You are given the following 20 bodily injury losses before the deductible is applied:

Loss	Number of Losses	Deductible	Policy Limit
750	3	200	[math]\infty[/math]
200	3	0	10,000
300	4	0	20,000
>10,000	6	0	10,000
400	4	300	[math]\infty[/math]

Past experience indicates that these losses follow a Pareto distribution with parameters [math]\alpha [/math] and [math]\theta = 10,000 [/math].

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

• Less than 2.0
• At least 2.0, but less than 3.0
• At least 3.0, but less than 4.0
• At least 4.0, but less than 5.0
• At least 5.0

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

Key: C

[[math]] \begin{aligned} &L = \left [ \frac{f (750)}{1 − F (200)}\right]^3 f (200)^3 f (300)^4 [1 − F (10, 000)]^6 \left[\frac{f (400)}{1 − F (300)} \right] \\ =& \left[ \frac{\alpha 10200^{\alpha}}{10750^{\alpha +1}}\right]^3 \left[ \frac{\alpha 10000^{\alpha}}{10200^{\alpha +1}}\right]^3 \left[ \frac{\alpha 10000^{\alpha}}{10300^{\alpha +1}}\right]^4\left[ \frac{\alpha 10000^{\alpha}}{20000^{\alpha +1}}\right]^6\left[ \frac{\alpha 10300^{\alpha}}{10400^{\alpha +1}}\right]^4 \\ &= \alpha^{14} 10200^{-3}10000^{13\alpha} 10300^{-4}10750^{-3\alpha -3} 20000^{-6\alpha}10400^{-4\alpha -4} \\ &\propto\alpha^{14}10000^{13\alpha}10750^{-3\alpha}20000^{-6\alpha}10400^{-4\alpha} \\ & \ln L = 14 \ln \alpha + 13\alpha \ln(10, 000) − 3\alpha \ln(10, 750) − 6\alpha \ln(20, 000) − 4\alpha \ln(10, 400) \\ &= 14 \ln \alpha − 4.5327 \alpha . \end{aligned} [[/math]]

The derivative is [math]14/\alpha − 4.5327 [/math] and setting it equal to zero gives [math]\hat{\alpha} = 3.089. [/math]

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