Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 22E from Chapter 6 from Devore's Probability and Statistics for Engineering and Science.
Problem 22E
Chapter:
Problem:
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 ... a. Use the method of moments to obtain an estimator of θ, and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of θ, and then compute the estimate for the given data.
Step-by-Step Solution
Step 1
We are given with probability density function of X (proportion of allotted time that a randomly selected student spends working on a certain aptitude test.\[f\left( {x;\theta } \right) = \left\{ {\begin{array}{*{20}{c}}{\left( {\theta + 1} \right){x^\theta }}&{0 \le x \le 1}\\0&{{\rm{Otherwise}}}\end{array}} \right.\]
Step 2: (a)
The expected value of the random variable\[\begin{array}{l}E\left( x \right) = \int\limits_0^1 {xf\left( x \right)dx} \\ = \int\limits_0^1 {x\left( {\theta + 1} \right){x^\theta }dx} \\ = \int\limits_0^1 {\left( {\theta + 1} \right){x^{\theta + 1}}dx} \\ = \dfrac{{\left( {\theta + 1} \right)}}{{\left( {\theta + 2} \right)}}\left[ {{x^{\theta + 2}}} \right]_0^1\\ = \dfrac{{\left( {\theta + 1} \right)}}{{\left( {\theta + 2} \right)}}\end{array}\]The estimator of $\theta$: \[\begin{array}{l}\dfrac{{\left( {\theta + 1} \right)}}{{\left( {\theta + 2} \right)}} = \bar X\\\theta + 1 = \bar X\left( {\theta + 2} \right)\\\theta \left( {1 - \bar X} \right) = 2\bar X - 1\\\theta = \dfrac{{2\bar X - 1}}{{1 - \bar X}}\end{array}\]Therefore, \[\theta = \dfrac{{2\bar X - 1}}{{1 - \bar X}}\]
Step 3:
Mean value of random variable:\[\begin{array}{l}\bar X = \dfrac{{\Sigma {x_i}}}{n}\\ = \dfrac{{\left[ \begin{array}{l}0.92 + 0.79 + 0.90 + 0.65 + 0.86\\ + 0.47 + 0.73 + 0.97 + 0.94 + 0.77\end{array} \right]}}{{10}}\\ = 0.8\end{array}\]Therefore, \[\bar X = 0.8\]
Step 4:
Estimate of the parameter $\theta$\[\begin{array}{l}\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \theta } = \dfrac{{2 \times \bar X - 1}}{{1 - \bar X}}\\ = \dfrac{{2 \times 0.8 - 1}}{{1 - 0.8}}\\ = \dfrac{{0.6}}{{0.3}}\\ = 3\end{array}\]Therefore, \[\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \theta } = 3\]
Step 5: (b)
Likelihood of $\theta$:\[\begin{array}{l}L = f\left( {{x_{,1}},{x_2},.....{x_{10}};\,\,\theta } \right)\\L = \left( {\theta + 1} \right)x_1^\theta \times \left( {\theta + 1} \right)x_2^\theta \times \left( {\theta + 1} \right)x_3^\theta \times \left( {\theta + 1} \right)x_4^\theta ...\left( {\theta + 1} \right)x_{10}^\theta \\L = {\left( {\theta + 1} \right)^{10}}\prod\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^\theta }} \\\log \left( L \right) = 10\ln \left( {\theta + 1} \right) + \ln \prod\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^\theta }} \\\log \left( L \right) = 10\ln \left( {\theta + 1} \right) + \theta \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \end{array}\]
Step 6:
Differentiate and equate to zero:\[\begin{array}{l}\dfrac{\begin{array}{l}\\d\left( {\log L} \right)\end{array}}{{d\theta }} = 0\\\dfrac{{10}}{{\left( {\theta + 1} \right)}} + \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} = 0\\\dfrac{{10}}{{\left( {\theta + 1} \right)}} = - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \\10 = - \theta \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \\\theta \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} = - 10 - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \\\theta = \dfrac{{ - 10 - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} }}{{\sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} }}\\\theta = \dfrac{{ - 10 - \left[ \begin{array}{l}\ln \left( {0.92} \right) + \ln \left( {0.79} \right) + \ln \left( {0.90} \right) + \ln \left( {0.65} \right) + \ln \left( {0.86} \right) + \\\ln \left( {0.47} \right) + \ln \left( {0.73} \right) + \ln \left( {0.97} \right) + \ln \left( {0.94} \right) + \ln \left( {0.77} \right)\end{array} \right]}}{{\left[ \begin{array}{l}\ln \left( {0.92} \right) + \ln \left( {0.79} \right) + \ln \left( {0.90} \right) + \ln \left( {0.65} \right) + \ln \left( {0.86} \right) + \\\ln \left( {0.47} \right) + \ln \left( {0.73} \right) + \ln \left( {0.97} \right) + \ln \left( {0.94} \right) + \ln \left( {0.77} \right)\end{array} \right]}}\\\theta = \dfrac{{ - 10 - \left[ \begin{array}{l} - 0.08338 - 0.23572 - 0.10536 - 0.43078 - 0.15082\\ - 0.75502 - 0.31471 - 0.03046 - 0.06188 - 0.26136\end{array} \right]}}{{\left[ \begin{array}{l} - 0.08338 - 0.23572 - 0.10536 - 0.43078 - 0.15082\\ - 0.75502 - 0.31471 - 0.03046 - 0.06188 - 0.26136\end{array} \right]}}\\\theta = \dfrac{{ - 10 + 2.4295}}{{ - 2.4295}}\\\theta = 3.12\end{array}\]Therefore, \[\theta = 3.12\]
We are given with probability density function of X (proportion of allotted time that a randomly selected student spends working on a certain aptitude test.\[f\left( {x;\theta } \right) = \left\{ {\begin{array}{*{20}{c}}{\left( {\theta + 1} \right){x^\theta }}&{0 \le x \le 1}\\0&{{\rm{Otherwise}}}\end{array}} \right.\]
Step 2: (a)
The expected value of the random variable\[\begin{array}{l}E\left( x \right) = \int\limits_0^1 {xf\left( x \right)dx} \\ = \int\limits_0^1 {x\left( {\theta + 1} \right){x^\theta }dx} \\ = \int\limits_0^1 {\left( {\theta + 1} \right){x^{\theta + 1}}dx} \\ = \dfrac{{\left( {\theta + 1} \right)}}{{\left( {\theta + 2} \right)}}\left[ {{x^{\theta + 2}}} \right]_0^1\\ = \dfrac{{\left( {\theta + 1} \right)}}{{\left( {\theta + 2} \right)}}\end{array}\]The estimator of $\theta$: \[\begin{array}{l}\dfrac{{\left( {\theta + 1} \right)}}{{\left( {\theta + 2} \right)}} = \bar X\\\theta + 1 = \bar X\left( {\theta + 2} \right)\\\theta \left( {1 - \bar X} \right) = 2\bar X - 1\\\theta = \dfrac{{2\bar X - 1}}{{1 - \bar X}}\end{array}\]Therefore, \[\theta = \dfrac{{2\bar X - 1}}{{1 - \bar X}}\]
Step 3:
Mean value of random variable:\[\begin{array}{l}\bar X = \dfrac{{\Sigma {x_i}}}{n}\\ = \dfrac{{\left[ \begin{array}{l}0.92 + 0.79 + 0.90 + 0.65 + 0.86\\ + 0.47 + 0.73 + 0.97 + 0.94 + 0.77\end{array} \right]}}{{10}}\\ = 0.8\end{array}\]Therefore, \[\bar X = 0.8\]
Step 4:
Estimate of the parameter $\theta$\[\begin{array}{l}\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \theta } = \dfrac{{2 \times \bar X - 1}}{{1 - \bar X}}\\ = \dfrac{{2 \times 0.8 - 1}}{{1 - 0.8}}\\ = \dfrac{{0.6}}{{0.3}}\\ = 3\end{array}\]Therefore, \[\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \theta } = 3\]
Step 5: (b)
Likelihood of $\theta$:\[\begin{array}{l}L = f\left( {{x_{,1}},{x_2},.....{x_{10}};\,\,\theta } \right)\\L = \left( {\theta + 1} \right)x_1^\theta \times \left( {\theta + 1} \right)x_2^\theta \times \left( {\theta + 1} \right)x_3^\theta \times \left( {\theta + 1} \right)x_4^\theta ...\left( {\theta + 1} \right)x_{10}^\theta \\L = {\left( {\theta + 1} \right)^{10}}\prod\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^\theta }} \\\log \left( L \right) = 10\ln \left( {\theta + 1} \right) + \ln \prod\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^\theta }} \\\log \left( L \right) = 10\ln \left( {\theta + 1} \right) + \theta \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \end{array}\]
Step 6:
Differentiate and equate to zero:\[\begin{array}{l}\dfrac{\begin{array}{l}\\d\left( {\log L} \right)\end{array}}{{d\theta }} = 0\\\dfrac{{10}}{{\left( {\theta + 1} \right)}} + \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} = 0\\\dfrac{{10}}{{\left( {\theta + 1} \right)}} = - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \\10 = - \theta \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \\\theta \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} = - 10 - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} \\\theta = \dfrac{{ - 10 - \sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} }}{{\sum\limits_{i = 1}^{10} {\ln \left( {{x_i}} \right)} }}\\\theta = \dfrac{{ - 10 - \left[ \begin{array}{l}\ln \left( {0.92} \right) + \ln \left( {0.79} \right) + \ln \left( {0.90} \right) + \ln \left( {0.65} \right) + \ln \left( {0.86} \right) + \\\ln \left( {0.47} \right) + \ln \left( {0.73} \right) + \ln \left( {0.97} \right) + \ln \left( {0.94} \right) + \ln \left( {0.77} \right)\end{array} \right]}}{{\left[ \begin{array}{l}\ln \left( {0.92} \right) + \ln \left( {0.79} \right) + \ln \left( {0.90} \right) + \ln \left( {0.65} \right) + \ln \left( {0.86} \right) + \\\ln \left( {0.47} \right) + \ln \left( {0.73} \right) + \ln \left( {0.97} \right) + \ln \left( {0.94} \right) + \ln \left( {0.77} \right)\end{array} \right]}}\\\theta = \dfrac{{ - 10 - \left[ \begin{array}{l} - 0.08338 - 0.23572 - 0.10536 - 0.43078 - 0.15082\\ - 0.75502 - 0.31471 - 0.03046 - 0.06188 - 0.26136\end{array} \right]}}{{\left[ \begin{array}{l} - 0.08338 - 0.23572 - 0.10536 - 0.43078 - 0.15082\\ - 0.75502 - 0.31471 - 0.03046 - 0.06188 - 0.26136\end{array} \right]}}\\\theta = \dfrac{{ - 10 + 2.4295}}{{ - 2.4295}}\\\theta = 3.12\end{array}\]Therefore, \[\theta = 3.12\]