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Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\) from the shifted exponential pdf

\({\rm{f(x;\lambda ,\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda (x - \theta )}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\). Taking \({\rm{\theta = 0}}\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \({\rm{4}}{\rm{.5}}\), in which the variable of interest was time headway in traffic flow and \({\rm{\theta = }}{\rm{.5}}\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \({\rm{\theta }}\) and \({\rm{\lambda }}\). b. If \({\rm{n = 10}}\) time headway observations are made, resulting in the values \({\rm{3}}{\rm{.11,}}{\rm{.64,2}}{\rm{.55,2}}{\rm{.20,5}}{\rm{.44,3}}{\rm{.42,10}}{\rm{.39,8}}{\rm{.93,17}}{\rm{.82}}\), and \({\rm{1}}{\rm{.30}}\), calculate the estimates of \({\rm{\theta }}\) and \({\rm{\lambda }}\).

Short Answer

Expert verified

(a) Maximum likelihood estimators are \({\rm{\hat \lambda = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - \hat \theta }}} \right)} }}\) and \({\rm{\hat \theta = min(}}{{\rm{X}}_{\rm{i}}}{\rm{)}}\).

(b) The estimates are \({\rm{\hat \theta = 0}}{\rm{.64}}\) and \({\rm{\hat \lambda = 0}}{\rm{.202}}\).

Step by step solution

01

Define exponential function

A function that increases or decays at a rate proportional to its present value is called an exponential function.

02

Explanation

(a) Allow joint pdf or pmb 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,

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

Look at the log likelihood function to determine the maximum.

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

03

Evaluating the maximum likelihood estimators

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

As a result, the derivative,

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

As a result, solving equation provides the maximum likelihood estimator \({\rm{\hat \lambda }}\).

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

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

\({\rm{\hat \lambda = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - \hat \theta }}} \right)} }}\)

The maximum likelihood estimator of parameter\({\rm{\theta }}\)is shown as\({\rm{\hat \theta }}\), with the estimator calculated as follows.

To find the maximum in terms\({\rm{\theta }}\)of the likelihood function,

\({{\rm{\lambda }}^{\rm{n}}}{{\rm{e}}^{{\rm{ - \lambda }}\sum\limits_{{\rm{i - 1}}}^{\rm{n}} {\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - \theta }}} \right)} }}{\rm{ = }}{{\rm{\lambda }}^{\rm{n}}}{{\rm{e}}^{{\rm{ - \lambda }}\sum\limits_{{\rm{i - 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{\rm{ \times }}{{\rm{e}}^{{\rm{ - n\lambda \theta }}}}{\rm{ }}\)

it's worth noting that\({\rm{\theta }}\)only appears in the term

\({{\rm{e}}^{{\rm{ - n\lambda \theta }}}}\)

Furthermore, the likelihood function is defined only for values where all\({{\rm{x}}_{\rm{i}}}\); are greater or equal to\({\rm{\theta }}\), and where the minimum value of all\({{\rm{x}}_{\rm{i}}}\); is greater or equal to\({\rm{\theta }}\).

Because\({\rm{\theta }}\)only exists in the specified term, and the exponent\({\rm{n\lambda \theta }}\)is positive, and the likelihood function is zero for\({\rm{\theta }}\)bigger than\({\rm{min}}\left( {{{\rm{x}}_{\rm{i}}}} \right)\), the highest value is attained when

\({\rm{\hat \theta = min(}}{{\rm{X}}_{\rm{i}}}{\rm{)}}\)

04

Explanation

(b) The minimum values - the maximum likelihood estimate of \({\rm{\theta }}\) is calculated using available data.

\({\rm{\hat \theta = 0}}{\rm{.64}}\)

The fact that\({\rm{n = 10}}\)and\({\rm{\hat \theta = 0}}{\rm{.64}}\),

\(\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} {\rm{ = 3}}{\rm{.11 + 0}}{\rm{.64 + \ldots + 1}}{\rm{.3 = 55}}{\rm{.8}}\)

The maximum likelihood estimates of\({\rm{\lambda }}\)is,

\(\begin{array}{c}{\rm{\hat \lambda = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - \hat \theta }}} \right)} }}\\{\rm{ = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} {\rm{ - n\hat \theta }}}}\\{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{55}}{\rm{.8 - 6}}{\rm{.4}}}}\\{\rm{ = 0}}{\rm{.202}}\end{array}\)

Therefore, \({\rm{\hat \lambda = 0}}{\rm{.202}}\) and \({\rm{\hat \theta = 0}}{\rm{.64}}\).

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Most popular questions from this chapter

We defined a negative binomial\({\rm{rv}}\)as the number of failures that occur before the\({\rm{rth}}\)success in a sequence of independent and identical success/failure trials. The probability mass function (\({\rm{pmf}}\)) of\({\rm{X}}\)is\({\rm{nb(x,r,p) = }}\)\(\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 1}}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\quad {\rm{x = 0,1,2, \ldots }}\)

a. Suppose that. Show that\({\rm{\hat p = (r - 1)/(X + r - 1)}}\)is an unbiased estimator for\({\rm{p}}\). (Hint: Write out\({\rm{E(\hat p)}}\)and cancel\({\rm{x + r - 1}}\)inside the sum.)

b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether\({\rm{(S)}}\)or not\({\rm{(F)}}\)they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate\({\rm{p = }}\)the true proportion who support the candidate.

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 \({\rm{n = 20}}\)?

The article from which the data in Exercise 1 was extracted also gave the accompanying strength observations for cylinders:

\(\begin{array}{l}\begin{array}{*{20}{r}}{{\rm{6}}{\rm{.1}}}&{{\rm{5}}{\rm{.8}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{7}}{\rm{.1}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{9}}{\rm{.2}}}&{{\rm{6}}{\rm{.6}}}&{{\rm{8}}{\rm{.3}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{8}}{\rm{.3}}}\\{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\\\begin{array}{*{20}{l}}{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\end{array}\)

Prior to obtaining data, denote the beam strengths by X1, … ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi ’s constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi ’s form a random sample (independent of the Xi ’s) from another distribution with mean m2 and standard deviation\({{\rm{\sigma }}_{\rm{2}}}\).

a. Use rules of expected value to show that \({\rm{\bar X - \bar Y}}\)is an unbiased estimator of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\). 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 \({{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}\)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 \({\rm{X - Y}}\) between beam strength and cylinder strength.

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.

Suppose the true average growth\({\rm{\mu }}\)of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is\({{\rm{\sigma }}^{\rm{2}}}\), whereas for the second type the variance is\({\rm{4}}{{\rm{\sigma }}^{\rm{2}}}{\rm{. Let }}{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{m}}}\)be\({\rm{m}}\)independent growth observations on the first type (so\({\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \sigma\hat 2}}\)$ ), and let\({{\rm{Y}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{Y}}_{\rm{n}}}\)be\({\rm{n}}\)independent growth observations on the second type\(\left( {{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{Y}}_{\rm{j}}}} \right){\rm{ = 4}}{{\rm{\sigma }}^{\rm{2}}}} \right)\)

a. Show that the estimator\({\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\)is unbiased for\({\rm{\mu }}\)(for\({\rm{0 < \delta < 1}}\), the estimator is a weighted average of the two individual sample means).

b. For fixed\({\rm{m}}\)and\({\rm{n}}\), compute\({\rm{V(\hat \mu ),}}\)and then find the value of\({\rm{\delta }}\)that minimizes\({\rm{V(\hat \mu )}}\). (Hint: Differentiate\({\rm{V(\hat \mu )}}\)with respect to\({\rm{\delta }}{\rm{.)}}\)

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