X_n$ are i.i.d. with density $\frac{1}{2} exp(-|x- \theta|)$, where $\theta \in \mathbb{R}$ is unknown. The goal is to compute $I_n(\theta)$ for $X_1,X_2, …, X_n$.
Since $X_1,X_2, …, X_n$ are i.i.d., $I_n(\theta) = n I(\theta)$ where $I(\theta)$ is the Fisher information for $X_1$.
Use the definition that $I(\theta) = -E_{\theta} \frac{\partial ^2}{\partial \theta ^2} log p_{\theta}(X)$, get $\frac{\partial}{\partial \theta} log p_{\theta}(X) = \frac{x – \theta}{|x – \theta|}$, and $\frac{\partial ^2}{\partial \theta ^2} log p_{\theta}(X) = \frac{(x – \theta)^2 – |x – \theta|^2}{|x – \theta|^3} = 0$, so $I_n(\theta) = n * 0 = 0$.
I have never seen a zero Fisher information so I am afraid I got it wrong. Any help would be appreciated!
Update: I'm now checking whether the smoothness condition is satisfied, which is used when deriving the formula for Fisher information.
Best Answer
You need to take distributional derivatives:$$ \frac{\partial^2}{\partial \theta^2}|x-\theta| =-\frac{\partial}{\partial \theta}\operatorname{sgn}(x-\theta) = 2\delta(x-\theta)$$ where $\delta$ is the Dirac delta. Then $I(\theta) = E(2\delta(X-\theta)) = 2p_\theta(\theta) = 1.$
EDIT
Ahh, yes, regularity conditions try to stop us from differentiating non-differentiable functions and the like. For your peace of mind (and also if you don't know about / like the distribution approach), you can instead you use the less-differentiated formula $$ I(\theta) = E\left[\left(\frac{\partial \log p_\theta(X)}{\partial \theta}\right)^2\right] = E(\operatorname{sgn}(X-\theta)^2) = E(1),$$ and get the same answer. Surely there's nothing bad going on with the cusp at the origin... the derivative doesn't even diverge there so it can easily be regulated and then removed at the end of the calculation with no effect.