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Chapter 4: Problem 13

Suppose a random sample of size 2 is obtained from a distribution that has pdf \(f(x)=2(1-x), 0

Short Answer

Expert verified
To solve the problem, one should perform double integration over the specified ranges followed by simplification of the obtained results. The probability of one sample observation being at least twice as large as the other is the sum of the two probabilities obtained.

Step by step solution

01

Understand the task

The probability density function \(f(x) = 2(1-x)\) for \(0 < x < 1\) is given. The task is to compute the probability that one random sample observation is at least twice as large as the other from the distribution.
02

Formulate the integral

The probability that one sample observation is twice the other involves two possible ways. The first case occurs when the greater observation lies in the interval \(0.5 < x < 1\) and the smaller observation lies in the interval \(0 < x < 0.5\). This can be expressed as the double integral: \(\int_{0.5}^{1}\int_{0}^{x/2}2(1-x)2(1-y) \, dy \, dx\). The second case is when the smaller observation lies in the interval \(0.5 < x < 1\) and the greater observation is at least twice the smaller, which lies in the interval \(0 < x < 1\). This is given by the double integral: \(\int_{0}^{0.5}\int_{2x}^{1}2(1-x)2(1-y) \, dy \, dx\).
03

Solve the integrals

Evaluate both double integrals separately and add the results to compute the total probability.
04

Simplify the result

Simplify the result obtained from the previous step to get a precise value of the probability.

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Key Concepts

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

Random Sample
In probability and statistics, a random sample is a subset of individual observations within a population, such that each observation has an equal chance of being selected. It's a fundamental concept because it ensures that the sample represents the population well, reducing the chance of bias. In the given exercise, a random sample of size 2 is taken from a distribution with a specified probability density function (pdf).

The importance of a random sample, especially in this context, is that it allows for the computation of probabilities regarding observations from the distribution without any predictability or fixed pattern. When the exercise mentions that one observation is at least twice as large as the other, we must consider all possible selections of two observations from the population that satisfy this condition, assuming each pair of observations is equally likely.
Double Integral
A double integral is a form of multiple integration that is used to calculate the volume under a surface in a two-dimensional space. Its application extends to calculating probabilities for continuous random variables in two dimensions. In the exercise, the double integral is utilized to determine the probability of one event occurring relative to another within specific boundaries, allowing for a precise computation in a continuous setting.

Double integrals are computed over a two-dimensional region, and in this exercise, they are evaluated over the ranges where one sample is at least twice the other. When formulating the problem, two cases are considered, and each case yields a separate double integral. Visualizing the region or area that these integrals represent can often aid in understanding the problem. Solving the double integrals separately and then adding them gives us the total probability we're looking for.
Probability Computation
The process of probability computation involves calculating the likelihood of various outcomes for random events. Probability can be assessed using various methods, including mathematical formulas and integrals for continuous probability distributions. In this exercise, the goal is to compute the probability that one random sample is at least twice as large as another.

To do this, we use the provided probability density function and integrate over the appropriate intervals. The probability density function (pdf) dictates the likelihood of each possible outcome within a continuous range. By integrating the pdf over a given range, we obtain the probability of an outcome falling within that range. The concept of probability computation is central to statistics, as it allows prediction and understanding of the chances of different events occurring. In practical terms, our double integrals are tools for this computation, ensuring we get an accurate measure of probability for the defined scenario.

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Most popular questions from this chapter

Verzani (2014), page 323 , presented a data set concerning the effect that different dosages of the drug AZT have on patients with HIV. The responses we consider are the p24 antigen levels of HIV patients after their treatment with AZT. Of the \(20 \mathrm{HIV}\) patients in the study, 10 were randomly assign the dosage of \(300 \mathrm{mg}\) of AZT while the other 10 were assigned \(600 \mathrm{mg}\). The hypotheses of interest are \(H_{0}: \Delta=0\) versus \(H_{1}: \Delta \neq 0\) where \(\Delta=\mu_{600}-\mu_{300}\) and \(\mu_{600}\) and \(\mu_{300}\) are the true mean p24 antigen levels under dosages of \(600 \mathrm{mg}\) and \(300 \mathrm{mg}\) of AZT, respectively. The data are given below but are also available in the file aztdoses. rda. \begin{tabular}{|l|llllllllll|} \hline \(300 \mathrm{mg}\) & 284 & 279 & 289 & 292 & 287 & 295 & 285 & 279 & 306 & 298 \\ \hline \(600 \mathrm{mg}\) & 298 & 307 & 297 & 279 & 291 & 335 & 299 & 300 & 306 & 291 \\ \hline \end{tabular} (a) Obtain comparison boxplots of the data. Identify outliers by patient. Comment on the comparison plots. (b) Compute the two-sample \(t\) -test and obtain the \(p\) -value. Are the data significant at the \(5 \%\) level of significance? (c) Obtain a point estimate of \(\Delta\) and a \(95 \%\) confidence interval for it. (d) Conclude in terms of the problem.

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a \(N(0,1)\) distribution. Then the probability that the random interval \(\bar{X} \pm t_{\alpha / 2, n-1}(s / \sqrt{n})\) traps \(\mu=0\) is \((1-\alpha)\). To verify this empirically, in this exercise, we simulate \(m\) such intervals and calculate the proportion that trap 0, which should be "close" to \((1-\alpha)\). (a) Set \(n=10\) and \(m=50\). Run the \(\mathrm{R}\) code mat=matrix (rnorm \((\mathrm{m} * \mathrm{n}), \mathrm{n} \overline{\mathrm{col}=\mathrm{n}})\) which generates \(m\) samples of size \(n\) from the \(N(0,1)\) distribution. Each row of the matrix mat contains a sample. For this matrix of samples, the function below computes the \((1-\alpha) 100 \%\) confidence intervals, returning them in a \(m \times 2\) matrix. Run this function on your generated matrix mat. What is the proportion of successful confidence intervals? (b) Run the following code which plots the intervals. Label the successful intervals. Comment on the variability of the lengths of the confidence intervals.

Suppose \(X\) is a random variable with the pdf \(f_{X}(x)=b^{-1} f((x-a) / b)\), where \(b \geq 0\). Suppose we can generate observations from \(f(z)\). Explain how we can generate observations from \(f_{X}(x)\).

Frequently, the bootstrap percentile confidence interval can be improved if the estimator \(\widehat{\theta}\) is standardized by an estimate of scale. To illustrate this, consider a bootstrap for a confidence interval for the mean. Let \(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\) be a bootstrap sample drawn from the sample \(x_{1}, x_{2}, \ldots, x_{n} .\) Consider the bootstrap pivot [analog of \((4.9 .13)]:\) $$ t^{*}=\frac{\bar{x}^{*}-\bar{x}}{s^{*} / \sqrt{n}} $$ where \(\bar{x}^{*}=n^{-1} \sum_{i=1}^{n} x_{i}^{*}\) and $$ s^{* 2}=(n-1)^{-1} \sum_{i=1}^{n}\left(x_{i}^{*}-\bar{x}^{*}\right)^{2} . $$ (a) Rewrite the percentile bootstrap confidence interval algorithm using the mean and collecting \(t_{j}^{*}\) for \(j=1,2, \ldots, B\). Form the interval $$ \left(\bar{x}-t^{*(1-\alpha / 2)} \frac{s}{\sqrt{n}}, \bar{x}-t^{*(\alpha / 2)} \frac{s}{\sqrt{n}}\right) $$ where \(t^{*(\gamma)}=t_{([\gamma * B])}^{*} ;\) that is, order the \(t_{j}^{*} \mathrm{~s}\) and pick off the quantiles. (b) Rewrite the \(\mathrm{R}\) program percentciboot.s and then use it to find a \(90 \%\) confidence interval for \(\mu\) for the data in Example 4.9.3. Use 3000 bootstraps. (c) Compare your confidence interval in the last part with the nonstandardized bootstrap confidence interval based on the program percentciboot.s.

Determine a method to generate random observations for the logistic pdf, (4.4.11). Write an R function that returns a random sample of observations from a logistic distribution. Use your function to generate 10,000 observations from this pdf. Then obtain a histogram (use hist \((x, p r=T)\), where \(x\) contains the observations). On this histogram overlay a plot of the pdf.
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