Fisher Information for parametrized Binomial

Question

I am trying to calculate the Fisher Information for a $Bin(n,p)$ with the following parametrization:

$$\theta=\text{log}\frac{p}{1-p}$$ which implies: $$p=\frac{e^\theta}{1+e^\theta}$$

I tried two methods but got two different answers:

$$f_\theta(x)={n\choose k}p^{x}(1-p)^{n-x}$$

In canonical form:

$$f_\theta(x)=e^{log{n\choose k}+x\theta+nlog(\frac{1}{1+e^\theta})}$$

then the fisher information is $$\frac{d^{2}(-n*log\frac{1}{1+e^\theta})}{d\theta^2}=\frac{-ne^\theta}{(1+e^\theta)^2}$$

When I calculate it directly(not using canonical form i.e taking the second derivative and taking the negative expected value) I get:

$$log f_\theta(x)=log {n\choose x}+x*log(\frac{e^\theta}{1+e^\theta})+(n-x)log(\frac{1}{1+e^\theta})$$ $$\frac{\partial}{\partial \theta}log f_\theta(x)=\frac{x(1-e^\theta)}{e^\theta}-(n-x)(1-e^\theta)$$

$$\frac{\partial^2}{\partial \theta^2}log f_\theta(x)=(n-x)e^\theta -\frac{x}{e^\theta}$$

$$E[-\frac{\partial^2}{\partial \theta^2}log f_\theta(x)]=\frac{n(1-e^\theta)}{1+e^\theta}$$ Which one is correct?

Answer 1

Use $f(g(x))'=f'(g(x))\cdot g'(x)$. So $$ \frac{\partial}{\partial \theta}\left(x\log(\frac{e^\theta}{1+e^\theta})\right) = x \frac{1+e^\theta}{e^\theta}\cdot \frac{\partial}{\partial \theta}\left(\frac{e^\theta}{1+e^\theta} \right) = x \frac{1+e^\theta}{e^\theta}\cdot \frac{\partial}{\partial \theta}\left(1-\frac{1}{1+e^\theta} \right) $$ $$ = x \frac{1+e^\theta}{e^\theta}\cdot \frac{1}{(1+e^\theta)^2} \cdot e^\theta = x\frac{1}{1+e^\theta}. $$ And the second derivative is $$ \frac{\partial}{\partial \theta}\left(x\frac{1}{1+e^\theta}\right) =- x \frac{1}{(1+e^\theta)^2} \cdot e^\theta. $$