You are given:

1. Losses follow a single-parameter Pareto distribution with density function:
   [[math]]f(x) = \frac{\alpha}{x^{\alpha+1}}, \, x\gt1, \, 0 \lt \alpha \lt \infty [[/math]]
2. A random sample of size five produced three losses with values 3, 6 and 14, and two losses exceeding 25.

Calculate the maximum likelihood estimate of [math]\alpha [/math]

• 0.25
• 0.30
• 0.34
• 0.38
• 0.42

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

Key: A

The distribution function is

[[math]]F(x)=\int_{1}^{x} \alpha t^{-\alpha-1} d t=-\left.t^{-\alpha}\right|_{1} ^{x}=1-x^{-\alpha}[[/math]]

. The likelihood function is

[[math]]\begin{aligned} L & =f(3) f(6) f(14)[1-F(25)]^{2} \\ & =\alpha 3^{-\alpha-1} \alpha 6^{-\alpha-1} \alpha 14^{-\alpha-1}\left(25^{-\alpha}\right)^{2} \\ & \propto \alpha^{3}[3(6)(14)(625)]^{-\alpha} . \end{aligned}[[/math]]

Taking [math]\log s[/math], differentiating, setting equal to zero, and solving:

[math]\ln L=3 \ln \alpha-\alpha \ln 157,500[/math] plus a constant

[math]d \ln L / d \alpha=3 \alpha^{-1}-\ln 157,500=0[/math]

[math]\hat{\alpha}=3 / \ln 157,500=0.2507[/math].

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