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]]

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