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Let\({\rm{X}}\)represent the error in making a measurement of a physical characteristic or property (e.g., the boiling point of a particular liquid). It is often reasonable to assume that\({\rm{E(X) = 0}}\)and that\({\rm{X}}\)has a normal distribution. Thus, the pdf of any particular measurement error is

\({\rm{f(x;\theta ) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \theta }}} }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\theta }}}}\quad {\rm{ - \yen < x < \yen}}\)

(Where we have used\({\rm{\theta }}\)in place of\({{\rm{\sigma }}^{\rm{2}}}\)). Now suppose that\({\rm{n}}\)independent measurements are made, resulting in measurement errors\({{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{.}}\)Obtain the mle of\({\rm{\theta }}\).

Short Answer

Expert verified

The solution is \({\rm{\hat \theta = }}\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} \)which the maximum likelihood estimator.

Step by step solution

01

Introduction

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

02

Explanation

Let random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)have joint pdf or pmb

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

Where, the parameters \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\) are unknown. When function \({\rm{f}}\)is a function of parameters\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\), it is called the

likelihood function

Values \({{\rm{\hat \theta }}_{\rm{i}}}\)that maximize the likelihood function are the maximum likelihood estimates (mle's), or equally values \({{\rm{\hat \theta }}_{\rm{i}}}\)for which

for every\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\). By substituting \({{\rm{x}}_{\rm{i}}}\)with\({{\rm{x}}_{\rm{i}}}\), the

maximum likelihood estimators

are obtained.

The likelihood function, because of the independence, becomes

\(\begin{aligned}{\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right)\\& = \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \times \theta }}} }}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{{\rm{x}}_{\rm{1}}^{\rm{2}}}}{{{\rm{2\theta }}}}} \right\}{\rm{ \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \times \theta }}} }}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{{\rm{x}}_{\rm{2}}^{\rm{2}}}}{{{\rm{2\theta }}}}} \right\}{\rm{ \times \ldots \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \times \theta }}} }}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{{\rm{x}}_{\rm{n}}^{\rm{2}}}}{{{\rm{2\theta }}}}} \right\}\\& = \prod\limits_{{\rm{i = 1}}}^{\rm{n}} {\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \times \theta }}} }}} {\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{{\rm{x}}_{\rm{i}}^{\rm{2}}}}{{{\rm{2\theta }}}}} \right\}\\&= (2\pi \sigma {{\rm{)}}^{{\rm{ - n/2}}}}{\rm{ \times exp}}\left\{ {{\rm{ - }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\frac{{{\rm{x}}_{\rm{i}}^{\rm{2}}}}{{{\rm{2\sigma }}}}} } \right\}\end{aligned}\)

In order to find maximum, look at the log likelihood function

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

By taking derivative of log likelihood function in respect to \({\rm{\theta }}\)and equating it to \({\rm{0}}\) the maximum likelihood estimator is obtained. Therefore, the derivative is

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

Therefore, the maximum likelihood estimator is obtained by solving equation

\(\begin{aligned}{\rm{ - }}\frac{{\rm{n}}}{{{\rm{2\hat \theta }}}}{\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\frac{{{\rm{x}}_{\rm{i}}^{\rm{2}}}}{{{\rm{2}}\widehat {{{\rm{\theta }}^{\rm{2}}}}}}} \\&= 0 - n\theta + \sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} \\ &= 0\end{aligned}\)
For\({\rm{\hat \theta }}\).

Therefore, the solution is\({\rm{\hat \theta = }}\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} \)which is the maximum likelihood estimator

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Most popular questions from this chapter

A diagnostic test for a certain disease is applied to\({\rm{n}}\)individuals known to not have the disease. Let\({\rm{X = }}\)the number among the\({\rm{n}}\)test results that are positive (indicating presence of the disease, so\({\rm{X}}\)is the number of false positives) and\({\rm{p = }}\)the probability that a disease-free individual's test result is positive (i.e.,\({\rm{p}}\)is the true proportion of test results from disease-free individuals that are positive). Assume that only\({\rm{X}}\)is available rather than the actual sequence of test results.

a. Derive the maximum likelihood estimator of\({\rm{p}}\). If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the estimate?

b. Is the estimator of part (a) unbiased?

c. If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the mle of the probability\({{\rm{(1 - p)}}^{\rm{5}}}\)that none of the next five tests done on disease-free individuals are positive?

An estimator \({\rm{\hat \theta }}\) is said to be consistent if for any \( \in {\rm{ > 0}}\), \({\rm{P(|\hat \theta - \theta |}} \ge \in {\rm{)}} \to {\rm{0}}\) as \({\rm{n}} \to \infty \). That is, \({\rm{\hat \theta }}\) is consistent if, as the sample size gets larger, it is less and less likely that \({\rm{\hat \theta }}\) will be further than \( \in \) from the true value of \({\rm{\theta }}\). Show that \({\rm{\bar X}}\) is a consistent estimator of \({\rm{\mu }}\) when \({{\rm{\sigma }}^{\rm{2}}}{\rm{ < }}\infty \) , by using Chebyshev’s inequality from Exercise \({\rm{44}}\) of Chapter \({\rm{3}}\). (Hint: The inequality can be rewritten in the form \({\rm{P}}\left( {\left| {{\rm{Y - }}{{\rm{\mu }}_{\rm{Y}}}} \right| \ge \in } \right) \le {\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{/}} \in \). Now identify \({\rm{Y}}\) with \({\rm{\bar X}}\).)

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 m denote the average gas usage during January by all houses in this area. Compute a point estimate of m. b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t 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?

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

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\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

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\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), 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 \({\rm{E(S) = }}\sqrt {{\rm{2/(n - 1)}}} {\rm{\Gamma (n/2)\sigma /\Gamma ((n - 1)/2)}}\)

Use this to obtain an unbiased estimator for \({\rm{\sigma }}\) of the form \({\rm{cS}}\). What is \({\rm{c}}\) when \({\rm{n = 20}}\)?

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