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A sample of \({\rm{n}}\) captured Pandemonium jet fighters results in serial numbers\({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \({\rm{\alpha }}\)and ending with\({\rm{\beta }}\), so that the total number of planes manufactured is \({\rm{\beta - \alpha + 1}}\) (e.g., if \({\rm{\alpha = 17}}\) and\({\rm{\beta = 29}}\), then \({\rm{29 - 17 + 1 = 13}}\)planes having serial numbers \({\rm{17,18,19, \ldots ,28,29}}\)were manufactured). However, the CIA does not know the values of \({\rm{\alpha }}\) or\({\rm{\beta }}\). A CIA statistician suggests using the estimator \({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ - min}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ + 1}}\)to estimate the total number of planes manufactured.

a. If\({\rm{n = 5, x\_}}\left\{ {\rm{1}} \right\}{\rm{ = 237, x\_}}\left\{ {\rm{2}} \right\}{\rm{ = 375, x\_}}\left\{ {\rm{3}} \right\}{\rm{ = 202, x\_}}\left\{ {\rm{4}} \right\}{\rm{ = 525,}}\)and\({{\rm{x}}_{\rm{5}}}{\rm{ = 418}}\), what is the corresponding estimate?

b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating\({\rm{\beta - \alpha + 1}}\)? Explain in one or two sentences.

Short Answer

Expert verified

a) The corresponding estimate is\({\rm{324}}\).

(b) If the sample includes both the plane with the highest serial number and the plane with the lowest serial number.

The greatest value can never exceed\({\rm{\beta }}\), and the smallest value can never be less than\({\rm{\alpha }}\).

Step by step solution

01

Definition

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

Finding corresponding estimate

(a)

Consider the given information and simplify,

\(\begin{array}{l}{\rm{n = 5}}\\{{\rm{x}}_{\rm{1}}}{\rm{ = 237}}\\{{\rm{x}}_{\rm{2}}}{\rm{ = 375}}\\{{\rm{x}}_{\rm{3}}}{\rm{ = 202}}\\{{\rm{x}}_{\rm{4}}}{\rm{ = 525}}\\{{\rm{x}}_{\rm{5}}}{\rm{ = 418}}\end{array}\)

We note that the smallest data value of the five data values is\({\rm{525}}\):

\({\rm{max}}\left( {{{\rm{x}}_{\rm{i}}}} \right){\rm{ = 525}}\)

We note that the largest data value of the five data values is\({\rm{202}}\):

\({\rm{min}}\left( {{{\rm{x}}_{\rm{i}}}} \right){\rm{ = 202}}\)

The value of the estimator then becomes:

\({\rm{max}}\left( {{{\rm{x}}_{\rm{i}}}} \right){\rm{ - min}}\left( {{{\rm{x}}_{\rm{i}}}} \right){\rm{ + 1 = 525 - 202 + 1 = 324}}\)

03

Finding corresponding estimate

(b)

The estimator will be exactly equal to the parameter if:

\(\begin{array}{l}{\rm{\beta = max}}\left( {{{\rm{x}}_{\rm{i}}}} \right)\\{\rm{\alpha = min}}\left( {{{\rm{x}}_{\rm{i}}}} \right)\end{array}\)

If the sample comprises the aircraft with the biggest serial number and the plane with the smallest serial number, the estimator is precisely equal to the parameter.

Because the maximum can never be higher than \({\rm{\beta }}\)and the minimum can never be lower than\({\rm{\alpha }}\), the difference between the maximum and minimum can never be more than\({\rm{\beta - \alpha + 1}}\), the estimate can never be greater than the real total.

If the sample comprises the aircraft with the biggest serial number and the plane with the smallest serial number, the estimator is precisely equal to the parameter.

Because the maximum can never be higher than \({\rm{\beta }}\)and the minimum can never be lower than\({\rm{\alpha }}\), the difference between the maximum and minimum can never be more than \({\rm{\beta - \alpha + 1}}\), the estimate can never be greater than the real total.

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Most popular questions from this chapter

At time \({\rm{t = 0}}\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \({\rm{\lambda }}\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \({\rm{\lambda }}\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential (\({\rm{\lambda }}\)) variables, which is exponential with parameter \({\rm{2\lambda }}\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential \({\rm{rv}}\) with parameter \({\rm{3\lambda }}\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \({\rm{25}}{\rm{.2,41}}{\rm{.7,51}}{\rm{.2,55}}{\rm{.5,59}}{\rm{.5,61}}{\rm{.8}}\) (from which you should calculate the times between successive births). Derive the mle of l. (Hint: The likelihood is a product of exponential terms.)

Of \({{\rm{n}}_{\rm{1}}}\)randomly selected male smokers, \({{\rm{X}}_{\rm{1}}}\) smoked filter cigarettes, whereas of \({{\rm{n}}_{\rm{2}}}\) randomly selected female smokers, \({{\rm{X}}_{\rm{2}}}\) smoked filter cigarettes. Let \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes.

a. Show that \({\rm{(}}{{\rm{X}}_{\rm{1}}}{\rm{/}}{{\rm{n}}_{\rm{1}}}{\rm{) - (}}{{\rm{X}}_{\rm{2}}}{\rm{/}}{{\rm{n}}_{\rm{2}}}{\rm{)}}\) is an unbiased estimator for \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\). (Hint: \({\rm{E(}}{{\rm{X}}_{\rm{i}}}{\rm{) = }}{{\rm{n}}_{\rm{i}}}{{\rm{p}}_{\rm{i}}}\) for \({\rm{i = 1,2}}\).)

b. What is the standard error of the estimator in part (a)?

c. How would you use the observed values \({{\rm{x}}_{\rm{1}}}\) and \({{\rm{x}}_{\rm{2}}}\) to estimate the standard error of your estimator?

d. If \({{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200, }}{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\), and \({{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\), use the estimator of part (a) to obtain an estimate of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

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

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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.

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