Suppose that the time to failure of an iten of equipment under specified conditions follows an exponential distribution with distribution function $\displaystyle F(t | \lambda) = 1 - exp(-\lambda t) (t > 0)$, where the expected time to failure $\displaystyle {\lambda}^{-1}$ is unknown. A life test is carried out in which items are used under conditions as close as possible to those specified. They are installed at different times, so at the end of the test the items have been installed for periods of length $\displaystyle u_1, ... ,u_n$. During these intervals, $\displaystyle m$ of the items have failed at times $\displaystyle t_1, ... ,t_m$ (numbering the items according to the time at which failure occurs), while the remaining $\displaystyle n - m$ failure times are censored at $\displaystyle u_{m+1}, ..., u_n$.

a) Show that the log-likelihood is given by

$\displaystyle m In\lambda - \lambda w$,

where $\displaystyle w$ denotes the sum of the times for which the $\displaystyle n$ items operated before the test ended (either $\displaystyle t_j (j = 1, ..., m)$) or $\displaystyle u_j (j = m+1, ... n))$. Show also that the log-likelihood is maximised when $\displaystyle \lambda$ equals $\displaystyle m/n$, i.e. that the maximum-likelihood estimate of $\displaystyle \lambda$ is $\displaystyle m/w$; find an expression for the maximum value.

b) The results of a life test on 10 items of coal mining equipment are given below, all times being measured from the day of installation of the first item. Assuming an exponential distribution of time to failure, calculate the log-likelihoods and the maximum-likelihood estimate of $\displaystyle \lambda$.

i) using the info available at the end of the test

ii) using only the data which were available on day 81 of the test.

Day of installation - 0 - 9 --- 11 - 21 - 40 - 50 - 50 - 51 - 52 - 60
Day of Failure ----- 2 - 128 - 62 - 98 - 73 - 77 - 64 - 75 - 66 - 97

dont quite understand the whole m and w thing... Can that just take any value? Up to 81 in the coal-mining example?