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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.

telemaco
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    I think it has to do with interchanging differentiation and expectation, which usually fails in the Uniform Distribution, but can't see why. – telemaco Nov 27 '15 at 04:53
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    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: https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Rao_bound. – Christoph Hanck Nov 27 '15 at 05:20
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    For uniform distributions like the one on $[0,\theta]$, there exist super-efficient estimators that converge faster than $\sqrt{n}$. – Xi'an Nov 27 '15 at 10:54
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    You would also need to keep track of the indicator function in the defition of the likelihood, which is $\theta^{-n}\mathbb{I}(\max_iX_i\leq\theta)$ – Christoph Hanck Nov 27 '15 at 13:17
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    Understanding this question is problematic. If "$x$" means "$X$" then the first equation is obviously false in general (because both expectations are the same but one of them is multiplied by $n$) whereas if "$x$" is something else, then the expectation on the right hand side is meaningless: where is the random variable? – whuber Nov 27 '15 at 16:07
  • Will it be true that under the uniform distribution, the score not necessarily has expectation equal to zero? – telemaco Nov 27 '15 at 16:20
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    Under the Uniform, the score function has an expectation different from $0$. – Xi'an Jan 25 '19 at 18:22
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    Related: https://stats.stackexchange.com/q/389065/119261. – StubbornAtom Jun 02 '20 at 14:35

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