If $X$ and $Y$ are uniform$(-1,1)$, how can I find the distribution of $W=X^2+Y^2$?

Question

If $Y$ and $X$ are independent uniform (-1,1) random variables, I would like to derive the distribution of $W=X^2+Y^2$.

At first I thought that I could use the CDF technique and a geometric argument, that is divide the area of a circle with radius $\sqrt{w}$, $0<w<2$ by $4$, the area of the $2$ by $2$ square that constitutes the support of $X$ and $Y$.

The problem with that approach is that the resulting pdf:

$$f_W (w)=\begin{cases} \pi /4,& 0<w<2 \\ 0, & \text{elsewhere} \end{cases} $$

does not integrate to $1$.

Could you please help me understand where I went wrong here and how to arrive at the right answer?

Thank you in advance.

Answer 1

Your geometric approach will work. Draw a picture. If $0\lt w\le 1$, we get a circle that if fully inside the square on which the joint density "lives," and as you saw things are straightforward.

If $1\lt w\lt 2$, we need to find the area of the part $D$ of the square that is in the disk of radius $\sqrt{w}$. Join the centre of the circle to the $8$ points where the circle meets the sides of the square. This divides our region $D$ into $8$ parts, $4$ isosceles triangles and $4$ circular sectors.

We find the combined area of these and divide by $4$, or equivalently find the area of a triangle and a circular sector, and add.

To find the area of a triangle, let us for example find the two places where $x^2+y^2=w$ meets the line $x=1$. This is at $y=\pm \sqrt{w-1}$. So the base of the triangle is $2\sqrt{w-1}$, and therefore the area is $\sqrt{w-1}$.

The central angle of each triangle is $2\arctan(\sqrt{w-1})$, and therefore the angle of each of the $4$ circular sectors is $\frac{\pi}{2}-\arctan(\sqrt{w-1})$. For the area, divide by $2$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align}&\color{#66f}{\large% \int_{-1}^{1}\int_{-1}^{1}\delta\pars{w - x^{2} - y^{2}}\,{\dd x \over 2}\, {\dd y \over 2}} \\[3mm]&={1 \over 4}\,2\int_{0}^{1}2\int_{0}^{1}\bracks{% {\delta\pars{y + \root{w - x^{2}}} \over 2\verts{y}} +{\delta\pars{y - \root{w - x^{2}}} \over 2\verts{y}}}\,\dd y\,\dd x \\[3mm]&=\left. \half\int_{0}^{1}{\dd x \over \root{w - x^{2}}}\, \right\vert_{w\ >\ x^{2}\,,\root{\vphantom{\Large A}w - x^{2}}\ <\ 1} =\left. \half\int_{0}^{1}{\dd x \over \root{w - x^{2}}}\, \right\vert_{w - 1\ <\ x^{2}\ <\ w} \\[3mm]&=\left\lbrace\begin{array}{lcl} \half\int_{0}^{\root{w}}{\dd x \over \root{w - x^{2}}} & \mbox{if} & 0 < w < 1 \\[2mm] \half\int_{\root{w - 1}}^{1}{\dd x \over \root{w - x^{2}}} & \mbox{if} & 1 \leq w < 2 \\[2mm] 0 && \mbox{otherwise} \end{array}\right. \\[3mm]&=\color{#66f}{\large\left\lbrace\begin{array}{lcl} {\pi \over 4} & \color{#000}{\mbox{if}} & 0 < w < 1 \\[2mm] \half\,\arcsin\pars{{2 \over w} - 1} & \color{#000}{\mbox{if}} & 1 \leq w < 2 \\[2mm] 0 && \color{#000}{\mbox{otherwise}} \end{array}\right.} \end{align}