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
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3-4   [\\(\> 80\\)]{.math .inline}g       71          82
      [\\(\< 80\\)]{.math .inline}g       60         441
                              total      131         523
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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
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Thus, a [$95\%$]{.math .inline} interval estimate is approximately [$[0.1,0.5]$]{.math .inline}.