# Godambe estimating equation (Proof)

Let $Y_1, \ldots ,Y_n$ be iid with density $f(y;\theta)$.

We assume that $\dfrac\partial{\partial\theta}\log f(y ; \theta)$ and $\dfrac{\partial^2}{\partial\theta^2}\log f(y ; \theta)$ exist for all $\theta$. Consider a class $G$ of real functions $g(Y;\theta)$ such that:

1) $E_\theta[g(Y;\theta)] = 0$.

2) $\dfrac{\partial{g(Y;\theta)}}{\partial\theta}$ exists, is negative and bounded for all $\theta$ and $Y$.

3) $E_\theta[g^2(Y;\theta)] < \infty,\,$ for all $\theta$.

The goal is to show that the score function (derivative of the log likelihood with respect to $\theta$) is a member of $G$ that minimizes:

$$\frac{E_{\theta}[g^2(Y;\theta)]}{(E_\theta[\partial_\theta g(Y;\theta)])^2}$$

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I can't help but read the first word of the title phonetically (almost). Are you frustrated? Or, what is it supposed to say? – cardinal Dec 18 '12 at 3:03
No, actually Godambe is the one who proposed this estimating equation in 1960. No frustration at all :) – Sam Dec 18 '12 at 3:06
Sorry. Bad joke. (+1) – cardinal Dec 18 '12 at 3:12

First, the formula $$\frac{E_{\theta} [g^2 (Y ; \theta)]}{(E_{\theta} [\partial_{\theta} g (Y ; \theta)])^2}$$ is the asymptotic variance of the estimator obtained by the estimating equation. Take $$g \left( y ; \theta \right) = \frac{\partial}{\partial \theta} \log f \left( y ; \theta \right)$$ (that's the score), then the numerator will be given by $$E_{\theta} \left(\left[ \frac{\partial}{\partial \theta} \log f \left( y ; \theta \right) \right]^2\right)$$ and the denominator will be given by $$\left( E_{\theta} \left[ \frac{\partial^2}{\partial \theta^2} \log f \left( y ; \theta \right) \right] \right)^2$$ Remember that the Fisher information is given by $$I \left( \theta \right) = E_{\theta} \left[ \frac{\partial}{\partial \theta} \log f \left( y ; \theta \right) \right]^2 = - E_{\theta} \left[ \frac{\partial^2}{\partial \theta^2} \log f \left( y ; \theta \right) \right]$$ Then the ratio simplifies to $\frac{1}{I \left( \theta \right)}$. This is the Cramér-Rao bound and therefore the variance of the most efficent estimator (the smallest possible variance in the class considered here).
@Sam Condition 2 is probably two strong. In any case, an additional assumption on the set of $\theta$'s being compact is required even for a weaker version. Condition 3 follows by Cauchy-Schwarz from the first condition. If you are still having difficulties, I think it is a good idea to ask a new question on these particular issues to clarify that. (question with a real-analysis tag) – Learner Dec 21 '12 at 0:43