An actuary determines that the claim size for a certain class of accidents is a random variable, [math]X[/math], with moment generating function

[[math]] M_X(t) =(1 − 2500t )^{-4}. [[/math]]

Calculate the standard deviation of the claim size for this class of accidents.

• 1,340
• 5,000
• 8,660
• 10,000
• 11,180

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

Solution: B

We are given that

[[math]]M_x(t) = \frac{1}{(1-2500t)^4}[[/math]]

for the claim size [math]X[/math] in a certain class of accidents. First, compute

[[math]] M_x^{'}(t) = \frac{(-4)(-2500)}{(1-2500t)^5} = \frac{10000}{(1-2500t)^5}, \, M_x^{''}(t) = \frac{(10 000)(−5)(−2500)}{(1-2500t)^6} = \frac{125000000}{(1-2500t)^6}. [[/math]]

Then

[[math]] \begin{align*} \operatorname{E}[X] = M_x^{'}(0) = 10,000 \\ \operatorname{E}[X^2] = M_x^{''}(0) = 125,000,000 \\ \operatorname{Var}[X] = E[X^2] - (E[X])^2 = 125,000,000 - (10,000)^2 = 25,000,000 \\ \sqrt{\operatorname{Var}[X]} = 5,000. \end{align*} [[/math]]

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