Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$X~U(0,1)$ 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.

share|cite|improve this question
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 $f_y(y)=\frac{\delta}{\delta y}F_y(y)=\frac{1}{2}y^{\frac{-1}{2}}$.

share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.