Solutions

10.41

This problem should have stated that $r$ is assumed known.

a.

The log likelihood for $p$ is

$$\log L(p) = + n (r \log p + \overline{x}\log(1-p))$$

The first and second partial derivatives with respoct to $p$ are

$$\frac{\partial}{\partial p} = \frac{n r}{p} - \frac{n \overline{x}}{1-p}$$

$$\frac{\partial^2}{\partial p^2} = - \frac{n r}{p^2} - \frac{n \overline{x}}{(1-p)^2}$$

So the Fisher information is $I_n(p) = \frac{n r}{p^2 (1-p)}$ and the score test statistic is

$$\sqrt{n} \frac{\frac{n r}{p} - \frac{n \overline{x}}{1-p}} {\sqrt... ...}} = \sqrt{\frac{n}{r}} \left(\frac{(1-p)r + p \overline{x}}{\sqrt{1-p}}\right)$$

b.

The mean is $\mu = r(1-p)/p$. The score statsitic can be written in terms of the mean as

$$\sqrt{\frac{n}{r}} \left(\frac{(1-p)r + p \overline{x}}{\sqrt{1-p}}\right) = \sqrt{n}\frac{\mu - \overline{x}}{\sqrt{\mu+\mu^2/r}}$$

A confidence interval is give by

$$C = \left\{\mu: \left\vert\sqrt{n}\frac{\mu - \overline{x}}{\sqrt{\mu+\mu^2/r}}\right\vert \le z_{\alpha/2}\right\}$$

The endpoints are the solutions to a quatriatic,

$$U,L = \frac{r(8\overline{x}+z_{\alpha/2}^2) \pm \sqrt{r z_{\alpha/2}^2}\sqrt{16r\overline{x}+16\overline{x}^2+rz_{\alpha/2}^2}}{8r-2z_{\alpha/2}^2}$$

To use the continuity corection, replace $\overline{x}$ with $\overline{x}+\frac{1}{2n}$ for the upper end point and $\overline{x}-\frac{1}{2n}$ for the lower end point.

Problem: Let $x_1,\dots,x_n$ be constants, and suppose

$$Y_i = \beta_1(1-e^{-\beta_2 x_i}) + \varepsilon_i$$

with the $\varepsilon_i$ independent $N(0.\sigma^2)$ ramdom variables.

a.

Find the normal equations for the least squares estimators of $\beta_1$ and $\beta_2$.

b.

Suppose $\beta_2$ is known. Find the least squares estimator for $\beta_1$ as a function of the data and $\beta_2$.

Solution:

a.

The mean response is $\mu_i(\beta) = \beta_1(1-e^{-\beta_2 x_i})$. So the partial derivatives are

$$\frac{\partial}{\partial\beta_1}\mu_i(\beta) = 1 - e^{-\beta_2 x_i}$$

$$\frac{\partial}{\partial\beta_2}\mu_i(\beta) = \beta_1 (1-e^{-\beta_2 x_i}) x_i$$

So the normal equations are

$$\sum_{i=1}^n \beta_1(1-e^{-\beta_2 x_i})^2 = \sum_{i=1}^n (1-e^{-\beta_2 x_i}) Y_i$$

$$\sum_{i=1}^n \beta_1^2 (1-e^{-\beta_2 x_i})^2 x_i = \sum_{i=1}^n \beta_1 (1-e^{-\beta_2 x_i}) x_i Y_i$$

b.

If $\beta_2$ is known then the least squares estimator for $\beta_1$ can be found by solving the first normal equation:

$$\widehat{\beta}_1 = \frac{\sum_{i=1}^n (1-e^{-\beta_2 x_i}) Y_i} {\sum_{i=1}^n (1-e^{-\beta_2 x_i})^2}$$

Problem: Let $x_1,\dots,x_n$ be constants, and suppose

$$Y_i = \beta_1 + \beta_2 x_i + \varepsilon_i$$

Let $y^*$ be a constant and let let $x^*$ satisfy

$$y^* = \beta_0 + \beta_1 x^*$$

that is, $x^*$ is the value of $x$ at which the mean response is $y^*$.

a.

Find the maximum likelihood estimator $\widehat{x}^*$ of $x^*$.

b.

Use the delta method to find the approximate sampling distribution of $\widehat{x}^*$.

Solution: This prolem should have explicitly assumed normal errors.

a.

Since $x^* = (y^* - \beta_1)/\beta_2$, the MLE is

$$\widehat{x}^* = \frac{y^* - \widehat{\beta}_1}{\widehat{\beta}_2}$$

by MLE invariance.

b.

The partial derivatives of the function $g(\beta_1,\beta_2) = (y^* - \beta_1)/\beta_2$ are

$$\frac{\partial}{\partial\beta_1} g(\beta_1,\beta_2) = -\frac{1}{\beta_2}$$

$$\frac{\partial}{\partial\beta_2} g(\beta_1,\beta_2) = -\frac{y^*-\beta_1}{\beta_2^2}$$

So for $\beta_2 ~= 0$ the variance of the approximate sampling distribution is

$$\widehat{\text{Var}}(\widehat{x}^*) = \nabla g \sigma^2 (X^T X)^{-1} \nabla g^T$$

$$= \frac{\frac{1}{n\beta_2^2}\sum x_i^2 - 2 \overline{x}\frac{y^*-... ...a_1}{\beta_2^3} + \frac{(y^*-\beta_1)^2}{\beta_2^4}}{\sum (x_i-\overline{x})^2}$$

$$= \frac{\frac{1}{n\beta_2^2}\sum (x_i-\overline{x})^2 + \frac{(y^* - \beta_1 - \beta_2\overline{x})^2}{\beta_2^4}}{\sum (x_i-\overline{x})^2}$$

$$= \frac{1}{n\beta_2^2} + \frac{(y^* - \beta_1 - \beta_2\overline{x})^2}{\beta_2^4\sum (x_i-\overline{x})^2}$$

So by the delta method $\widehat{x}^* \sim$   AN$(x^*, \widehat{\text{Var}}(\widehat{x}^*))$. The approximation is reasonably good if $\beta_2$ is far from zero, but the actual mean and variance of $\widehat{x}^*$ do not exist.