Mixed Distributions and Deductibles | Six Minutes: Actuarial Exam P, Part 17

Six Minutes: Actuarial Exam P, Part 17: Mixed Distributions and Deductibles: Piecewise discrete-continuous CDFs; then Deductibles A mixed distribution has point masses plus a continuous density, so its CDF is written in pieces with jumps at the atoms. We build such a CDF, read probabilities from it, and compute the mean two ways. Then: An ordinary deductible d turns a loss X into a payment Y = max(X - d, 0), and the expected payment per loss is the integral of (x - d) f(x) from d upward. A shortcut using E[X] minus E[min(X, d)] gives the same answer and doubles as a check. Chapters 0:00 Introduction 0:22 What a mixed distribution is 0:42 The piecewise CDF 1:08 Expectation two ways 1:37 Worked example 2:13 The trap 2:36 Recap 3:01 The idea 3:28 A shortcut 4:00 Worked example 5:02 Trap 5:35 Recap 5:52 Thanks for watching Practice problems 1. X = 0 with probability 0.4, and otherwise X is exponential with mean 5. Find P(X ≤ 5). Answer: 0.4 + 0.6(1 − e^(−1)) ≈ 0.779. 2. X = 2 with probability 1/3, and otherwise X is uniform on (2, 8). Find E[X]. Answer: E[X] = (1/3)(2) + (2/3)(5) = 4. 3. A loss X is exponential with mean 5. With an ordinary deductible of 1, the expected payment per loss is 5 e^{-0.2} ≈ 4.09. 4. A loss X is uniform on (0, 12) with an ordinary deductible of 3. The expected payment per loss is (12-3)^2 / (2·12) = 81/24 = 3.375. Six Minutes @SixMinuteClass: One technique, about six minutes. The narrator's voice is synthetic (computer-generated). Independent educational video. Not affiliated with or endorsed by the Society of Actuaries. © 2026 Six Minutes · @SixMinuteClass