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Chapter 6: Q30E (page 274)

At time \({\rm{t = 0, 20}}\) identical components are tested. The lifetime distribution of each is exponential with parameter \({\rm{\lambda }}\). The experimenter then leaves the test facility unmonitored. On his return \({\rm{24}}\) hours later, the experimenter immediately terminates the test after noticing that \({\rm{y = 15}}\) of the \({\rm{20}}\) components are still in operation (so \({\rm{5}}\) have failed). Derive the mle of \({\rm{\lambda }}\). (Hint: Let \({\rm{Y = }}\) the number that survive \({\rm{24}}\) hours. Then \({\rm{Y}} \sim {\rm{Bin(n,p)}}\). What is the mle of \({\rm{p}}\)? Now notice that \({\rm{p = P(}}{{\rm{X}}_{\rm{i}}} \ge {\rm{24)}}\), where \({{\rm{X}}_{\rm{i}}}\) is exponentially distributed. This relates \({\rm{\lambda }}\) to \({\rm{p}}\), so the former can be estimated once the latter has been.)

Short Answer

Expert verified

The values are \({\rm{\hat \lambda = - }}\frac{{{\rm{ln}}\frac{{\rm{Y}}}{{\rm{n}}}}}{{{\rm{24}}}}\) and \({\rm{\hat \lambda = 0}}{\rm{.012}}\).

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

X is a random variable that can be used with pdf,

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda x}}}}}&{{\rm{,x}} \ge {\rm{0}}}\\{\rm{0}}&{{\rm{,x < 0}}}\end{array}} \right.\)

With parameter\({\rm{\lambda }}\), it is said to have an exponential distribution.

With parameters n and p, the random variable Y has a Binomial distribution. Because of this,

\({\rm{p = P}}\left( {{{\rm{X}}_{\rm{i}}} \ge {\rm{24}}} \right)\)

The probability value must be approximated, as stated in the clue - keep in mind that the parameter\({\rm{\lambda }}\)is unknown!

The maximum likelihood estimator of the Binomial distribution's parameter p is,

\({\rm{\hat p = }}\frac{{\rm{Y}}}{{\rm{n}}}\)

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

03

Evaluating the maximum likelihood estimators

The maximum likelihood estimator of parameter\({\rm{\lambda }}\)will thus be a function of the maximum likelihood estimator\(\widehat {\rm{p}}\), according to the invariance principle.

The\(\widehat {\rm{p}}\)function can be deduced from,

\(\begin{array}{c}{\rm{p = P(X}} \ge {\rm{24)}}\\{\rm{ = 1 - P(X < 24)}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{{\rm{24}}} {\rm{\lambda }} {{\rm{e}}^{{\rm{ - \lambda x}}}}{\rm{dx}}\\{\rm{ = 1 - }}\left. {{\rm{\lambda }}\frac{{{\rm{ - 1}}}}{{\rm{\lambda }}}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right|_{\rm{0}}^{{\rm{24}}}\\{\rm{ = 1 + }}\left( {{{\rm{e}}^{{\rm{ - 24\lambda }}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{ - 24\lambda }}}}\end{array}\)

As a result,\({\rm{\lambda }}\)maximum likelihood estimator is,

\(\begin{array}{c}{{\rm{e}}^{{\rm{ - 24\hat \lambda }}}}{\rm{ = \hat p}}\\{\rm{ - 24\hat \lambda = ln}}\frac{{\rm{Y}}}{{\rm{n}}}\end{array}\)

Last but not least, the estimator can be expressed as,

\({\rm{\hat \lambda = - }}\frac{{{\rm{ln}}\frac{{\rm{Y}}}{{\rm{n}}}}}{{{\rm{24}}}}\)

For\({\rm{n = 20}}\)and\({\rm{y = 15}}\), the maximum probability estimate is as follows:

\(\begin{array}{c}{\rm{\hat \lambda = - }}\frac{{{\rm{ln}}\frac{{\rm{Y}}}{{\rm{n}}}}}{{{\rm{24}}}}\\{\rm{ = - }}\frac{{{\rm{ln}}\frac{{{\rm{15}}}}{{{\rm{20}}}}}}{{{\rm{24}}}}\\{\rm{ = 0}}{\rm{.012}}\end{array}\)

Therefore, \({\rm{\hat \lambda = - }}\frac{{{\rm{ln}}\frac{{\rm{Y}}}{{\rm{n}}}}}{{{\rm{24}}}}\) and \({\rm{\hat \lambda = 0}}{\rm{.012}}\).

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Most popular questions from this chapter

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a pdf\({\rm{f(x)}}\)that is symmetric about\({\rm{\mu }}\), so that\({\rm{\backslash widetildeX}}\)is an unbiased estimator of\({\rm{\mu }}\). If\({\rm{n}}\)is large, it can be shown that\({\rm{V (\tilde X)\gg 1/}}\left( {{\rm{4n(f(\mu )}}{{\rm{)}}^{\rm{2}}}} \right)\).

a. Compare\({\rm{V(\backslash widetildeX)}}\)to\({\rm{V(\bar X)}}\)when the underlying distribution is normal.

b. When the underlying pdf is Cauchy (see Example 6.7),\({\rm{V(\bar X) = \yen}}\), so\({\rm{\bar X}}\)is a terrible estimator. What is\({\rm{V(\tilde X)}}\)in this case when\({\rm{n}}\)is large?

Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a pdf that is symmetric about \({\rm{\mu }}\). An estimator for \({\rm{\mu }}\) that has been found to perform well for a variety of underlying distributions is the Hodges鈥揕ehmann estimator. To define it, first compute for each \({\rm{i}} \le {\rm{j}}\) and each \({\rm{j = 1,2, \ldots ,n}}\) the pairwise average \({{\rm{\bar X}}_{{\rm{i,j}}}}{\rm{ = }}\left( {{{\rm{X}}_{\rm{i}}}{\rm{ + }}{{\rm{X}}_{\rm{j}}}} \right){\rm{/2}}\). Then the estimator is \({\rm{\hat \mu = }}\) the median of the \({{\rm{\bar X}}_{{\rm{i,j}}}}{\rm{'s}}\). Compute the value of this estimate using the data of Exercise \({\rm{44}}\) of Chapter \({\rm{1}}\). (Hint: Construct a square table with the \({{\rm{x}}_{\rm{i}}}{\rm{'s}}\) listed on the left margin and on top. Then compute averages on and above the diagonal.)

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, 鈥� ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi 鈥檚 constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi 鈥檚 form a random sample (independent of the Xi 鈥檚) 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.

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)

Consider randomly selecting \({\rm{n}}\) segments of pipe and determining the corrosion loss (mm) in the wall thickness for each one. Denote these corrosion losses by \({{\rm{Y}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{Y}}_{\rm{n}}}\). The article 鈥淎 Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains鈥� (Reliability Engr. and System Safety (\({\rm{(2013:270 - 279)}}\)) proposes a linear corrosion model: \({{\rm{Y}}_{\rm{i}}}{\rm{ = }}{{\rm{t}}_{\rm{i}}}{\rm{R}}\), where \({{\rm{t}}_{\rm{i}}}\) is the age of the pipe and \({\rm{R}}\), the corrosion rate, is exponentially distributed with parameter \({\rm{\lambda }}\). Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). (Hint: If \({\rm{c > 0}}\) and \({\rm{X}}\) has an exponential distribution, so does \({\rm{cX}}\).)
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