Revision as of 01:11, 28 June 2024 by Admin (Created page with "'''Solution: B''' We condition on <math>Y=y</math> then the expected value of <math>|X-y|</math> equals <math display = "block"> \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 display ="block"> \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>")
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]]