What is the Fisher information for the parameter $\theta$ in a uniform distribution with likelihood $f(X,\theta)=\frac1\theta 1\{0\le x\le\theta\}$?

If X is U[$0$,$\theta$], then the likelihood is given by $f(X,\theta) = \dfrac{1}{\theta}\mathbb{1}\{0\leq x \leq \theta\}$. The definition of Fisher information is $I(\theta) = \mathbb{E} \left[ \left(\dfrac{d \log(f(X,\theta))}{d\theta} \right)^2 \right]$. How can this be calculated when $\log f(X,\theta)$ is not defined for $\theta < X$? I understand that we also have $f(X,\theta) = 0$ for $\theta < X$ but can we ignore this when taking the expectation? If so, why?

• I'm not sure, but I think one chooses to define the log of the density only on the support of the density. Commented Nov 13, 2017 at 22:41
• I think that makes sense. I suppose we can see the random variable $X$ as a function from $X: \Omega \rightarrow [0,\theta]$, in which case $\log f(X,\theta)$ is well defined. Does that work?
– bri
Commented Nov 15, 2017 at 10:48
• Yes that is one thing you can do. Commented Nov 15, 2017 at 13:44
• See this answer for why Fisher information is not defined here in the usual sense. Commented May 15, 2020 at 16:17
• Thanks for the pointer. So as I've defined it, $I(\theta)$ does exist but it is not an 'interesting' quantity for the reasons outlined in that answer.
– bri
Commented May 18, 2020 at 8:31

it is $$n^2/\theta$$.
We get this from calculating the log-likelihood first which is $$-n \log(\theta)$$, then taking its derivative, we will get $$\frac{-n}{\theta}$$. squaring it and take its expectations we will have $$\frac{n^2}{\theta^2}$$