# Finding Fisher information

#### Atstovas

##### New member
Let $X$ distribution belongs for the family$$\displaystyle \mathcal{P}\{P_{\theta}, \theta \in \Theta \}$$. We need to find Fisher information$$\displaystyle I(\theta)$$ according$$\displaystyle n$$ simple sample, when $$\displaystyle P_{\theta}$$ is$$\displaystyle N(\mu,\sigma^2)$$ distribution, $$\displaystyle \theta=(\mu,\sigma^2)^T.$$

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?