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Chapter 6: Q37SE (page 275)

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

Short Answer

Expert verified

The value is \({\rm{c = 1}}{\rm{.0132}}\).

Step by step solution

01

Define standard deviation

The standard deviation is a metric that indicates how much variation (such as spread, dispersion, and spread) there is from the mean. A "typical" divergence from the mean is indicated by the standard deviation. Since it returns to the data set's original units of measure, it's a common measure of variability.

02

Explanation

It is well knowledge that,

\({\rm{E(S) = }}\frac{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{\rm{\sigma }}\)

As a result, the following must be true in order to generate an unbiased estimator\({\rm{\theta }}\):

\({\rm{E(cS) = \sigma }}\)

Property is the reason.

\({\rm{E(cS) = cE(S)}}\)

and because the\({\rm{E(S)}}\)is well-known, it's easy to see that c has to be,

\({\rm{c = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}\)

\({\rm{cS}}\)must be an unbiased estimator in order for it. Take a look.

\(\begin{array}{c}{\rm{E(cS) = cE(S)}}\\{\rm{ = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}{\rm{ \times }}\frac{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{\rm{\sigma }}\\{\rm{ = \sigma }}\end{array}\)

03

Evaluating the values

The c becomes when\({\rm{n = 20}}\)is substituted in the formula for c.

\(\begin{array}{c}{\rm{c = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{20 - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{20 - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{{\rm{20}}}}{{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{\Gamma (9}}{\rm{.5)}}}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{19}}}}} {\rm{ \times \Gamma (10)}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{8}}{\rm{.5 \times 7}}{\rm{.5 \times \ldots \times 1}}{\rm{.5 \times 0}}{\rm{.5 \times \Gamma (0}}{\rm{.5)}}}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{19}}}}} {\rm{ \times (10 - 1)!}}}}\\{\rm{ = }}\frac{{{\rm{8}}{\rm{.5 \times 7}}{\rm{.5 \times \ldots \times 1}}{\rm{.5 \times 0}}{\rm{.5 \times \pi }}}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{19}}}}} {\rm{ \times 9!}}}}\\{\rm{ = 1}}{\rm{.0132}}\end{array}\)

  1. ; the gamma function's characteristics!

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Most popular questions from this chapter

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.

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.)

a. Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a uniform distribution on \({\rm{(0,\theta )}}\). Then the mle of \({\rm{\theta }}\) is \({\rm{\hat \theta = Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\). Use the fact that \({\rm{Y}} \le {\rm{y}}\) if each \({{\rm{X}}_{\rm{i}}} \le {\rm{y}}\) to derive the cdf of Y. Then show that the pdf of \({\rm{Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\) is \({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}}&{{\rm{0}} \le {\rm{y}} \le {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

b. Use the result of part (a) to show that the mle is biased but that \({\rm{(n + 1)}}\)\({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{/n}}\) is unbiased.

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 \({\rm{cX}}\).)

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 \({\rm{\lambda }}\).
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