Problem 1

If we have a non-linear function of some random variable, in the delta method we use a Taylor approximation of the function centered around the random variables expected value such that the approximate variance of the function of the random variable is easily computed.

Let [\(\operatorname{g}(\hat\pi) = \log(\hat\pi/(1-\hat\pi))\)]{.math .inline}. A linear approximation of [\(\operatorname{g}\)]{.math .inline} is given by [\[\operatorname{\hat g}(\hat\pi) = \operatorname{g}(\pi) +\operatorname{g'}(\pi)(\hat\pi - \pi),\]]{.math .display} the derivative of [\(\operatorname{g}\)]{.math .inline} is given by [\[\operatorname{g'}(\pi) = \frac{1}{\pi(1-\pi)},\]]{.math .display} and the variance of [\(\operatorname{\hat g}(\hat\pi)\)]{.math .inline} is given by [\[\begin{aligned} \operatorname{Var}(\operatorname{\hat g}(\hat \pi)) &= \left(\operatorname{g'}(\pi)\right)^2 \operatorname{Var}(\hat\pi)\\ &= \frac{1}{\pi^2(1-\pi)^2} \frac{\pi(1-\pi)}{n}\\ &= \frac{1}{\pi(1-\pi)} \frac{1}{n}. \end{aligned}\]]{.math .display}

Since we do not know [\(\pi\)]{.math .inline}, we approximate it with [\(\hat\pi\)]{.math .inline}, thus [\[\sigma^2(\log(\hat\pi/(1-\hat\pi))) = \frac{1}{\hat\pi(1-\hat\pi)} \frac{1}{n}.\]]{.math .display} # Problem 2

Consider data from a retrospective study on the relationship between daily alcohol consumption and the onset of esophagus cancer.

                                      cancer   no cancer

3-4   [\\(\> 80\\)]{.math .inline}g       71          82
      [\\(\< 80\\)]{.math .inline}g       60         441
                              total      131         523

Part (a)

In a retrospective study, the table draws samples from the conditional probabilities [\(\operatorname{P}(X = i | Y = j)\)]{.math .inline} and we denote [\(\operatorname{P}(X = 1 | Y = j)\)]{.math .inline} by [\(p_j\)]{.math .inline}.

The ML estimator of [\(p_j\)]{.math .inline} from the data in a retrospective study is given by [\[\hat p_j = \frac{n_{1 j}}{n_{+j}}.\]]{.math .display}

In a retrospective study, an estimator of [\(\sigma(\log \hat\theta)\)]{.math .inline} is given by [\[\hat\sigma(\log \hat\theta) = \left[ \left(\frac{1}{\hat{p}_1} + \frac{1}{1-\hat{p}_1}\right)\frac{1}{n_{+1}} + \left(\frac{1}{\hat{p}_2} + \frac{1}{1-\hat{p}_2}\right)\frac{1}{n_{+2}} + \right]^{1/2}.\]]{.math .display}

Part (b)

It does not depend on the sampling scheme. When asymptotic normality holds, replacing parameters with statistics invokes the likelihood principle.

When we substitute the MLE [\(\hat p_j\)]{.math .inline} into the equation for [\(\hat \sigma(\log \hat \theta)\)]{.math .inline}, we get the result [\[\hat\sigma(\log \hat\theta) = \left(\frac{1}{n_{1 1}} + \frac{1}{n_{2 1}} + \frac{1}{n_{1 2}} + \frac{1}{n_{2 2}} \right)^{1/2},\]]{.math .display} which is the same as for retrospective studies and cross-sectional studies.

Part (c)

Recall [\(\theta = \frac{p_1 / (1 - p_1)}{p_2 / (1 - p_2)}\)]{.math .inline}. If we replace [\(p_j\)]{.math .inline} by its ML estimator [\(\hat p_j\)]{.math .inline}, then [\[\hat p_1 = \frac{n_{1 1}}{n_{+1}} = \frac{71}{131} \approx 0.542\]]{.math .display} and [\[\hat p_2 = \frac{n_{1 2}}{n_{+2}} = \frac{82}{523} \approx 0.157.\]]{.math .display} Thus, by the invariance property of MLEs, [\[\hat\theta = \frac{\hat p_1 / (1 - \hat p_1)}{\hat p_2 / (1 - \hat p_2)} \approx \frac{0.542 / 0.458}{0.157 / 0.843} \approx 6.354\]]{.math .display} and therefore [\(\log \theta \approx \log 6.354 \approx 1.850\)]{.math .inline}. Next, we need to find the standard deviation of the [\(\log \hat\theta\)]{.math .inline}, [\[\hat\sigma(\log \hat\theta) = \left[ \left(\frac{1}{0.542} + \frac{1}{0.458}\right)\frac{1}{131} + \left(\frac{1}{0.157} + \frac{1}{0.843}\right)\frac{1}{523} \right]^{1/2} \approx 0.213.\]]{.math .display}

Thus, a [\(95\%\)]{.math .inline} confidence interval for [\(\log \theta\)]{.math .inline} is [\[\log \hat \theta \pm 1.96 \hat\sigma(\log \hat \theta).\]]{.math .display} Plugging in these computed values, we get the [\(95\%\)]{.math .inline} confidence interval [\[[1.850 - 0.417,1.850 + 0.417] = [1.433,2.267].\]]{.math .display}

Part (d)

Observe that [\[\hat\gamma = \frac{\hat\theta-1}{\hat\theta+1}.\]]{.math .display} The MLE [\(\hat\theta \approx 6.354\)]{.math .inline}. Thus, [\(\hat\gamma \approx \frac{6.354-1}{6.354+1} \approx 0.728\)]{.math .inline}.

We estimate that there is a large size, positive association between daily alcohol consumption and the onset of cancer.

Problem 3

For the counts array, we populated it with the values [\((7,7,2,3,2,8,3,7,1,5,4,9,2,8,9,14)\)]{.math .inline} and ran the code.

##       2.5%        25%        50%        75%      97.5% 
## 0.09879317 0.24392244 0.31425372 0.38217439 0.50166928

Thus, a [\(95\%\)]{.math .inline} interval estimate is approximately [\([0.1,0.5]\)]{.math .inline}.