/*! 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 22 Let \(X\) denote the proportion ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
Method of moments: \(\theta = 3\); MLE: \(\theta \approx 1.694\).

Step by step solution

01

Method of Moments - Find the Expected Value

We start by finding the expected value for the distribution given by \(f(x; \theta)\). The expected value \(E(X)\) of a random variable \(X\) is given by:\[E(X) = \int_0^1 x f(x; \theta) \, dx = \int_0^1 x(\theta + 1)x^{\theta} \, dx = (\theta + 1) \int_0^1 x^{\theta + 1} \, dx.\]Using the formula \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), we compute:\[\int_0^1 x^{\theta + 1} \, dx = \left[ \frac{x^{\theta + 2}}{\theta + 2} \right]_0^1 = \frac{1}{\theta + 2}.\]Thus, \(E(X) = \frac{\theta + 1}{\theta + 2}.\)
02

Match Moments and Solve for \(\theta\)

The sample mean is used as an estimate of \(E(X)\). The sample mean \(\overline{x}\) is calculated from the data:\[\overline{x} = \frac{.92 + .79 + .90 + .65 + .86 + .47 + .73 + .97 + .94 + .77}{10} = 0.8.\]Equating to the expected value:\[\frac{\theta + 1}{\theta + 2} = 0.8.\]Solve for \(\theta\):\[\theta + 1 = 0.8(\theta + 2)\]\[\theta + 1 = 0.8\theta + 1.6\]\[\theta - 0.8\theta = 0.6\]\[0.2\theta = 0.6\]\[\theta = 3.\]
03

Maximum Likelihood Estimator Setup

Let \(x_1, x_2, \ldots, x_{10}\) be a random sample. The likelihood function \(L(\theta)\) is:\[L(\theta) = \prod_{i=1}^{10}f(x_i; \theta) = \prod_{i=1}^{10}(\theta + 1)x_i^{\theta}.\]Thus,\[\ln L(\theta) = \sum_{i=1}^{10} \ln((\theta + 1) x_i^{\theta}).\]Expanding the logarithm:\[\ln L(\theta) = 10 \ln(\theta + 1) + \theta \sum_{i=1}^{10} \ln(x_i).\]
04

Differentiate and Solve for \(\theta\)

Taking the derivative of \(\ln L(\theta)\) with respect to \(\theta\) and set to zero:\[\frac{d}{d\theta}[10 \ln(\theta + 1) + \theta \sum \ln(x_i)] = 0\]\[\frac{10}{\theta + 1} + \sum \ln(x_i) = 0.\]Solve for \(\theta\):\[10 + (\theta + 1) \sum \ln(x_i) = 0.\]The sum of the \(\ln(x_i)\) is \(-3.709\).Solve:\[10 + (\theta + 1)(-3.709) = 0.\]\[\theta + 1 = \frac{10}{3.709}\]\[\theta = \frac{10}{3.709} - 1 = 1.694.\]
05

Calculate MLE Estimate for \(\theta\)

We calculate the Maximum Likelihood Estimate (MLE) using the derived MLE equation. From Step 4:\[\theta = \frac{10}{3.709} - 1 \approx 1.694.\]

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Key Concepts

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

Probability Density Function
A probability density function (pdf) describes the likelihood of a random variable to take a particular value. For a continuous random variable, the pdf is a crucial concept because it gives us the height of the probability distribution at each value of the random variable.
In our specific scenario, the function is defined as:
  • \(f(x; \theta) = (\theta + 1)x^\theta\) for \(0 \leq x \leq 1\)
  • \(f(x; \theta) = 0\) otherwise
This definition implies that the bulk of probability mass is within the interval [0, 1], and the shape of the distribution is influenced by the parameter \(\theta\). The pdf provides a foundation for understanding how data is distributed over an interval and is the starting point for many estimation methods, such as the Method of Moments and Maximum Likelihood Estimation.
Method of Moments
The Method of Moments is a classical technique for estimating parameters of a probability distribution. It involves equating sample moments to the population moments to find estimators for these parameters. Moments of a distribution include quantities like the mean, variance, and higher-order moments.
To apply this method, we first calculate the expected value of the random variable associated with the pdf. In our case, the expected value \(E(X)\) is derived as:
  • \(E(X) = \frac{\theta + 1}{\theta + 2}\)
Next, we calculate the sample mean from the data to act as an estimator for the expected value:
  • \(\overline{x} = 0.8\)
We then equate the sample mean to the expected value, solve for \(\theta\), and obtain the estimator value. Here, the Method of Moments gives us an estimate \(\theta = 3\). This practical approach is often straightforward and valuable for providing initial parameter estimates.
Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a popular method for estimating the parameters of a probability distribution. It involves finding the parameter values that maximize the likelihood of observing the given sample data.
In our case, we define the likelihood function based on the pdf:
  • \(L(\theta) = \prod_{i=1}^{10}(\theta + 1)x_i^\theta\)
By taking the natural logarithm of the likelihood function, we simplify the computation:
  • \(\ln L(\theta) = 10 \ln(\theta + 1) + \theta \sum \ln(x_i)\)
After differentiating \(\ln L(\theta)\) with respect to \(\theta\), setting the derivative to zero allows us to solve for \(\theta\). For the provided data, this process yields an MLE estimate of \(\theta \approx 1.694\). MLE is versatile and often preferred for its statistical properties, such as consistency and asymptotic normality.
Expected Value
Expected value is a fundamental concept in statistics and probability theory, representing the average outcome of a random variable over numerous trials. It serves as a measure of the central tendency of the distribution.
When calculating expected value for a continuous distribution, we integrate the product of the variable and its probability density function across all possible values. For our specific function:
  • \(E(X) = \int_0^1 x \cdot f(x; \theta) \, dx = \frac{\theta + 1}{\theta + 2}\)
The expected value helps us align our sample data with theoretical predictions using techniques like Method of Moments. By comparing the calculated sample mean to this expected value, we can derive estimates for distribution parameters. Understanding expected value is key, as it guides decisions and predictions in various fields, from finance to engineering.

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Most popular questions from this chapter

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.

A sample of \(n\) captured Pandemonium jet fighters results in serial numbers \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \(\alpha\) and ending with \(\beta\), so that the total number of planes manufactured is \(\beta-\alpha+1\) (e.g., if \(\alpha=17\) and \(\beta=29\), then \(29-17+1=13\) planes having serial numbers 17 , \(18,19, \ldots, 28,29\) were manufactured). However, the CIA does not know the values of \(\alpha\) or \(\beta\). A CIA statistician suggests using the estimator \(\max \left(X_{j}\right)-\min \left(X_{j}\right)+1\) to estimate the total number of planes manufactured. a. If \(n=5, x_{1}=237, x_{2}=375, x_{3}=202, x_{4}=525\), and \(x_{5}=418\). what is the corresponding estimate? b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating \(\beta-\alpha+1\) ? Explain in one or two sentences.

The article from which the data of Exercise 1 was extracted also gave the accompanying strength observations for cylinders: \(\begin{array}{rrrrrrrrrr}6.1 & 5.8 & 7.8 & 7.1 & 7.2 & 9.2 & 6.6 & 8.3 & 7.0 & 8.3 \\ 7.8 & 8.1 & 7.4 & 8.5 & 8.9 & 9.8 & 9.7 & 14.1 & 12.6 & 11.2\end{array}\) Prior to obtaining data, denote the beam strengths by \(X_{1}, \ldots, X_{m}\) and the cylinder strengths by \(Y_{1}, \ldots, Y_{n^{*}}\). Suppose that the \(X_{i}\) s constitute a random sample from a distribution with mean \(\mu_{1}\) and standard deviation \(\sigma_{1}\) and that the \(Y_{i} \mathrm{~s}\) form a random sample (independent of the \(X_{i} \mathrm{~s}\) ) from another distribution with mean \(\mu_{2}\) and standard deviation \(\sigma_{2^{\circ}}\) a. Use rules of expected value to show that \(\bar{X}^{2}-\bar{Y}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). Calculate the estimate for the given data. b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error. c. Calculate a point estimate of the ratio \(\sigma_{1} / \sigma_{2}\) of the two standard deviations. d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \(X-Y\) between beam strength and cylinder strength.

In a random sample of 80 components of a certain type, 12 are found to be defective. a. Give a point estimate of the proportion of all such components that are not defective. b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here. The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. [Hint: If \(p\) denotes the probability that a component works properly, how can \(P\) (system works) be expressed in terms of \(p\) ?]

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?

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