I need help with the following exercise:
A point is uniformly taken from the intervall $(0,1)$, let $X$ denote the lenght of the shorter one of the two resulting intervalls and $Y$ the lenght of the other.
Calculate $\Bbb E(X^2)$, $\Bbb E(Y^2)$, $\Bbb E(\frac{X^2}{Y^2}).$
Now again a point $X$ is taken from $(0,1)$ with uniform distribution, then we take a point $Y$ uniformly from $(0,X)$
What is the joint density function of $X$ and $Y$.
Thank you in advance !
Given $U$, uniformly distributed over $[0,1]$, we have $$ X=\min(U,1-U),\qquad Y=\max(U,1-U) =1-X$$ and $$\mathbb{E}[X^2] = \int_{0}^{1}\min(x,1-x)^2\,dx = 2\int_{0}^{1/2}x^2\,dx=\frac{1}{12},$$ $$\mathbb{E}[Y^2] = \int_{0}^{1}\max(x,1-x)^2\,dx = 2\int_{0}^{1/2}(1-x)^2\,dx=\frac{7}{12},$$ $$\mathbb{E}\left[\frac{X^2}{Y^2}\right] = \int_{0}^{1/2}\frac{x^2}{(1-x)^2}\,dx + \int_{1/2}^{1}\frac{(1-x)^2}{x^2}\,dx = 3-4\log 2.$$ At last, if $X=U$ and $Y=X V$ (with $V$ uniformly distributed over $[0,1]$ and independent from $U$), the joint density $f(x,y)$ is supported on the triangle $(0,1)^2\cap\{y\leq x\}$ and given by $f(x,y)=\frac{1}{x}$.