distribution of $X^2 + Y^2$ 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?
 A: Set $Z=X^2+Y^2$. Since $X$ and $Y$ are continuous and independent random variables,
$$
f_Z(z)=\int_{-\infty}^\infty f_{X^2}(x)f_{Y^2}(z-x)\mathrm dx.
$$
Observe that $f_{X^2}(x)f_{Y^2}(z-x)=0$ if $x\not\in[\max\{0,z-1\},\min\{1,z\}]$. So we have that
$$
f_Z(z)=
\begin{cases}
\int_0^zf_{X^2}(x)f_{Y^2}(z-x)\mathrm dx&\text{if }0\le z\le1,\\
\int_{z-1}^1f_{X^2}(x)f_{Y^2}(z-x)\mathrm dx&\text{if }1\le z\le2.
\end{cases}
$$
Now
$$
\frac14\int_0^z\frac1{\sqrt{x(z-x)}}\mathrm dx=\frac\pi4
$$
and
$$
\frac14\int_{z-1}^1\frac1{\sqrt{x(z-x)}}\mathrm dx=\frac12\arctan\biggl(\frac1{\sqrt{z-1}}\biggr)-\frac12\arctan(\sqrt{z-1})
$$
since
$$
\biggl(2\arctan\biggl(\frac{\sqrt x}{\sqrt{z-x}}\biggr)\biggr)'=\frac{1}{\sqrt{x(z-x)}}
$$
(I used Integral Calculator to find this antiderivative).
A: You can view this one geometrically: For each $r > 0$, what is the area of the set $\{(x,y) \in [0,1] \times [0,1]: x^2 + y^2 > r\}$?
A: 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.
