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Let \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\) represent a random sample from the Rayleigh distribution with density function given in Exercise \({\rm{15}}\). Determine a. The maximum likelihood estimator of \({\rm{\theta }}\), and then calculate the estimate for the vibratory stress data given in that exercise. Is this estimator the same as the unbiased estimator suggested in Exercise \({\rm{15}}\)? b. The mle of the median of the vibratory stress distribution. (Hint: First express the median in terms of \({\rm{\theta }}\).)

Short Answer

Expert verified

(a) Maximum likelihood estimator is \({\rm{\hat \theta = }}\frac{{\rm{1}}}{{{\rm{2n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} \).

(b) The mle of median is \({\rm{\hat \sim x = }}\sqrt {{\rm{1}}{\rm{.3863 \times \hat \theta }}} \).

Step by step solution

01

Define equations

A mathematical language that asserts that two algebraic expressions must be equal in nature is known as an equation.

02

Explanation

(a) The pdf of Rayleigh distribution is,

\({\rm{f(x;\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}{\rm{, x > 0}}\)

Allow joint pdf or pmf for random variables\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\).

\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right){\rm{, n,m}} \in {\rm{N}}\)

where\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\)are unknown parameters. The likelihood function is defined as a function of parameters\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\)where function f is a function of parameter. The maximum likelihood estimates (mle's), or values\(\widehat {{{\rm{\theta }}_{\rm{i}}}}\)for which the likelihood function is maximised, are the maximum likelihood estimates,

\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{{\rm{\hat \theta }}}_{\rm{1}}}{\rm{,}}{{{\rm{\hat \theta }}}_{\rm{2}}}{\rm{, \ldots ,}}{{{\rm{\hat \theta }}}_{\rm{m}}}} \right) \ge {\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right)\)

As,\({\rm{i = 1,2, \ldots ,m}}\)for every\({{\rm{\theta }}_{\rm{i}}}\). Maximum likelihood estimators are derived by replacing\({{\rm{X}}_{\rm{i}}}\)with\({{\rm{x}}_{\rm{i}}}\).

Because of the independence, the likelihood function becomes,

\(\begin{array}{c}{\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{x}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right){\rm{ = }}\frac{{{{\rm{x}}_{\rm{1}}}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - x}}_{\rm{1}}^{\rm{2}}{\rm{/(2\theta )}}}}{\rm{ \times }}\frac{{{{\rm{x}}_{\rm{2}}}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - x}}_{\rm{2}}^{\rm{2}}{\rm{/(2\theta )}}}}{\rm{ \times \ldots \times }}\frac{{{{\rm{x}}_{\rm{n}}}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - x}}_{\rm{n}}^{\rm{2}}{\rm{/(2\theta )}}}}\\{\rm{ = }}\left( {{{\rm{x}}_{\rm{1}}}{\rm{ \times }}{{\rm{x}}_{\rm{2}}}{\rm{ \times \ldots \times }}{{\rm{x}}_{\rm{n}}}} \right){\rm{ \times }}\frac{{\rm{1}}}{{{{\rm{\theta }}^{\rm{n}}}}}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2\theta }}}}} \right\}\end{array}\)

03

Evaluating the maximum likelihood estimators

Look at the log likelihood function to determine the maximum.

\(\begin{array}{c}{\rm{lnf}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{x}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right){\rm{ = ln}}\left( {\left( {{{\rm{x}}_{\rm{1}}}{\rm{ \times }}{{\rm{x}}_{\rm{2}}}{\rm{ \times \ldots \times }}{{\rm{x}}_{\rm{n}}}} \right){\rm{ \times }}\frac{{\rm{1}}}{{{{\rm{\theta }}^{\rm{n}}}}}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2\theta }}}}} \right\}} \right)\\{\rm{ = ln}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{ \times }}{{\rm{x}}_{\rm{2}}}{\rm{ \times \ldots \times }}{{\rm{x}}_{\rm{n}}}} \right){\rm{ - nln\theta - }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2\theta }}}}\end{array}\)

The maximum likelihood estimator is generated by taking the derivative of the log likelihood function in regard to\({\rm{\theta }}\)and equating it to\({\rm{0}}\).

As a result, the derivative,

\(\begin{array}{c}\frac{{\rm{d}}}{{{\rm{d\theta }}}}{\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{x}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right){\rm{ = }}\frac{{\rm{d}}}{{{\rm{d\theta }}}}\left( {{\rm{ln}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{ \times }}{{\rm{x}}_{\rm{2}}}{\rm{ \times \ldots \times }}{{\rm{x}}_{\rm{n}}}} \right){\rm{ - nln\theta - }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2\theta }}}}} \right)\\{\rm{ = 0 - n \times }}\frac{{\rm{1}}}{{\rm{\theta }}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\theta }}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} \\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\theta }}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - n \times }}\frac{{\rm{1}}}{{\rm{\theta }}}\end{array}\)

As a result, solving equation provides the maximum likelihood estimator.

\(\begin{array}{c}\frac{{\rm{1}}}{{{\rm{2}}{{{\rm{\hat \theta }}}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - n \times }}\frac{{\rm{1}}}{{{\rm{\hat \theta }}}}{\rm{ = 0}}\\{\rm{n\hat \theta = }}\frac{{\rm{1}}}{{\rm{2}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} \end{array}\)

For\({\rm{\theta }}\). Hence, the maximum likelihood estimator is,

\({\rm{\hat \theta = }}\frac{{\rm{1}}}{{{\rm{2n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} \)

04

Explanation

(b) The median can be calculated by equating the cdf of the Rayleigh distribution with \({\rm{0}}{\rm{.5}}\), or equally in terms of \({\rm{\theta }}\),

\({\rm{F(x;\theta ) = 0}}{\rm{.5}}\)

For\({\rm{x > 0}}\), the cdf is,

\(\begin{array}{c}{\rm{ F(x;\theta ) = P(X}} \le {\rm{x)}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{x}} {\rm{f}} {\rm{(t;\theta )dt}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\end{array}\)

where this cdf can be found (you can solve integral and obtain it as well). Hence,

\(\begin{array}{c}{\rm{F(x;\theta ) = 0}}{\rm{.5}}\\{\rm{1 - }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}{\rm{ = 0}}{\rm{.5}}\\{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}{\rm{ = 0}}{\rm{.5}}\\\frac{{{\rm{ - }}{{\rm{x}}^{\rm{2}}}}}{{{\rm{2\theta }}}}{\rm{ = ln0}}{\rm{.5}}\\{{\rm{x}}^{\rm{2}}}{\rm{ = - 2\theta \times - 0}}{\rm{.6931}}\\{\rm{x = }}\sqrt {{\rm{1}}{\rm{.3863 \times \theta }}} \end{array}\)

Let\({\rm{i = 1,2,}}......{\rm{,n}}\)be the maximum likelihood estimates of the parameters\(\widehat {{{\rm{\theta }}_{\rm{i}}}}{\rm{,i = 1,2,}}......{\rm{,n}}\). The function of the mle's\(\widehat {{{\rm{\theta }}_{\rm{i}}}}\); is the mle of any function of parameters\(\widehat {{{\rm{\theta }}_{\rm{i}}}}\).

Because of the invariance principle and the fact that the median is a function of\({\rm{\theta }}\), the maximum likelihood estimator of the median is a function\(\widehat {\rm{\theta }}\)of or equally is,

\({\rm{\hat \sim x = }}\sqrt {{\rm{1}}{\rm{.3863 \times \hat \theta }}} \).

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Most popular questions from this chapter

The article cited in Example 1.2 also gave the accompanying strength observations for cylinders:

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

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14.1

12.6

11.2


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d. If \({\rm{1}}{\rm{.5}}\) thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the \({\rm{1}}{\rm{.5}}\) thousand gallons is expected to be left at the end of the week? (Hint: Let \({\rm{h(x) = }}\) amount left when demand \({\rm{ = x}}\).)

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape (鈥淥xygen Consumption and Ventilation During Escape from an Offshore Platform,鈥 Ergonomics, 1997: 281鈥292):

389 356 359 363 375 424 325 394 402

373 373 370 364 366 364 325 339 393

392 369 374 359 356 403 334 397

a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare?

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d. What are the values of \(\bar x\)and \(\tilde x\), when the observations are re expressed in minutes?

Once an individual has been infected with a certain disease, let \({\rm{X}}\) represent the time (days) that elapses before the individual becomes infectious. The article proposes a Weibull distribution with \({\rm{\alpha = 2}}{\rm{.2}}\), \({\rm{\beta = 1}}{\rm{.1}}\), and \({\rm{\gamma = 0}}{\rm{.5}}\).

a. Calculate \({\rm{P(1 < X < 2)}}\).

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The article 鈥淎 Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants鈥 (Lubric. Engr., 1984: 75鈥83) reportedthe following data on oxidation-induction time (min) for various commercial oils:

87 103 130 160 180 195 132 145 211 105 145

153 152 138 87 99 93 119 129

a. Calculate the sample variance and standard deviation.

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