 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1/5/b/ii/solution

Include the negative atom as well as the continuous part in the expected value:

E [Y] = \frac{1}{2} (- 1) + \frac{1}{2} \int_{0}^{\infty} y e^{- y} d y = - \frac{1}{2} + \frac{1}{2} = 0.

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