Revision as of 10:18, 1 May 2023 by Admin (Created page with "'''Solution: B''' Denote the insurance payment by the random variable <math>Y</math>. Then <math display = "block"> Y = \begin{cases} 0, \quad 0 < X \leq C \\ X-C, \quad...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Exercise


May 01'23

Answer

Solution: B

Denote the insurance payment by the random variable [math]Y[/math]. Then

[[math]] Y = \begin{cases} 0, \quad 0 \lt X \leq C \\ X-C, \quad C \lt X \lt 1 \end{cases} [[/math]]

Now we are given that

[[math]] \begin{align*} 0.64 = \operatorname{P}[Y \lt 0.5 ) = \operatorname{P}[ 0 \lt X \lt 0.5 + C ) &= \int_0^{0.5 + C} 2x dx \\ &= x^2 \Big |_0^{0.5 + C} \\ &= (0.5 + C)^2. \end{align*} [[/math]]

Therefore, solving for [math]C[/math], we find [math]C = ±0.8 − 0.5[/math]. Finally, since [math]0 \lt C \lt 1 [/math] , we conclude that [math]C = 0.3[/math].

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

00