Revision as of 02:03, 28 June 2024 by Admin (Created page with "'''Solution: A''' For fixed <math>y \in [0,1]</math>, we have <math>E[X^y] = \frac{1}{y+1}</math> and thus <math display = "block"> E[X^Y] = \int_{0}^1 \frac{1}{y+1} \, dy = \log(2). </math>")
Exercise
Jun 28'24
Answer
Solution: A
For fixed [math]y \in [0,1][/math], we have [math]E[X^y] = \frac{1}{y+1}[/math] and thus
[[math]]
E[X^Y] = \int_{0}^1 \frac{1}{y+1} \, dy = \log(2).
[[/math]]