The proportion [math]X[/math] of yearly dental claims that exceed 200 is a random variable with probability density function

[[math]] f(x) = \begin{cases} 60 x^3 (1−x)^2, \, 0 \lt x \lt 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate [math]\operatorname{Var}[X/(1 – X)][/math].

• 149/900
• 10/7
• 6
• 8
• 10

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

Solution: C

[[math]] \operatorname{E}(\frac{X}{1-X}) = 60 \int_0^1 \frac{x}{1-x} x^3 (1-x)^2 dx = 60 \int_0^1 x^4(1-x) dx = 60(x^5/5 - x^6/6) \Big |_0^1 = 60(1/5 -1/6) = 2 [[/math]]

[[math]] \operatorname{E}[(\frac{X}{1-X})^2] = 60 \int_0^1 \frac{x^2}{(1-x)^2} x^3 (1-x)^2 dx = 60 \int_0^1 x^5 dx = 60(x^6/6 ) \Big |_0^1 = 60(1/6) = 10 [[/math]]

[[math]] \operatorname{Var}\left( \frac{X}{1-X}\right) = 10-2^2 = 6. [[/math]]

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