91Ó°ÊÓ

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a uniform distribution on the interval \([0, \theta]\), so that $$ f(x)= \begin{cases}\frac{1}{\theta} & 0 \leq x \leq \theta \\ 0 & \text { otherwise }\end{cases} $$ Then if \(Y=\max \left(X_{i}\right)\), by the first proposition in Section \(5.5, U=Y / \theta\) has density function $$ f_{U}(u)=\left\\{\begin{array}{cl} n u^{n-1} & 0 \leq u \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Use \(f_{U}(u)\) to verify that $$ P\left[(\alpha / 2)^{1 / n} \leq \frac{Y}{\theta} \leq(1-\alpha / 2)^{1 / n}\right]=1-\alpha $$ and use this to derive a \(100(1-\alpha) \%\) CI for \(\theta\). b. Verify that \(P\left(\alpha^{1 / n} \leq Y / \theta \leq 1\right)=1-\alpha\), and derive a \(100(1-\alpha) \%\) CI for \(\theta\) based on this probability statement. c. Which of the two intervals derived previously is shorter? If your waiting time for a morning bus is uniformly distributed and observed waiting times are \(x_{1}=4.2, x_{2}=3.5, x_{3}=1.7\), \(x_{4}=1.2\), and \(x_{5}=2.4\), derive a 95\% CI for \(\theta\) by using the shorter of the two intervals.

Step by step solution

01

Understanding the Given Distribution and Variable Transformation

The problem gives a random sample from a uniform distribution on the interval \([0, \theta]\). The maximum of the sample \(Y = \max(X_i)\) is transformed into \(U = \frac{Y}{\theta}\), which has a derived density function \(f_U(u) = nu^{n-1}\) for \(0 \leq u \leq 1\). We use this knowledge to find a confidence interval for \(\theta\).

02

Confidence Interval Derivation for Part (a)

To verify and derive the confidence interval (CI), we calculate \(P((\alpha/2)^{1/n} \leq U \leq (1-\alpha/2)^{1/n})\). Integrating the density function, we get:\[P((\alpha/2)^{1/n} \leq U \leq (1-\alpha/2)^{1/n}) = \int_{(\alpha/2)^{1/n}}^{(1-\alpha/2)^{1/n}} n u^{n-1} \, du\]This integral evaluates to:\[[(1-\alpha/2)^{1/n}]^n - [(\alpha/2)^{1/n}]^n = 1 - \alpha\]Thus, the \(100(1-\alpha)\%\) CI for \(\theta\) using this method is:\[\frac{Y}{(1-\alpha/2)^{1/n}} \leq \theta \leq \frac{Y}{(\alpha/2)^{1/n}}\]

03

Confidence Interval Derivation for Part (b)

For \(P(\alpha^{1/n} \leq Y/\theta \leq 1) = 1 - \alpha\), use the cumulative method:\[P(\alpha^{1/n} \leq U \leq 1) = \int_{\alpha^{1/n}}^{1} n u^{n-1} \, du\]Evaluating it gives:\[1 - (\alpha^{1/n})^n = 1 - \alpha\]Hence, the \(100(1-\alpha)\%\) CI for \(\theta\) is:\[Y \leq \theta \leq \frac{Y}{\alpha^{1/n}}\]

04

Compare CI Lengths for Parts (a) and (b)

To find which interval is shorter, compare:1. From Step 2: \(\frac{Y}{(1-\alpha/2)^{1/n}} - \frac{Y}{(\alpha/2)^{1/n}}\)2. From Step 3: \(\frac{Y}{\alpha^{1/n}} - Y\)Interval 2 (Step 3) is usually shorter as it begins at \(Y\), giving more precision.

05

Application of the CI Using Given Data

Given observed values \(x_1=4.2, x_2=3.5, x_3=1.7, x_4=1.2, x_5=2.4\), calculate \(Y = \max(x_i) = 4.2\). For a 95% CI using the shorter interval (Step 3):\[\frac{Y}{\alpha^{1/n}} = \frac{4.2}{0.05^{1/5}} \approx 9.27\]Thus, the 95% CI for \(\theta\) using the shorter interval is \([4.2, 9.27]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximum Likelihood Estimation

Maximum Likelihood Estimation (MLE) is a method used in statistics to estimate the parameters of a probability distribution. Essentially, it seeks the parameter values that make the observed data most probable. Suppose you are working with a uniform distribution on the interval \([0, \theta]\). The likelihood function for a sample \(X_1, X_2, \ldots, X_n\) becomes maximal when \(\theta\) is equal to the maximum observed value, \(Y = \max(X_i)\). In our exercise, this is why \(Y\) is used extensively in calculating the confidence interval for \(\theta\). MLE provides a point estimate of the parameter, and such estimates are often foundational for constructing confidence intervals.

Probability Density Function

The Probability Density Function (PDF) represents how the probability of a continuous random variable is distributed over its possible values. For a uniform distribution, the PDF is constant over its interval, meaning every outcome is equally likely within that range. In the given problem, the PDF is defined as: - \(f(x) = \frac{1}{\theta}\) for \(0 \leq x \leq \theta\) - 0 otherwise.For the derived variable \(U = \frac{Y}{\theta}\), the PDF is: - \(f_U(u) = nu^{n-1}\) for \(0 \leq u \leq 1\) - 0 otherwise.This mathematical function is crucial because it allows the calculation of probabilities and, consequently, the confidence intervals for \(\theta\). By integrating the PDF over a given range, we can find the probabilities required to inform our confidence interval calculations.

Uniform Distribution

Uniform Distribution is a type of probability distribution in which every outcome in the interval has an equal likelihood of occurring. This distribution is one of the simplest to understand and is used often to model situations where we assume all values are equally likely. In our exercise, we deal with a continuous uniform distribution defined over the interval \([0, \theta]\). This means that any value within this range is equally likely to be sampled. This simplicity makes the uniform distribution an excellent starting point for statistical analysis and exercises involving maximum likelihood estimation, as seen in this problem. With uniform distributions, the maximum observed value \(Y\) becomes a natural estimator for \(\theta\), further simplifying statistical computations and interpretations. Understanding uniform distribution can significantly enhance your grasp of foundational statistical methods and their applications.