Losses follow an exponential distribution with mean 1. Two independent losses are observed.

Calculate the probability that either of the losses is more than twice the other.

• 1/6
• 1/4
• 1/3
• 1/2
• 2/3

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

Solution: E

If the first loss is [math]X[/math], then [math]\operatorname{P}(X \gt x) = e^{-x} [/math]; the probability that the second loss is more than twice [math]X[/math], would be [math]\operatorname{P}(X \gt 2x) = e^{-2x} [/math]. Thus, the probability that the second loss is more than twice the first loss is

[[math]]\int_0^{\infty} e^{-2x} e^{-x} dx = \frac{1}{3}[[/math]]

due to independence. By symmetry, the probability that the first loss 0 is more than twice the second loss is also 1/3. Thus, the answer is 2/3.

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