Revision as of 21:36, 4 May 2023 by Admin (Created page with "'''Solution: E''' Let <math>X_J, X_K,</math> and <math>X_L</math> represent annual losses for cities J, K, and L, respectively. Then <math>X = X_J + X_K + X_L</math> and due...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Exercise


May 04'23

Answer

Solution: E

Let [math]X_J, X_K,[/math] and [math]X_L[/math] represent annual losses for cities J, K, and L, respectively. Then [math]X = X_J + X_K + X_L[/math] and due to independence

[[math]] M(t) = M_J(t)M_K(t) M_L(t) = (1-2t)^{-3} (1-2t)^{-2.5} (1-2t)^{-4.5} = (1-2t)^{-10}. [[/math]]

Therefore,

[[math]] M'(t) = 20(1-2t)^{-11}, \, M^{''}(t) = 440(1-2t)^{-12}, \, M^{'''}(t) = 10560(1-2t)^{-13}, \, \operatorname{E}[X^3] = M^{'''}(0) = 10560. [[/math]]

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

00