Question

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.)

Short answer

The value is \({\rm{\hat \lambda = 0}}{\rm{.0436}}\).

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

As, 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.

Denote with

\({{\rm{X}}_{\rm{1}}}\)= time till the first birth,

\({{\rm{X}}_{\rm{2}}}{\rm{ = }}\) time between the first and second births,

.

.

\({{\rm{X}}_{\rm{n}}}{\rm{ = }}\) time between the \({{\rm{n}}^{{\rm{th }}}}\) and \({{\rm{(n - 1)}}^{{\rm{th}}}}\) births,

As, \({{\rm{X}}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\) are independent and exponentially distributed random variables with appropriate parameters \({\rm{ - }}{{\rm{X}}_{\rm{1}}}\) with parameter \({\rm{\lambda }}\), \({{\rm{X}}_{\rm{2}}}\) with parameter \({\rm{2\lambda }}\),..., \({{\rm{X}}_{\rm{n}}}\) with parameter \({\rm{n\lambda }}\).

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

As, \({{\rm{X}}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\) are independent and exponentially distributed random variables with appropriate parameters \({\rm{ - }}{{\rm{X}}_{\rm{1}}}\) with parameter \({\rm{\lambda }}\), \({{\rm{X}}_{\rm{2}}}\)

Evaluating the maximum likelihood estimate

Because of the independence, the likelihood function becomes,

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

Look at the log likelihood function to determine the maximum.

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

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,

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

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

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

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

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

Evaluating the value

We need the values x, which are the times between successive births, based on the serial birth times. The following formula is used to calculate the values:

\({{\rm{x}}_{\rm{1}}}{\rm{ = 25}}{\rm{.2}}\) the time of the first birth;

\({{\rm{x}}_{\rm{2}}}{\rm{ = 41}}{\rm{.7 - 25}}{\rm{.2 = 16}}{\rm{.5}}\) the period of time between the first and second births;

\({{\rm{x}}_{\rm{3}}}{\rm{ = 51}}{\rm{.2 - 41}}{\rm{.7 = 9}}{\rm{.5}}\) the time between the second and third births;

\({{\rm{x}}_{\rm{4}}}{\rm{ = 55}}{\rm{.5 - 51 = 4}}{\rm{.3}}\) the time between the third and fourth births;

\({{\rm{x}}_{\rm{5}}}{\rm{ = 59}}{\rm{.5 - 55}}{\rm{.5 = 4}}\) the time between the fourth and fifth births;

\({{\rm{x}}_{\rm{6}}}{\rm{ = 61}}{\rm{.8 - 59}}{\rm{.5 = 2}}{\rm{.3}}\) the time between the fifth and sixth births;

As a result, the total can now be calculated as follows:

\(\begin{array}{c}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{i}} {{\rm{x}}_{\rm{i}}}{\rm{ = 1 \times }}{{\rm{x}}_{\rm{1}}}{\rm{ + 2 \times }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + 6 \times }}{{\rm{x}}_{\rm{6}}}\\{\rm{ = 1 \times 25}}{\rm{.2 + 2 \times 16}}{\rm{.5 + \ldots + 6 \times 2}}{\rm{.3}}\\{\rm{ = 137}}{\rm{.7}}\end{array}\)

Last but not least, the maximum likelihood estimates of\({\rm{\lambda }}\)is,

\(\begin{array}{c}{\rm{\hat \lambda = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{i}} {{\rm{x}}_{\rm{i}}}}}\\{\rm{ = }}\frac{{\rm{6}}}{{{\rm{137}}{\rm{.7}}}}\\{\rm{ = 0}}{\rm{.0436}}\end{array}\)

Therefore, \({\rm{\hat \lambda = 0}}{\rm{.0436}}\).