I am asked to find the fisher information contained in $X_1 \sim N(\theta_1, \theta_2)$ (ie: two unknown parameters, only one observation). How would I find the Fisher information here?
I know that with a sample $X_1,X_2,\ldots,X_n $~$N(\mu,\sigma^2)$ and $\sigma^2=1$, Fisher's information is given by :
$$
-E(\frac{d^2}{d\mu^2} \ln f(x))=1/\sigma^2.
$$
Though this is the case with one paramter and I am not sure how it would map on to the case with two parameters. I imagine there is some use of a Hessian but I am not sure what to do.
Best Answer
It will be the expected value of the Hessian matrix of $\ln f(x;\mu, \sigma^2)$. Specifically for the normal distribution, you can check that it will a diagonal matrix. The $\mathcal{I}_{11}$ you have already calculated. For the second diagonal term $$ \ln f(x;\mu, \sigma)=-\frac{1}{2}\ln(2 \sigma^2)+\frac{1}{2\sigma^2}(x-\mu)^2, $$ $$ l'_{\sigma^2} = - \frac{1}{2\sigma^2} - \frac{1}{2\sigma^4}(x-\mu)^2, $$ hence $$ \mathcal{I}_{22}= -\mathbb{E}[l''_{\sigma^2}] = - \mathbb{E} [ \frac{1}{2\sigma^4} - \frac{1}{\sigma^6}(x-\mu)^2] = -\frac{1}{2\sigma^4} + \frac{2}{\sigma^4} = \frac{1}{2\sigma^4} . $$ And for the non-diagonal terms $$ \mathcal{I}_{22}= -\mathbb{E}[l''_{\sigma^2,\mu}] = - \mathbb{E}\frac{2(x-\mu)}{2\sigma^4} = 0. $$