Individuals purchase both collision and liability insurance on their automobiles. The value of the insured’s automobile is V. Assume the loss L on an automobile claim is a random variable with cumulative distribution function

[[math]] F(l) = \begin{cases} \frac{3}{4}\left(\frac{l}{V}\right)^3, \, 0 ≤ l \lt V \\ 1-\frac{1}{10}e^{\frac{-(l-V)}{V}}, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate the probability that the loss on a randomly selected claim is greater than the value of the automobile.

• 0.00
• 0.10
• 0.25
• 0.75
• 0.90

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

Step 1: Define the Probability to be Calculated

The problem asks for the probability that the loss [math]L[/math] on a randomly selected claim is greater than the value of the automobile [math]V[/math]. This can be expressed as [math]\operatorname{P}(L \gt V)[/math]. Using the properties of cumulative distribution functions (CDFs), we know that [math]\operatorname{P}(L \gt V) = 1 - \operatorname{P}(L \leq V)[/math]. Since the CDF, [math]F(l)[/math], is defined as [math]\operatorname{P}(L \leq l)[/math], we can write this as:

[[math]]\operatorname{P}(L \gt V) = 1 - F(V)[[/math]]

Step 2: Identify the Relevant CDF Segment for [math]F(V)[/math]

The given cumulative distribution function [math]F(l)[/math] is defined piecewise:

[[math]] F(l) = \begin{cases} \frac{3}{4}\left(\frac{l}{V}\right)^3, & 0 \leq l \lt V \\ 1-\frac{1}{10}e^{\frac{-(l-V)}{V}}, & \text{otherwise} \end{cases} [[/math]]

To calculate [math]F(V)[/math], we need to determine which part of the definition applies when [math]l = V[/math]. The "otherwise" condition covers [math]l \geq V[/math]. Therefore, we use the second expression for [math]F(l)[/math] where [math]l = V[/math]:

[[math]]F(l) = 1-\frac{1}{10}e^{\frac{-(l-V)}{V}} \quad \text{for } l \geq V[[/math]]

Step 3: Calculate the Value of [math]F(V)[/math]

Substitute [math]l = V[/math] into the appropriate CDF segment identified in Step 2:

[[math]] F(V) = 1-\frac{1}{10}e^{\frac{-(V-V)}{V}} [[/math]]

Simplify the exponent:

[[math]] \frac{-(V-V)}{V} = \frac{-0}{V} = 0 [[/math]]

So, the expression for [math]F(V)[/math] becomes:

[[math]] F(V) = 1-\frac{1}{10}e^0 [[/math]]

Recall that [math]e^0 = 1[/math]. Therefore:

[[math]] F(V) = 1-\frac{1}{10}(1) = 1 - 0.10 = 0.90 [[/math]]

Step 4: Calculate [math]\operatorname{P}(L \gt V)[/math]

Now, using the relationship established in Step 1, [math]\operatorname{P}(L \gt V) = 1 - F(V)[/math], and the calculated value of [math]F(V)[/math] from Step 3:

[[math]] \operatorname{P}(L \gt V) = 1 - 0.90 = 0.10 [[/math]]

Thus, the probability that the loss on a randomly selected claim is greater than the value of the automobile is 0.10.

Key Insights

• The probability [math]\operatorname{P}(X \gt x)[/math] can be calculated as [math]1 - F(x)[/math] when [math]F(x)[/math] is the cumulative distribution function [math]\operatorname{P}(X \leq x)[/math].
• For piecewise-defined functions, it is crucial to correctly identify which segment of the function applies for the specific input value. In this case, for [math]F(V)[/math], the "otherwise" condition covers [math]l \geq V[/math].
• Any non-zero number raised to the power of zero is 1 (e.g., [math]e^0 = 1[/math]), which is a common simplification in probability and actuarial calculations.

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Solution: B

[[math]] \operatorname{P}(X \gt V) = 1-\operatorname{P}( X \leq V ) = 1-F(V) = 1- (1-\frac{1}{10}e^{-\frac{V-V}{V}}) = 0.10 [[/math]]

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