For what values of [math]a[/math] and [math]b[/math] is the function [math]f(x,y)[/math] given below a valid joint density function?

• [math]2a + b = 6 [/math]
• [math]2a + b = 6 [/math] and both [math]a [/math] and [math]b[/math] must be non-negative
• [math]a + b = 2 [/math] and both [math]a [/math] and [math]b[/math] must be non-negative
• For all non-negative [math]a[/math] and [math]b[/math]
• There are no values [math]a[/math] and [math]b[/math] that give a valid joint density

Two loss variables [math]L_1[/math] and [math]L_2[/math] have a strictly increasing continuous joint cumulative distribution [math]F(x,y)[/math]. Which of the following expressions represent the probability that both losses exceed their 95^th percentile?

• [math]F(F_X^{-1}(0.95), F_Y^{-1}(0.95))[/math] with [math]F_X(x) = F(x,\infty)[/math] and [math]F_Y(y) = F(\infty, y)[/math]
• [math]F(F_X^{-1}(0.95), F_Y^{-1}(0.95)) - 0.9.[/math] with [math]F_X(x) = F(x,\infty)[/math] and [math]F_Y(y) = F(\infty, y)[/math]
• [math]F(F_X(0.95), F_Y(0.95)) - 0.9.[/math] with [math]F_X(x) = F(x,\infty)[/math] and [math]F_Y(y) = F(\infty, y)[/math]
• [math]0.05^2[/math]
• [math]F(F_X^{-1}(0.95), F_Y^{-1}(0.95)) - 0.9.[/math] with [math]F_X(x) = F(\infty, x)[/math] and [math]F_Y(y) = F(y, \infty)[/math]

The joint density function of the logarithm of the two random variables [math]X[/math] and [math]Y[/math] equals [math]f(x,y)[/math]. Which of the following expressions represent the joint density function of [math]X[/math] and [math]Y[/math]?

• [math]y^{-1}x^{-1}f(\ln(x),\ln(y))[/math]
• [math]f(\ln(x),\ln(y))[/math]
• [math]y^{-1}x^{-1}f(x,y)[/math]
• [math]e^x e^yf(e^x,e^y)[/math]
• [math]x^{-1}f(\ln(x),x) + y^{-1}f(x,\ln(y))[/math]

The joint distribution of claim frequency and claim severity, conditional on the claim frequency being positive, is given below:

If the probability that the claim frequency is zero is 0.75, determine the variance of the claim frequency.

• 0.2475
• 0.5123
• 0.6625
• 0.8833
• 1

A machine consists of two components, whose lifetimes have the joint density function

The machine operates until both components fail. Calculate the expected operational time of the machine.

• 1.7
• 2.5
• 3.3
• 5.0
• 6.7

A client spends [math]X[/math] minutes in an insurance agent’s waiting room and [math]Y[/math] minutes meeting with the agent. The joint density function of [math]X[/math] and [math]Y[/math] can be modeled by

Determine which of the following expressions represents the probability that a client spends less than 60 minutes at the agent’s office.

• [math]\frac{1}{800}\int_0^{40}\int_0^{20}e^{-x/40}e^{-y/20} dy dx[/math]
• [math]\frac{1}{800} \int_0^{40}\int_0^{20-x}e^{-x/40}e^{-y/20} dy dx[/math]
• [math]\frac{1}{800} \int_0^{20}\int_0^{40-x}e^{-x/40}e^{-y/20} dy dx[/math]
• [math] \frac{1}{800} \int_0^{60}\int_0^{60}e^{-x/40}e^{-y/20} dy dx[/math]
• [math]\frac{1}{800} \int_0^{60}\int_0^{60-x}e^{-x/40}e^{-y/20} dy dx [/math]

An insurance company will cover losses incurred from tornadoes in a single calendar year. However, the insurer will only cover losses for a maximum of three separate tornadoes during this timeframe. Let [math]X[/math] be the number of tornadoes that result in at least 50 million in losses, and let [math]Y[/math] be the total number of tornadoes. The joint probability function for [math]X[/math] and [math]Y[/math] is

where [math]c[/math] is a constant.

Calculate the expected number of tornadoes that result in fewer than 50 million in losses.

• 0.19
• 0.28
• 0.76
• 1.00
• 1.10

A device runs until either of two components fails, at which point the device stops running. The joint density function of the lifetimes of the two components, both measured in hours, is

Calculate the probability that the device fails during its first hour of operation.

• 0.125
• 0.141
• 0.391
• 0.625
• 0.875

A device contains two components. The device fails if either component fails. The joint density function of the lifetimes of the components, measured in hours, is [math]f(s,t)[/math], where [math]0 \lt s \lt 1[/math] and [math]0 \lt t \lt 1 [/math].

Determine which of the following represents the probability that the device fails during the first half hour of operation.

• [math]\int_{0}^{0.5}\int_{0}^{0.5}f(s,t)\,ds\,dt[/math]
• [math]\int_0^1\int_0^{0.5}f(s,t)\,ds\,dt[/math]
• [math]\int_{0.5}^1\int_{0.5}^1f(s,t)\,ds\,dt[/math]
• [math]\int_{0}^{0.5}\int_{0}^1f(s,t)\,ds\,dt + \int_{0}^1\int_{0}^{0.5}f(s,t)\,ds\,dt[/math]
• [math]\int_{0}^{0.5}\int_{0.5}^1f(s,t)\,ds\,dt + \int_{0}^1\int_0^{0.5}f(s,t)\,ds\,dt[/math]

The future lifetimes (in months) of two components of a machine have the following joint density function:

Determine which of the following represents the probability that both components are still functioning 20 months from now.

• [math]\frac{6}{125000}\int_{0}^{20}\int_{0}^{20}(50-x-y)\,ds\,dt[/math]
• [math]\frac{6}{125000}\int_{20}^{30}\int_{20}^{50-x}(50-x-y)\,ds\,dt[/math]
• [math]\frac{6}{125000}\int_{20}^{30}\int_{20}^{50-x-y}(50-x-y)\,ds\,dt[/math]
• [math]\frac{6}{125000}\int_{20}^{50}\int_{20}^{50-x}(50-x-y)\,ds\,dt [/math]
• [math]\frac{6}{125000}\int_{20}^{50}\int_{20}^{50-x-y}(50-x-y)\,ds\,dt [/math]