Solutions

10.30 (b)

For the Huber M-estimator $\psi(-\infty) = -k$ and $\psi(\infty) = k$, so $\eta = 1/(1+1) = 1/2$ and the breakdown is 50%.

The formula for the breakdown given in this problem is only applicable to monotone $\psi$ functions. For redescending $\psi$ functions the estimating equation need not have a unique root. To resolve this one can specify that an estimator should be determined using a local root finding procedure starting at, say, the sample median. In this case the M-estimator inherits the 50% breakdown of the median. See Huber, pages 53-55, for a more complete discussion.

Problem: Consider the setting of Problem 10.31. Derive an expression for $-2 \log \Lambda$, where $\Lambda$ is the likelihood ratio test statistic, and find the approximate distribution of this quantity under the null hypothesis.

Solution: The restricted likelihood corresponds to $n_1+n_2$ Bernoulli trials with $S_1+S_2$ successes and common success probability $p$, so the MLE of $p$ is $\widehat{p} = (S_1+S_2)/(n_1+n_2)$. The unrestricted likelihood consists of two independent sets of Bernoulli trials with success probabilities $p_1$ and $p_2$, and the correpsonding MLS's are $\widehat{p}_1 = S_1/n_1$ and $\widehat{p}_2 = S_2/n_2$. The likelihood ratio statistic is therefore

$$\Lambda = \frac{\widehat{p}^{S_1+S_2}(1-\widehat{p})^{F_1+F_2}} {... ...hat{p}_1}\right)^{F_1} \left(\frac{1-\widehat{p}}{1-\widehat{p}_2}\right)^{F_2}$$

and

$$-2\log\Lambda = 2 \left(S_1 \log\frac{\widehat{p}_1}{\widehat{p}}... ...hat{p}_1}{1-\widehat{p}} + F_2 \log\frac{1-\widehat{p}_2}{1-\widehat{p}}\right)$$

The restricted parameter space under the null hypothesis is one-dimensional and the unrestricted parameter space is two-dimensional. Thus under the null hypothesis the distribution of $-2 \log \Lambda$ is approximately $\chi_1^2$.

10.38

The log likelihood for a random sample from a Gamma( $\alpha,\beta$) distribution is

$$\log L(\beta) = n \left(-\log \Gamma(\alpha) -\alpha \log\beta + (\alpha-1) \frac{1}{n}\sum\log X_i - \overline{X}/\beta\right)$$

So the score function is

$$V_n(\beta) = \frac{\partial}{\partial\beta}\log L(\beta) = n \lef... ...rac{\overline{X}}{\beta^2} \right) = n \frac{\overline{x}-\alpha\beta}{\beta^2}$$

and the Fisher information is

$$I_n(\beta) = - n E\left[\frac{\alpha}{\beta^2}-\frac{2\overline{X... ... \frac{2\alpha\beta}{\beta^3}-n\frac{\alpha}{\beta^2} = n\frac{\alpha}{\beta^2}$$

So the score statistic is

$$\frac{V_n(\beta)}{\sqrt{I_n(\beta)}} = \sqrt{n}\frac{\overline{X}... ...t{\alpha}\beta} = \sqrt{n}\frac{\overline{X}-\alpha\beta}{\sqrt{\alpha\beta^2}}$$