I have read this derivation. $$L(f)\equiv{lng(f|DX)}=nln(f)+(N-n)ln(1-f)+const \;(4.69) $$ expand L(f) in a power series about $\hat{f}$.The first terms as $$L(f) = L(\hat{f}) - \frac{(f-\hat{f})^2}{2\sigma^2}+... \;(4.70)$$ , where $$\sigma^2 \equiv\frac{\hat{f}(1-\hat{f})}{N},\; where \;n = Nf$$

But I can not get that result like 4.70. For $$L'(\hat{f})=\frac{(n-n\hat{f} - N\hat{f}+n\hat{f})}{\hat{f}(1-\hat{f})}=\frac{N(f-\hat{f})}{\hat{f}(1-\hat{f})}$$, So I will give my result with $$L(f)=L(\hat{f})+\frac{(f-\hat{f})^2}{\frac{\hat{f}(1-\hat{f})}{N}}+...\;(4.71)$$ I don't think (4.71) is the same as 4.70. So i posted this thread. Where am i wrong? Thanks.

  • Your Taylor series $(4.7.1)$ seems missing $(f-\hat{f})$. Shoud it be $L(f)=L(\hat{f})+\frac{L'(\hat{f})}{1!}(f-\hat{f})?$ And are you sure the Taylor series just expanded to the first derivative? – Deep North Oct 19 '17 at 5:23
  • Yes, I have tried. – hello_god Oct 19 '17 at 14:52

Your equation for the first derivative is incorrect, since f should be replaced by f-hat, giving a first derivative of zero. You need to use the second derivative to get Jaynes's result.

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