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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 \(\sum X_{i}^{2}\) (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 108.95.

Step by step solution

01

Understand the given information

The Rayleigh distribution is given with the probability density function (pdf) \( f(x; \theta) = \frac{x}{\theta} e^{-x^2 / (2\theta)} \) for \( x > 0 \). It is given that the expected value \( E(X^2) = 2\theta \). We are tasked with finding an unbiased estimator for \( \theta \) using the sum of squared observations from the distribution.
02

Establish the unbiased estimator

To find an unbiased estimator, we start with \( E\left( X^2 \right) = 2\theta \). If we sum all squared observations from a sample, \( \sum_{i=1}^{n} X_i^2 \), the expected value becomes \( E\left( \sum_{i=1}^{n} X_i^2 \right) = nE(X^2) = 2n\theta \). Therefore, the unbiased estimator for \( \theta \) is \( \hat{\theta} = \frac{1}{2n} \sum_{i=1}^{n} X_i^2 \). Ensuring this estimator is unbiased, we have \( E\left( \hat{\theta} \right) = E\left( \frac{1}{2n} \sum_{i=1}^{n} X_i^2 \right) = \frac{1}{2n} \cdot 2n\theta = \theta \). Thus, \( \hat{\theta} \) is unbiased.
03

Sum the squared data points

Given the vibratory stress observations \( 16.88, 10.23, 4.59, 6.66, 13.68, 14.23, 19.87, 9.40, 6.51, \) and \( 10.95 \), compute their squares and sum them. \( 16.88^2 + 10.23^2 + 4.59^2 + 6.66^2 + 13.68^2 + 14.23^2 + 19.87^2 + 9.40^2 + 6.51^2 + 10.95^2 = 972.5944 + 104.6529 + 21.0681 + 44.3556 + 187.1424 + 202.6129 + 394.8609 + 88.3600 + 42.4401 + 119.9025 = 2178.9907 \).
04

Calculate the estimator

Substitute the sum \( 2178.9907 \) into the formula \( \hat{\theta} = \frac{1}{2n} \sum_{i=1}^{n} X_i^2 \) with \( n = 10 \), obtaining \( \hat{\theta} = \frac{1}{20} \times 2178.9907 = 108.949535 \). Hence, \( \theta \) is estimated to be approximately \( 108.95 \).

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Key Concepts

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

Rayleigh Distribution
The Rayleigh distribution is a continuous probability distribution, which is often used in applications involving random vectors with magnitude and direction. It's particularly important in fields like mechanical engineering and telecommunications.
This distribution is defined by the probability density function (pdf) given by \( f(x; \theta) = \frac{x}{\theta} e^{-x^2 / (2\theta)} \)
where \( x > 0 \) and \( \theta \) is a parameter known as the scale parameter. The Rayleigh distribution is typically used to model the magnitude of a vector that has normally distributed components, which makes it useful for problem contexts such as wind speeds, signal integrity, or mechanical stress vibrations.
Probability Density Function
The probability density function (pdf) is a key concept when working with continuous probability distributions such as the Rayleigh distribution. It describes the likelihood of a random variable to take on a particular value. For Rayleigh, the pdf is
\[ f(x; \theta) = \frac{x}{\theta} e^{-x^2 / (2\theta)} \]
Here, \( \theta \) is the scale parameter and dictates how "spread out" the distribution is.
Understanding the pdf provides valuable insight into how values are distributed and can be used to compute probabilities over various intervals. For instance, the likelihood of the vibratory stress on a turbine blade exceeding a certain level under specific conditions can be deduced using this function.
Expected Value
The expected value, often called the mean, is a measure of the center of a distribution. In the context of the Rayleigh distribution, it provides us a way to determine the average outcome we might expect for squared observations.
Given for the Rayleigh distribution, the expected value of the square of the random variable \( X \) is \( E(X^2) = 2\theta \). This implies that if we take many samples from a Rayleigh distribution, the average of these squared values should approximate \( 2\theta \).
Expected value computations are intuitively useful because they help us relate sample data to theoretical insights, like those needed for estimating parameters such as \( \theta \).
Estimator Calculation
A critical task in statistics is finding unbiased estimators, which are statistical measures that reflect a true parameter value without systematic error. From a sample, unbiased estimator for \( \theta \) can be derived from the Rayleigh distribution based on the sum of squared observations.The formula for the estimator in the exercise is\[ \hat{\theta} = \frac{1}{2n} \sum_{i=1}^{n} X_i^2 \]where \( n \) is the sample size. To ensure that this estimator is unbiased, we establish that its expected value also equals \( \theta \). Processing sample data using this unbiased method gives us a calculated estimate that holds true average accuracy.Using this approach with the provided observations, we find that \( \hat{\theta} = 108.95 \), which means that the estimator closely mirrors the actual parameter value of the sample's underlying distribution.

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Most popular questions from this chapter

Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86, x_{6}=.47\), \(x_{7}=.73, x_{8}=.97, x_{9}=.94, x_{10}=.77\). a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

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. 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? [Hint: \(\left.\sum x_{i}^{2}=1860.94 .\right]\) 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.

As an example of a situation in which several different statistics could reasonably be used to calculate a point estimate, consider a population of \(N\) invoices. Associated with each invoice is its "book value," the recorded amount of that invoice. Let \(T\) denote the total book value, a known amount. Some of these book values are erroneous. An audit will be carried out by randomly selecting \(n\) invoices and determining the audited (correct) value for each one. Suppose that the sample gives the following results (in dollars). Let $$ \begin{aligned} &\bar{Y}=\text { sample mean book value } \\ &\bar{X}=\text { sample mean audited value } \\ &\bar{D}=\text { sample mean error } \end{aligned} $$ Propose three different statistics for estimating the total audited (i.e., correct) value-one involving just \(N\) and \(\bar{X}\), \(\frac{\text { another involving } T, N \text {, and } \bar{D} \text {, and the last involving } T \text { and }}{}\) \(\bar{X} / \bar{Y}\). If \(N=5000\) and \(T=1,761,300\), calculate the three corresponding point estimates. (The article "Statistical Models and Analysis in Auditing," Statistical Science, 1989: 2-33). discusses properties of these estimators.)

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are \(103,156,118,89,125,147,122,109,138,99\). Let \(\mu\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu\). b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

A random sample of \(n\) bike helmets manufactured by a certain company is selected. Let \(X=\) the number among the \(n\) that are flawed, and let \(p=P\) (flawed). Assume that only \(X\) is observed, rather than the sequence of \(S\) s and \(F\) 's. a. Derive the maximum likelihood estimator of \(p\). If \(n=20\) and \(x=3\), what is the estimate? b. Is the estimator of part (a) unbiased? c. If \(n=20\) and \(x=3\), what is the mle of the probability \((1-p)^{5}\) that none of the next five helmets examined is flawed?

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