You are given:

1. At time 4 hours, there are 5 working light bulbs.
2. The 5 bulbs are observed for [math]p[/math] more hours.
3. Three light bulbs burn out at times 5, 9, and 13 hours, while the remaining light bulbs are still working at time 4 + [math]p[/math] hours.
4. The distribution of failure times is uniform on (0, [math]\omega[/math] ) .
5. The maximum likelihood estimate of [math]\omega [/math] is 29.

Calculate [math]p[/math].

• Less than 10
• At least 10, but less than 12
• At least 12, but less than 14
• At least 14, but less than 16
• At least 16

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: D

[[math]] \begin{aligned} L(\omega) &= \frac{\frac{1}{\omega}\frac{1}{\omega}\frac{1}{\omega}\left(\frac{\omega-4-p}{\omega}\right)^2}{(\frac{\omega-4}{\omega})^5} = \frac{(\omega-4-p)^2}{(\omega-4)^5} \\ l(\omega) &= 2 \ln(\omega − 4 − p ) − 5\ln(\omega − 4), l ^{'}(\omega ) = \frac{2}{\omega - 4 - p} - \frac{5}{\omega -4} = 0 \\ 0 &= l^{'}(29) = \frac{2}{25-p} - \frac{5}{25} \Rightarrow p = 15. \end{aligned} [[/math]]

The denominator in the likelihood function is S(4) to the power of five to reflect the fact that it is known that each observation is greater than 4.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.