Final answer:
The density function of U to the power of 1/2, where U is uniformly distributed on [0, 1], is found by transforming the random variable and differentiating its CDF. The resulting density function is 2v for v in [0, 1].
Step-by-step explanation:
The question asks to find the density function of U1/2 when U is uniformly distributed on the interval [0, 1]. To solve this, we use the method of transformation of the random variable. If U is uniform on [0, 1], the density function of U is f(U) = 1 for U in [0, 1] and 0 otherwise. The transformation V = U1/2 implies that U = V2. Hence, the cumulative distribution function (CDF) of V, denoted by FV(v), is FV(v) = P(V ≤ v) = P(U1/2 ≤ v) = P(U ≤ v2). The derivative of the CDF FV(v) with respect to v gives us the density function of V, fV(v), which is 2v for v in [0, 1] because f(U) = 1 and by the chain rule, fV(v) = f(U) · |dU/dv| = 1 · |2v| = 2v.