I know that \(\displaystyle I(\theta)= -\mathbb{E} l''(\theta). \)

Here\(\displaystyle l(\theta)=log L(\theta).\)

Also I know \(\displaystyle f(x; \mu, \sigma^2)= \frac{1}{\sqrt{2\pi \sigma^2}} \exp \Big(

- \frac{(x-\mu)^2}{2\sigma^2}\Big)\)

and the likelihood function is

\(\displaystyle L(\mu,\sigma^2|x_i)= \prod_{i=1}^n f(x_i;\mu,\sigma^2).\)

So I can find

\(\displaystyle l(\theta)=log L(\theta)=-\frac{n}{2}log(2\pi \sigma^2)-\frac{\sum_{i=1}^n(x_i-\mu)^2}{2\sigma^2}.\)

Is it right? But what is next? I need to find \(\displaystyle l''(\theta)\) but how to do this?