Question

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

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

Define exponential function

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

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 \({{\rm{\theta }}_{\rm{i}}}\).

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 \({\rm{\hat \lambda = 0}}{\rm{.012}}\).