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Exercise
May 09'23
Answer
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.