Let $X$ distribution belongs for the family[MATH] \mathcal{P}\{P_{\theta}, \theta \in \Theta \}[/MATH]. We need to find Fisher information[MATH] I(\theta)[/MATH] according[MATH] n[/MATH] simple sample, when [MATH]P_{\theta}[/MATH] is[MATH] N(\mu,\sigma^2)[/MATH] distribution, [MATH]\theta=(\mu,\sigma^2)^T.[/MATH]
I know that [MATH] I(\theta)= -\mathbb{E} l''(\theta). [/MATH]Here[MATH] l(\theta)=log L(\theta).[/MATH]
Also I know [MATH]f(x; \mu, \sigma^2)= \frac{1}{\sqrt{2\pi \sigma^2}} \exp \Big( - \frac{(x-\mu)^2}{2\sigma^2}\Big)[/MATH]
and the likelihood function is
[MATH]L(\mu,\sigma^2|x_i)= \prod_{i=1}^n f(x_i;\mu,\sigma^2).[/MATH]
So I can find
[MATH]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}.[/MATH]
Is it right? But what is next? I need to find [MATH]l''(\theta)[/MATH] but how to do this?
I know that [MATH] I(\theta)= -\mathbb{E} l''(\theta). [/MATH]Here[MATH] l(\theta)=log L(\theta).[/MATH]
Also I know [MATH]f(x; \mu, \sigma^2)= \frac{1}{\sqrt{2\pi \sigma^2}} \exp \Big( - \frac{(x-\mu)^2}{2\sigma^2}\Big)[/MATH]
and the likelihood function is
[MATH]L(\mu,\sigma^2|x_i)= \prod_{i=1}^n f(x_i;\mu,\sigma^2).[/MATH]
So I can find
[MATH]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}.[/MATH]
Is it right? But what is next? I need to find [MATH]l''(\theta)[/MATH] but how to do this?