Let $Y = U^3$
$P(Y \leq y) = P(U^3 \leq y) = P(U \leq y^{1/3}) = ?$
Am I approaching this question right? I am kinda stucked here and am wondering how do I determine the distribution for $U^3$.
2026-03-28 22:28:01.1774736881
Uniform distribution [0, 1], what is the distribution of $U^3$?
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$P(Y\leq y)=P(U^3\leq y)=P(U\leq y^{1/3})=\int_0^{y^{1/3}}dx=y^{1/3}.$ If you want the pdf, just differentiate and you get $f_Y(y)=\frac{1}{3}y^{-2/3}$ when $0\leq y \leq 1$. Otherwise $0$.