 Solution

The likelihood is $L(\theta ;x)={\prod }_{i=1}^{n}f(x_i;\theta )$, and a maximum likelihood estimator is any maximizer $\hat{\theta }\in \arg {\max }_{\theta }L(\theta ;X)$. Write $\ell =\log L$ and score $U={\nabla }_{\theta }\ell$. Fisher information for the sample iswhere the equality uses the usual differentiability and interchange assumptions. For independent identically distributed observations $I_n=n{I}_1$. Under standard interior, identifiability and regularity assumptions,Without those assumptions the large-sample normal assertion need not hold, for example at a boundary parameter.