Exercise


Jun 28'24

Answer

Solution: B

We condition on [math]Y=y[/math] then the expected value of [math]|X-y|[/math] equals

[[math]] \int_{0}^y y-x \, dx + \int_{y}^1 x-y \, dx = \int_{0}^{y} w \, dw + \int_{0}^{1-y} w \, dw = \frac{y^2}{2} + \frac{(1-y)^2}{2}. [[/math]]

Hence the expected value of |X-Y| equals

[[math]] \int_{0}^1 \frac{y^2}{2} + \frac{(1-y)^2}{2} \, dy = 2\int_{0}^1 \frac{y^2}{2} dy = \frac{1}{3}. [[/math]]

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