excans:7644f73378: Difference between revisions
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(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>") |
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\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}. | \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> | </math> | ||
{{AI-certified-answer}} | |||
Latest revision as of 14:45, 20 July 2025
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]]
Answer was verified by AI