TITLE: distribution of $X^2 + Y^2$
QUESTION [8 upvotes]: Suppose $X$ and $Y$ are independent uniform distributions between $(0,1)$. What is the distribution of $X^2 + Y^2$?
I derived that the pdf of $X^2$ is $\frac{1}{2\sqrt{x}}$ for $0\leq x \leq 1$. How can I continue from here?

REPLY [4 votes]: We have
$$
f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)=1\quad;\text{ for }0\le x\le1,\ 0\le y\le1
$$
Let $U=X^2+Y^2$ and $V=X^2$, then $Y=\sqrt{U-V}$ and $X=\sqrt{V}$. The Jacobian is $\dfrac{1}{4\sqrt{v(u-v)}}$. The corresponding regions are $\text{R : }0\le\sqrt{v}\le1$ and $0\le\sqrt{u-v}\le1$. The joint pdf of $\ U$ and $V$ is
$$
f_{U,V}(u,v)=f_{X,Y}(x,y)\cdot|J|=\dfrac{1}{4\sqrt{v(u-v)}}
$$
and the marginal pdf of $U$ is
$$
f_U(u)=\int_R f_{U,V}(u,v)\ dv=\left\{ 
  \begin{array}{l l}
    \dfrac14\int_{v=0}^u \dfrac{1}{\sqrt{v(u-v)}}\ dv&\quad;\text{ for }0\le u\le1\\
\\\\\\\\
    \dfrac14\int_{v=u-1}^1 \dfrac{1}{\sqrt{v(u-v)}}\ dv&\quad;\text{ for }1\le u\le2.
  \end{array} \right.
$$
To evaluate the integral, write
$$\dfrac{1}{\sqrt{v(u-v)}}=\frac1u\left(\dfrac{\sqrt{u-v}}{\sqrt{v}}+\dfrac{\sqrt{v}}{\sqrt{u-v}}\right)$$
and you can use this technique.