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$X~U(0,1)$ Find $E[\sqrt {X}]$ and the probability density function of $Y$ defined as $Y=X^2$.

I know that $E[Y]=E[g(X)] = \int_a^b g(x)f(x) dx$ I don't know why this question does not meet the quality standards.

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up vote 2 down vote accepted

$\mathbb{E}(\sqrt{X})=\int_{-\infty}^\infty \sqrt{x} f(x) dx = \int_0^1 \sqrt{x} dx = \frac{2}{3} x^{\frac{3}{2}}|_0^1=\frac{2}{3}$ If $Y=X^2$ then $F_Y(y)=\mathbb{P}\{Y\leq y\}=\mathbb{P}\{X^2\leq y\}=\mathbb{P}\{X\leq \sqrt{y}\}=F_X(\sqrt{y})=\sqrt{y}$. Therefore $f_y(y)=\frac{\delta}{\delta y}F_y(y)=\frac{1}{2}y^{\frac{-1}{2}}$.

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The expression for $F_Y$ and $f_Y$ is for $y\in [0,1]$ only. – Stefan Hansen Jan 16 '13 at 21:34
what do you mean? – mathemagician Jan 16 '13 at 21:41
@mathemagician I think what Stefan means is $f_y(y)=\frac{1}{2}y^{-1/2}\mathbb{I}(0\leq y\leq 1)$. – Patrick Li Jan 16 '13 at 21:52
oh that goes without saying lol – mathemagician Jan 16 '13 at 22:06

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