Stat 203 Lecture 17
Exercise. \(\mathcal{I}(\zeta) = E[U(\zeta)] = \text{var}[U(\zeta)]\).
A Taylor series expansion of the log-likelihood around \(\zeta = \hat\zeta\) shows that
\[ \text{var}[\hat\zeta] \approx 1/\mathcal{I}(\zeta). \]
For regression models, the log-likelihood function is
\[ \ell (\beta_0, \beta_1, \ldots, \beta_p; y) = \sum\limits_{i=1}^n \log\mathcal{P}(y_i; \mu_i, \phi). \]
The score functions have the form:
\[ U(\beta_j) = \frac{\partial \ell(\beta_0, \beta_1, \ldots, \beta_p ; y)}{\partial\beta_j} = \sum\limits_{i=1}^n \frac{\partial \log\mathcal{P}(y_i; \mu_i, \phi)}{\partial\mu_i} \frac{\partial\mu_i}{\partial\beta_j}. \]
quilpie
Recall that \(\mu = \text{Pr}(y=1)\) is the probability that the 10mm rain threshold is exceeded. A (bad) direct linear model could be
\[ \mu = \beta_0 + \beta_1 x. \]
Possible:
\[ \log \frac{\mu}{1-\mu} = \eta = \beta_0 + \beta_1 x. \]
When we have more than one parameter:
\[ \mathcal{J}_{jk}(\beta) = - \frac{\partial U(\beta_j)}{\partial \beta_k} = - \frac{dU(\beta_j)}{d\mu} \frac{\partial \mu}{\partial \beta_k}. \]
The expected information is \(\mathcal{I}_{jk}(\beta) = E[\mathcal{J}_{jk}(\beta)]\).
The expected information relating to \(\beta_j\) is \(\mathcal{I}_{jj}(\beta)\).
quilpie
We find:
\[ \begin{align*} \mathcal{J}_{00}(\beta) &= - \frac{\partial U(\beta_0)}{\partial\beta_0} = - \frac{d U(\beta_0)}{d\mu} \frac{\partial \mu}{\partial \beta_0} &= \sum\limits_{i=1}^n \mu_i (1-\mu_i); \\ \mathcal{J}_{11}(\beta) &= - \frac{\partial U(\beta_1)}{\partial\beta_1} = - \frac{d U(\beta_1)}{d\mu} \frac{\partial \mu}{\partial \beta_1} &= \sum\limits_{i=1}^n \mu_i (1-\mu_i)x_i^2;\\ \mathcal{J}_{01}(\beta) = \mathcal{J}_{10}(\beta) &= - \frac{\partial U(\beta_1)}{\partial\beta_0} = - \frac{d U(\beta_1)}{d\mu} \frac{\partial \mu}{\partial \beta_0} &= \sum\limits_{i=1}^n \mu_i (1-\mu_i)x_i.\\ \end{align*} \]
Similarly:
\[ \text{var}[\hat{\beta}_j] \approx 1/\mathcal{I}_{jj}(\beta), \]
which means that \(\text{se}(\hat{\beta}_j) \approx 1/\mathcal{I}_{jj}(\hat{\beta})^{1/2}\).
The MLE of \(\zeta\), denoted \(\hat{\zeta}\), has the following properties.
The Poisson distribution has the probability function
\[ \mathcal{P}(y; \mu) = \frac{\exp(-\mu)\mu^y}{y!} \]
for \(\mu < \infty\) and where \(y\) is a nonnegative integer. Initially, consider estimating the mean \(\mu\) for the Poisson distribution, based on a sample \(y_1, y_2, \ldots, y_n\).