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QUESTION The life-time of a type of electric bulb can be assumed
to have an exponential distribution with mean
$\displaystyle\frac{1}{\lambda}$.  8 such bulbs are tested.

In each of the cases below write down the likelihood function and
hence estimate $\lambda$ by the method of maximum likelihood.

\begin{description}
\item[(i)]
the bulbs fail at times 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5
thousand hours;

\item[(ii)]
the first five bulbs to fail do so at times 1.1, 1.3, 1.5, 1.7,
1.9 thousand hours, the remainder are still working after 2.0
thousand hours;

\item[(iii)]
three bulbs fail between 1.0 and 1.6 thousand hours, three between
1.6 and t.t thousand hours and two between 2.2 and 2.8 thousand
hours.

\end{description}



ANSWER
\begin{description}
  \item[(i)]
   $L(\lambda)=\lambda e^{-1.1\lambda}\lambda e^{-1.3\lambda}\lambda e^{-1.5\lambda}\lambda
   e^{-1.7\lambda}\lambda e^{-1.9\lambda} e^{-2.1\lambda}\lambda
   e^{-2.3\lambda}\lambda
   e^{-2.5\lambda}=\lambda^8 e^{_14.4\lambda}\\
   \ln \lambda=8\ln \lambda-14.4 \lambda\\
   \frac{\partial \ln L(\lambda)}{\partial
   \lambda}=\frac{8}{\lambda}-14.4$ therefore
   $\hat{\lambda}=\frac{8}{14.4}=0.556$

  \item[(ii)]
   $L(\lambda)=\lambda e^{-1.1\lambda}\lambda e^{-1.3\lambda}\lambda e^{-1.5\lambda}\lambda
   e^{-1.7\lambda}\lambda e^{-1.9\lambda} (e^{-2.0
   \lambda})^3=\lambda^5 e^{_13.5\lambda}$\\
   from above $\hat {\lambda}=\frac{5}{13,5}=0.370$

  \item[(iii)]
  $L(\lambda)=(e^{-\lambda}-e^{1.6\lambda})^3(e^{-1.6\lambda}-e^{-2.2\lambda})^3((e^{-2.2\lambda}-e^{-2.8\lambda})^3=e^{-12.2
  \lambda}(1-e^{-0.6\lambda})^8\\
  \ln L(\lambda)=-12.2\lambda+8 \ln (1-e^{-0.6\lambda})\\
  \frac{\partial \ln L(\lambda)}{\partial \lambda}=-12.2 +\frac{8
  \times 0.6 e^{-0.6\lambda}}{1-e^{-0.6\lambda}}\\
  12.2=\frac{4.8e^{-0.6 \hat{\lambda}}}{1-e^{-0.6
  \hat{\lambda}}}=17e^{-0.6\hat{\lambda}}\\
  e^{0.6\hat{\lambda}}=\frac{17}{12.2}\ \ \hat{\lambda}=0.553$

  \end{description}


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