Let $U$ be $~U [0,1] $and let $Y = U^{\frac{1}{2}}$ Let $U$ be $\sim \mathcal{U}[0,1]$ and let $Y = U^{1/2}$.
I'm having trouble finding the $E(Y)$. How do I go about doing this? 
 A: Hint: (1) Start with computing the distribution function of $Y$, we have for $y \in [0,1]$:
$$ F_Y(y) = \def\P{\mathbb P}\P(Y \le y) = \P(U^{1/2} \le y) = \P(U \le y^2) = \P(U \in [0,y^2])$$
Now use that $U$ is uniformly distributed on $[0,1]$ and hence the probiability of an interval is its lenght.
(2) The density of $U$ is given by $f_Y(y) = F_Y'(y)$.
(3) Now use 
$$ \mathbb E(Y) = \int_0^1 y \, d\P_Y = \int_0^1 yf_Y(y)\, dy $$
A: Answer using CDF transformation method:
$$F_{Y}(y) = P(\sqrt(U) <y)$$
$$F_{Y}(y)=P(U<y^2) = \frac{y^2-0}{1-0} = y^2 ; 0<y<1$$
$$f_{Y}(y) = \frac{dF_{Y}(y)}{dy} = 2y; 0<y<1$$
$$E(Y) = \int_{0}^{1} y\cdot 2y dy$$
$$ = \frac{2y^3}{3}|_{0}^{1}$$
$$ = \frac{2}{3}$$
Thanks
Satish
A: *

*The probability density of a uniform distribution:$$U\sim\mathcal{U}[0,1] \iff f_U(u) = \mathbf{1}_{[0,1]}\!(u) = \begin{cases}1 & u\in [0,1]\\ 0 & \text{elsewhere}\end{cases}$$

*The definition of expectation of a function of a random variable:$$E(g(U)) = \int_{-\infty}^{\infty} g(u) f_U(u)\;\mathrm{d} u$$
So, putting it together: $$\begin{align}\therefore \operatorname{E}(Y) & = \operatorname{E}(\sqrt{U}) \\ & = \int_0^1 \sqrt{u}\;\mathrm{d} u\end{align}$$
