I am trying to understand the proof for this theorem from the book Casella and Berger (2nd ed.) page 336. $W(X)$ if any estimator for samples $X_1,\ldots,X_n$ on distribution $f(X|\theta)$. I notice that there is no assumption of unbiasedness for $W(X)$ till equation 7.3.8 which reads $$E_{\theta}\left(\frac{\partial}{\partial\theta}log f(X|\theta)\right)=\frac{d}{d\theta}E_{\theta}[1]=0$$ This makes sense according to the argument given in the book where they substitute $W(X)=1$ to come to the above result. But I fail to see this equation holding in general. More specifically, if I think of a uniform distribution between 0 and $\theta$ (where $\theta$ parameterize the distribution), then $f(x|\theta)=\frac{1}{\theta}$. And hence, $E_{\theta}\left(\frac{\partial}{\partial\theta}log f(X|\theta)\right)=\int_{\mathcal{X}}\frac{\partial}{\partial{\theta}}log f(x|\theta)f(x|\theta)dx=\int_{\mathcal{X}}\frac{\partial}{\partial{\theta}}f(x|\theta)dx=-\int_{0}^{\theta}\frac{dx}{\theta^2}\neq0$

What am I not understanding?

edit: After πr8 suggestion, I tried it on exponential and normal distribution and it does give 0. Is discontinuity at boundary of the example I gave the reason?

  • 1
    $\begingroup$ there will be extra smoothness assumptions somewhere - which conditions do they place on $f$? $\endgroup$
    – πr8
    Jul 17 '20 at 12:34
  • $\begingroup$ The smoothness condition is the interchangeability of integration and differentiation, which seems to be satisfied for the example I gave, isn't it? except the boundaries, is that the issue? $\endgroup$
    – manav
    Jul 17 '20 at 12:39
  • 1
    $\begingroup$ yes, the boundaries are the issue $\endgroup$
    – πr8
    Jul 17 '20 at 12:59
  • 2
    $\begingroup$ The uniform distribution has a support depending on $\theta$, thus a regularity condition is violated. That is to say the CR bound/inequality is not applicable here. $\endgroup$ Jul 17 '20 at 13:09
  • $\begingroup$ thanyou @πr8 and StubbornAtom. regularity conditions, which I pushed under the carpet came back to bite me. Thanks for the hints. $\endgroup$
    – manav
    Jul 17 '20 at 13:15

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