# Efron Hastie CASI exercise 4.3

This is an exercise from Computer Age Statistical Inference by Efron and Hastie, student edition of the book.

Draw a schematic graph of $$\dot{l}_{x}(\theta)$$ versus $$\theta$$. Use it to justify (4.25) $$\left[\hat{\theta}\dot{=}\theta + \frac{\dot{l}_{x}(\theta)/n}{\ddot{l}_{x}(\theta)/n}\right]$$

I am unsure of what is being asked here. The score $$\dot{l}_{x}(\theta)$$ has been defined in several ways across the chapter. For example, in equation 4.14 as $$\frac{\partial}{\partial\theta}\log f_{\theta}(x)$$, equation 4.16 with $$\dot{l}_{x}(\theta)\sim (0, \mathcal{I}_{\theta})$$ (no distribution in here), equation 4.26 with $$\dot{l}_{x}(\theta)/n \dot{\sim} \mathcal{N}(0, \mathcal{I}_{\theta}/n)$$, and in the specific case of the Gaussian distribution, equation 4.30 gives $$\dot{l}_{x}(\theta)=\frac{1}{\sigma^{2}}\sum_{i=1}^{n}(x_{i}-\theta)$$.

I made plots for equation 4.26 and equation 4.30 (with a single observation):

Additionally, what is there to justify equation 4.25? I think the point is that if we find the $$\hat{\theta}$$ so that $$\dot{l}_{x}(\theta)=0$$, then the estimate is unbiased. If we don't find the exact value, some form of spread on the bias is given in inverse proportion to $$\ddot{l}_{x}(\theta)$$. Is this interpretation correct? Is it the justification that the book is asking for?