Exercise
May 14'23
Answer
Key: A
These observations are truncated at 500. The contribution to the likelihood function is
[[math]]
\frac{f(x)}{1-F(500)} = \frac{\theta^{-1}e^{-x/\theta}}{e^{-500/\theta}}.
[[/math]]
Then the likelihood function is
[[math]]
\begin{aligned}
&L(\theta) = \frac{\theta^{-1}e^{-600/\theta}\theta^{-1}e^{-700/\theta}\theta^{-1}e^{-900/\theta}}{\left( e^{-500/\theta}\right)^3} = \theta^{-3}e^{-700/\theta} \\
& l (\theta ) = \ln L(\theta ) = −3\ln \theta − 700\theta^{−1} \\
& l^{'}(\theta) = -3\theta^{-1} + 700\theta^{-2} = 0 \\
& \theta = 700 / 3 = 233.33.
\end{aligned}
[[/math]]