/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Let \(X_{1}, X_{2}, \ldots, X_{n... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from a Rayleigh distribution with pdf $$ f(x ; \theta)=\frac{x}{\theta} e^{-x^{2} /(2 \theta)} \quad x>0 $$ a. It can be shown that \(E\left(X^{2}\right)=2 \theta\). Use this fact to construct an unbiased estimator of \(\theta\) based on \(\Sigma X_{i}^{2}\) (and use rules of expected value to show that it is unbiased). b. Estimate \(\theta\) from the following \(n=10\) observations on vibratory stress of a turbine blade under specified conditions: \(\begin{array}{lllll}16.88 & 10.23 & 4.59 & 6.66 & 13.68 \\ 14.23 & 19.87 & 9.40 & 6.51 & 10.95\end{array}\)

Short Answer

Expert verified
The unbiased estimator for \(\theta\) is approximately 74.51.

Step by step solution

01

Understand the Given Information

We have a random sample from a Rayleigh distribution with the probability density function (pdf) given by \( f(x; \theta)=\frac{x}{\theta} e^{-x^{2} /(2 \theta)} \), where \(x > 0\). We know that the expected value \(E\left(X^2\right) = 2 \theta\).
02

Find the Unbiased Estimator

To construct an unbiased estimator for \(\theta\), we use the given expected value. Since \(E(X^2) = 2\theta\), the expectation of the sample sum is \(E\left(\sum_{i=1}^{n} X_i^2\right) = n \times 2\theta \). Thus, \(E\left(\frac{1}{2n} \sum_{i=1}^{n} X_i^2\right) = \theta\). Therefore, the unbiased estimator for \(\theta\) is \(\hat{\theta} = \frac{1}{2n}\sum_{i=1}^{n} X_i^2\).
03

Calculate the Unbiased Estimator Using Given Data

First, calculate the sum of squares of the given data points:\[ S = 16.88^2 + 10.23^2 + 4.59^2 + 6.66^2 + 13.68^2 + 14.23^2 + 19.87^2 + 9.40^2 + 6.51^2 + 10.95^2 \]Find \( rac{1}{2n} S\) using \(n = 10\).Calculate each square: - \(16.88^2 = 285.2544\)- \(10.23^2 = 104.6529\)- \(4.59^2 = 21.0681\)- \(6.66^2 = 44.3556\)- \(13.68^2 = 187.1424\)- \(14.23^2 = 202.6129\)- \(19.87^2 = 394.4569\)- \(9.40^2 = 88.36\)- \(6.51^2 = 42.4401\)- \(10.95^2 = 119.9025\)Sum these values: \[ S = 285.2544 + 104.6529 + 21.0681 + 44.3556 + 187.1424 + 202.6129 + 394.4569 + 88.36 + 42.4401 + 119.9025 = 1490.246 \]So, \(\hat{\theta} = \frac{1}{2 \times 10} \times 1490.246 = 74.5123\).
04

Conclusion

The unbiased estimate of \(\theta\) based on the given data is approximately \(74.5123\).

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Key Concepts

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

Unbiased Estimator
An unbiased estimator is a statistical term that refers to an estimator whose expected value is equal to the parameter it estimates. In simpler terms, it is a formula or rule that, on average, provides the true value of the parameter in question over many samples. For a Rayleigh distribution, finding an unbiased estimator is crucial for accurately determining the parameter \(\theta\). In the context of Rayleigh distribution, the exercise provides us with the expected value \(E\left(X^2\right) = 2\theta\), which can be used to construct an unbiased estimator for \(\theta\). Given data from a sample, the general idea is to manipulate this expectation so that the parameter \(\theta\) can be isolated and estimated. Consider a sample \(X_1, X_2, \ldots, X_n\) from a Rayleigh distribution. We know \(E\left(X^2\right) = 2\theta\), and therefore \(E\left(\sum_{i=1}^{n} X_i^2\right) = n \times 2\theta\). Solving for \(\theta\) gives us \(E\left(\frac{1}{2n} \sum_{i=1}^{n} X_i^2\right) = \theta\). Thus, \(\hat{\theta} = \frac{1}{2n} \sum_{i=1}^{n} X_i^2\) serves as an unbiased estimator. This means that on average, \(\hat{\theta}\) will provide a correct estimate of \(\theta\), assuming a sufficiently large number of samples.
Expected Value
The expected value in probability and statistics is essentially the average or mean of all possible outcomes of a random variable, weighted by their respective probabilities. For continuous distributions, the expected value is calculated using an integral of the product of a function and its probability density function (pdf). In this exercise, we are given the expected value \(E\left(X^2\right) = 2\theta\) for the Rayleigh distribution. This information is vital because it allows us to find relationships between statistical summaries of data and the parameters of the distribution. Expected value plays a key role in finding unbiased estimators because it ensures that the derived estimator matches the parameter across numerous samples. It serves as a bridge, distributing the actual process of collecting data and forming reasonable estimates of underlying distribution parameters. In essence, it helps in making predictions and making sense of data by providing what kind of results one can expect before actual data collection.
Probability Density Function
A Probability Density Function (pdf) describes the likelihood of a random variable to take on a certain value. For continuous random variables, the pdf provides a mathematical function such that the area under its curve (and above the x-axis) over an interval represents the probability that the variable falls within this interval. The pdf of a Rayleigh distribution is given by \( f(x ; \theta)=\frac{x}{\theta} e^{-x^{2} /(2 \theta)} \) for \(x > 0\). This ensures that for a Rayleigh-distributed random variable, it is more likely to exist around the area where the function achieves higher values. The parameter \(\theta\) in the pdf controls the spread of the distribution; adjusting \(\theta\) affects the scale of the distribution on the x-axis. Understanding the form of the pdf is essential because it enables us to calculate different probabilities and expected values associated with the distribution. Also, in practical applications like estimating parameters (\(\theta\) in this case), the pdf is combined with data to form an accurate representation of the real-world phenomena being studied. Therefore, a sound understanding of pdf is foundational to exploring more advanced concepts in probability and statistics.

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Most popular questions from this chapter

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of \(n\) students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of type I and 50 are of type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a 0,1 , or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let \(p\) denote the proportion of honor- code violators (i.e., the probability of a randomly selected student being a violator), and let \(\lambda=P(\) yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\). a. Let \(Y\) denote the number of yes responses, so \(Y \sim\) Bin \((n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hint: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) b. Use the fact that \(E(Y / n)=\lambda\) to show that your estimator \(\hat{p}\) is unbiased. c. If there were 70 type I and 30 type II cards, what would be your estimator for \(p\) ?

Of \(n_{1}\) randomly selected male smokers, \(X_{1}\) smoked filter cigarettes, whereas of \(n_{2}\) randomly selected female smokers, \(X_{2}\) smoked filter cigarettes. Let \(p_{1}\) and \(p_{2}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes. a. Show that \(\left(X_{1} / n_{1}\right)-\left(X_{2} / n_{2}\right)\) is an unbiased estimator for \(p_{1}-p_{2}\). [Hint: \(E\left(X_{i}\right)=n_{i} p_{i}\) for \(i=1,2\).] b. What is the standard error of the estimator in part (a)? c. How would you use the observed values \(x_{1}\) and \(x_{2}\) to estimate the standard error of your estimator? d. If \(n_{1}=n_{2}=200, x_{1}=127\), and \(x_{2}=176\), use the estimator of part (a) to obtain an estimate of \(p_{1}-p_{2}\). e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

Consider a random sample \(X_{1}, \ldots, X_{n}\) from the pdf $$ f(x ; \theta)=.5(1+\theta x) \quad-1 \leq x \leq 1 $$ where \(-1 \leq \theta \leq 1\) (this distribution arises in particle physics). Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator of \(\theta\).

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2. $$ \begin{array}{rrrrrrr} 5.9 & 7.2 & 7.3 & 6.3 & 8.1 & 6.8 & 7.0 \\ 7.6 & 6.8 & 6.5 & 7.0 & 6.3 & 7.9 & 9.0 \\ 8.2 & 8.7 & 7.8 & 9.7 & 7.4 & 7.7 & 9.7 \\ 7.8 & 7.7 & 11.6 & 11.3 & 11.8 & 10.7 & \end{array} $$ a. Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used. [Hint: \(\left.\Sigma x_{i}=219.8 .\right]\) b. Calculate a point estimate of the strength value that separates the weakest \(50 \%\) of all such beams from the strongest \(50 \%\), and state which estimator you used. c. Calculate and interpret a point estimate of the population standard deviation \(\sigma\). Which estimator did you use? d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds \(10 \mathrm{MPa}\). [Hint: Think of an observation as a "success" if it exceeds 10.] e. Calculate a point estimate of the population coefficient of variation \(\sigma / \mu\), and state which estimator you used.

When the sample standard deviation \(S\) is based on a random sample from a normal population distribution, it can be shown that $$ E(S)=\sqrt{2 /(n-1)} \Gamma(n / 2) \sigma / \Gamma((n-1) / 2) $$ Use this to obtain an unbiased estimator for \(\sigma\) of the form \(c S\). What is \(c\) when \(n=20\) ?

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