Hints: Score functions and observed information matrix

Note that \(\beta^T x_l = \beta_1 x_{l1} + \ldots + \beta_k x_{lk} + \ldots + \beta_p x_{lp}\), so that \(\frac{\partial \beta^T x_l}{\partial \beta_k} = x_{lk}\). To simplify some later calculations you can denote \(\theta_l := \exp(\beta^T x_l)\), and calculate \(\frac{\partial \theta_l}{\partial \beta_k}\). In (a) use this to calculate the \(k\)th element of \(U(\beta)\), \(\frac{\partial \log L(\beta)}{\partial \beta_k}\). Similarly, in (b) you calculate the element of \(I(\beta)\) in the \(k\)th row and \(m\)th column, namely \(-\frac{\partial \log L(\beta)}{\partial \beta_k \partial \beta_m}\). (The final expression in (b) is not particularly simple)