91Ó°ÊÓ

Let the random variables be IID and defined as\({{\rm{X}}_{\rm{1}}}{\rm{ \ldots }}{{\rm{X}}_{\rm{n}}}\).Every one of these random variables is assumed to be a sample from the same exponential, with the same \(\beta \) , \({X_i} \sim \exp \left( \beta \right)\)

It denotes the time between two successive events. \(\beta \) denotes the rate of occurrence of events per unit time.

The PMF is:

\(f\left( x \right) = \beta {e^{ - \beta x}},x > 0\)

The cumulative distribution function of \({{\rm{X}}_{\rm{1}}}{\rm{ \ldots }}{{\rm{X}}_{\rm{n}}}\)is:

\(F\left( x \right) = 1 - {e^{ - \beta x}},x > 0\)

For an experiment with two outcomes, which occurs n times with probability p, we would use binomial distribution. Let k be the favorable events out of n trials.

\(\Pr \left( {K = k} \right) = {}^n{C_k}{p^k}{\left( {1 - p} \right)^{n - k}} \ldots \left( 1 \right)\)

Here, p is the probability of success in any one trial.

The probability that a random lamp fails, given that the variable is exponentially distributed with mean\(\beta\),

\(\begin{array}{c}p = \int\limits_0^T {f\left( {t|\beta } \right)} \\ = 1 - {e^{ - \beta t}}\end{array}\)

The maximum likelihood function is given by:

\(L\left( {\theta |{x_1} \ldots {x_n}} \right) = \prod\limits_{i = 1}^n f \left( {{x_i};{\theta _1} \ldots {\theta _n}} \right)\)

The function is differentiated with respect to the parameter and equated with 0.

Let us calculate MLE to estimate the p parameter of a binomial distribution.

Applying likelihood to equation (1)-

\(\ln L\left( {p|X = x} \right) = \ln \left( {{}^n{C_x}} \right) + x\ln p + \left( {n - x} \right)\ln \left( {1 - p} \right)\)

Differentiating it to find the value of p that maximises the log function, we equate it with 0.

\(\begin{array}{c}\frac{\partial }{{\partial p}}\ln L\left( {p|X = x} \right) = 0\\\frac{\partial }{{\partial p}}\ln \left( {{}^n{C_x}} \right) + x\ln p + \left( {n - x} \right)\ln \left( {1 - p} \right) = 0\end{array}\)

Further,

\(\begin{array}{c}\frac{x}{p} - \frac{{\left( {n - x} \right)}}{{\left( {1 - p} \right)}} = 0\\x\left( {1 - p} \right) - p\left( {n - x} \right) = 0\\\widehat p = \frac{x}{n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{array}\)

Therefore, MLE of p is \(\frac{x}{n}\) .

Put \(\widehat p = 1 - {e^{ - \beta T}}\) and further solve for the parameter of exponential distribution,

\(\begin{array}{l}1 - {e^{ - \beta T}} = \frac{x}{n}\\{e^{ - \beta T}} = \frac{{n - x}}{n}\end{array}\)

By taking log on both sides,

\(\begin{array}{l}\ln \left( {{e^{ - \beta T}}} \right) = \ln \left( {\frac{{n - x}}{n}} \right)\\ - \beta T = \ln \left( {\frac{{n - x}}{n}} \right)\\\widehat \beta = \frac{{ - \ln \left( {\frac{{n - x}}{n}} \right)}}{T}\end{array}\)

Therefore, the MLE is \(\widehat \beta = \frac{{ - \ln \left( {\frac{{n - x}}{n}} \right)}}{T}\) .