A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable with mean 2. Annual claims are modeled by an exponential random variable with mean 1. Premiums and claims are independent. Let [math]X[/math] denote the ratio of claims to premiums, and let [math]f[/math] be the density function of [math]X[/math].

Determine [math]f(x)[/math], where it is positive.

• [math]\frac{1}{2x+1}[/math]
• [math]\frac{2}{(2x+1)^2}[/math]
• [math]e^{-x}[/math]
• [math]2e^{-2x}[/math]
• [math]xe^{-x}[/math]

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

Solution: B

Let

• [math]u[/math] be the annual claims
• [math]v[/math] be the annual premiums
• [math]f(x)[/math] be the density function of [math]X[/math]
• [math]F(x)[/math] be the distribution function of [math]X[/math]

Then since [math]U[/math] and [math]V[/math] are independent

[[math]] g(u,v) = (e^{-u})(\frac{1}{2}e^{-v/2}) = \frac{1}{2}e^{-u}e^{-v/2}, \, 0 \lt u \lt \infty, \, 0 \lt v \lt \infty [[/math]]

and

[[math]] \begin{align*} F(x) = P[X \leq x] = P[ \frac{u}{v} \leq x ] &= P[ U \leq Vx] \\ &= \int_0^{\infty}\int_0^{vx} g(u,v) du dv = \int_0^{\infty}\int_0^{vx} \frac{1}{2} e^{-u}e^{-v/2} du dv \\ &= \int_0^{\infty} -\frac{1}{2} e^{-u}e^{-v/2} \Big |_0^{vx} dv = \int_0^{\infty} \left( -\frac{1}{2} e^{-vx} e^{-v/2} + \frac{1}{2} e^{-v/2}\right) dv \\ &= \int_0^{\infty} \left( -\frac{1}{2} e^{-v(x+1/2)} + \frac{1}{2} e^{-v/2} \right )dv \\ &= \left [ \frac{1}{2x+1} e^{-v(x + 1/2)} - e^{-v/2} \right ]_0^{\infty} \\ &= -\frac{1}{2x+1} + 1 \end{align*} [[/math]]

Finally ,

[[math]] f(x) = F^{'}(x) = \frac{2}{(2x+1)^2}. [[/math]]

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