Answer 9.29 **Exercise 9.29**Let \( Y_1, Y_2, \ldots, Y_n \) denote a random sample of size \( n \) from a power family distribution (see Exercise 6.17). Then the methods of Section 6.7 imply that \( Y_{(n)} = \max(Y_1, Y_2, \ldots, Y_n) \) has the distribution function given by\[F_{(n)}(y) = \begin{cases} 0, & y < 0, \\\left(\frac{y}{\theta}\right)^{\alpha n}, & 0 \leq y \leq \theta, \\1, & y > \theta.\end{cases}\]Use the method described in Exercise 9.26 to show that \( Y_{(n)} \) is a consistent estimator of \( \theta \).---**Explanation:**- The notation \( Y_{(n)} \) refers to the maximum value in the sample \( Y_1, Y_2, \ldots, Y_n \).- The distribution function \( F_{(n)}(y) \) provides probabilities for different ranges of \( y \): - \( F_{(n)}(y) = 0 \) when \( y < 0 \), indicating no probability below zero. - \( F_{(n)}(y) = \left(\frac{y}{\theta}\right)^{\alpha n} \) for \( 0 \leq y \leq \theta \), showing the probability distribution within this range. - \( F_{(n)}(y) = 1 \) when \( y > \theta \), indicating certainty above \( \theta \).- The exercise asks for demonstration of the consistency of \( Y_{(n)} \) as an estimator for \( \theta \), using techniques from a previous exercise (9.26). **Exercise 9.26**It is sometimes relatively easy to establish consistency or lack of consistency by appealing directly to Definition 9.2, evaluating \( P(|\hat{\theta}_n - \theta| \leq \epsilon) \) directly, and then showing that \[\lim_{n \to \infty} P(|\hat{\theta}_n - \theta| \leq \epsilon) = 1\]Let \( Y_1, Y_2, \ldots, Y_n \) denote a random sample of size \( n \) from a uniform distribution on the interval \( (0, \theta) \). If \( Y_{(n)} = \max(Y_1, Y_2, \ldots, Y_n) \), we showed in Exercise 6.74 that the probability distribution function of \( Y_{(n)} \) is given by\[F_{(n)}(y) = \begin{cases} 0, & y < 0, \\(y/\theta)^n, & 0 \leq y \leq \theta, \\1, & y > \theta.\end{cases}\]**a** For each \( n \geq 1 \) and every \( \epsilon > 0 \), it follows that \[P(|Y_{(n)} - \theta| \leq \epsilon) = P(\theta - \epsilon \leq Y_{(n)} \leq \theta + \epsilon).\]If \( \epsilon > \theta \), verify that \( P(\theta - \epsilon \leq Y_{(n)} \leq \theta + \epsilon) = 1 \) and that, for every positive \( \epsilon < \theta \), we obtain \[P(\theta - \epsilon \leq Y_{(n)} \leq \theta + \epsilon) = 1 - [(\theta - \epsilon)/\theta]^n.\]**b** Using the result from part (a), show that \( Y_{(n)} \) is a consistent estimator for \( \theta \) by showing that, for every \( \epsilon > 0 \), \[\lim_{n \to \infty} P(|Y_{(n)} - \theta| \leq

Given that, let Y1,Y2,......,Yn denotes a random sample of size n from a power family distribution,…

*9.29 Let Y₁, Y₂,..., Y, denote a random sample of size n from a power family distribution (see Exercise 6.17). Then the methods of Section 6.7 imply that Y(n) = max(Y₁, Y₂, ..., Y₁) has the distribution function given by 0, y < 0, F(n) (y) (y/0), 0≤ y ≤0, - fre 1, y > 0. Use the method described in Exercise 9.26 to show that Y() is a consistent estimator of 0.