If I want to compute the CRLB for iid uniform on $[0,\theta]$. I need in the denominator this expression: $E_\theta\left[\left(\frac{\partial \log f(X)}{\partial \theta}\right)^2\right]=nE_\theta\left[\left(\frac{\partial \log f(x)}{\partial \theta}\right)^2\right]$.

Notation edit: $X\equiv(X_1,...,X_n)$ and $x\equiv x_1$. In this case we have:

\begin{align} f(X) &=\theta^{-n} \tag{1} \\ \log f(X) &=-n\log \theta \tag{2} \\ \frac{\partial \log f(X)}{\partial \theta} &=\frac{-n}{\theta} \tag{3} \\ \left(\frac{\partial \log f(X)}{\partial \theta}\right)^2 &=\frac{n^2}{\theta^2} \tag{4} \\ E_\theta\left[\left(\frac{\partial \log f(X)}{\partial \theta}\right)^2\right] &=\frac{n^2}{\theta^2}\neq \frac{n}{\theta^2}=nE_\theta\left[\left(\frac{\partial \log f(x)}{\partial \theta}\right)^2\right] \tag{5} \end{align}

I can't see where I am making a mistake.

  • 1
    $\begingroup$ I think it has to do with interchanging differentiation and expectation, which usually fails in the Uniform Distribution, but can't see why. $\endgroup$
    – telemaco
    Nov 27 '15 at 4:53
  • 5
    $\begingroup$ The CRLB does not apply for the uniform distribution, because the support of the distribution depends on the parameter $\theta$, one of the required regularity conditions: en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Rao_bound. $\endgroup$ Nov 27 '15 at 5:20
  • 4
    $\begingroup$ For uniform distributions like the one on $[0,\theta]$, there exist super-efficient estimators that converge faster than $\sqrt{n}$. $\endgroup$
    – Xi'an
    Nov 27 '15 at 10:54
  • 2
    $\begingroup$ You would also need to keep track of the indicator function in the defition of the likelihood, which is $\theta^{-n}\mathbb{I}(\max_iX_i\leq\theta)$ $\endgroup$ Nov 27 '15 at 13:17
  • 2
    $\begingroup$ Under the Uniform, the score function has an expectation different from $0$. $\endgroup$
    – Xi'an
    Jan 25 '19 at 18:22