# Distribution of $\sqrt{X^2+Y^2}$ when $X,Y$ are independent $U(0,1)$ variables

As a routine exercise, I am trying to find the distribution of $$\sqrt{X^2+Y^2}$$ where $$X$$ and $$Y$$ are independent $$U(0,1)$$ random variables.

The joint density of $$(X,Y)$$ is $$f_{X,Y}(x,y)=\mathbf 1_{0

Transforming to polar coordinates $$(X,Y)\to(Z,\Theta)$$ such that $$X=Z\cos\Theta\qquad\text{ and }\qquad Y=Z\sin\Theta$$

So, $$z=\sqrt{x^2+y^2}$$ and $$0< x,y<1\implies0< z<\sqrt 2$$.

When $$0< z<1$$, we have $$0< \cos\theta<1,\,0<\sin\theta<1$$ so that $$0<\theta<\frac{\pi}{2}$$.

When $$1< z<\sqrt 2$$, we have $$z\cos\theta<\implies\theta>\cos^{-1}\left(\frac{1}{z}\right)$$, as $$\cos\theta$$ is decreasing on $$\theta\in\left[0,\frac{\pi}{2}\right]$$; and $$z\sin\theta<1\implies\theta<\sin^{-1}\left(\frac{1}{z}\right)$$, as $$\sin\theta$$ is increasing on $$\theta\in\left[0,\frac{\pi}{2}\right]$$.

So, for $$1< z<\sqrt 2$$, we have $$\cos^{-1}\left(\frac{1}{z}\right)<\theta<\sin^{-1}\left(\frac{1}{z}\right)$$.

The absolute value of jacobian of transformation is $$|J|=z$$

Thus the joint density of $$(Z,\Theta)$$ is given by

$$f_{Z,\Theta}(z,\theta)=z\mathbf 1_{\{z\in(0,1),\,\theta\in\left(0,\pi/2\right)\}\bigcup\{z\in(1,\sqrt2),\,\theta\in\left(\cos^{-1}\left(1/z\right),\sin^{-1}\left(1/z\right)\right)\}}$$

Integrating out $$\theta$$, we obtain the pdf of $$Z$$ as

$$f_Z(z)=\frac{\pi z}{2}\mathbf 1_{0

Is my reasoning above correct? In any case, I would like to avoid this method and instead try to find the cdf of $$Z$$ directly. But I couldn't find the desired areas while evaluating $$\mathrm{Pr}(Y\le \sqrt{z^2-X^2})$$ geometrically.

EDIT.

I tried finding the distribution function of $$Z$$ as

\begin{align} F_Z(z)&=\Pr(Z\le z) \\&=\Pr(X^2+Y^2\le z^2) \\&=\iint_{x^2+y^2\le z^2}\mathbf1_{0

Mathematica says this should reduce to

$$F_Z(z)=\begin{cases}0 &,\text{ if }z<0\\ \frac{\pi z^2}{4} &,\text{ if } 0< z<1\\ \sqrt{z^2-1}+\frac{z^2}{2}\left(\sin^{-1}\left(\frac{1}{z}\right)-\sin^{-1}\left(\frac{\sqrt{z^2-1}}{z}\right)\right) &,\text{ if }1< z<\sqrt 2\\ 1 &,\text{ if }z>\sqrt 2 \end{cases}$$

which looks like the correct expression. Differentiating $$F_Z$$ for the case $$1< z<\sqrt 2$$ though brings up an expression which doesn't readily simplify to the pdf I already obtained.

Finally, I think I have the correct pictures for the CDF:

For $$0 :

And for $$1 : Shaded portions are supposed to indicate the area of the region $$\left\{(x,y):0

The picture immediately yields

\begin{align} F_Z(z)&=\Pr\left(-\sqrt{z^2-X^2}\le Y\le\sqrt{z^2-X^2}\right) \\&=\begin{cases}\frac{\pi z^2}{4} &,\text{ if } 0

, as I had previously found.

• To find the CDF directly, use indicator functions. For $z\ge 0,$ $$\Pr(\sqrt{X^2+Y^2}\le z)=\int_0^1\int_0^1\mathcal{I}(x^2+y^2\le z^2)\mathrm{d}x\mathrm{d}y.$$ The rest is purely algebraic manipulation. (Edit: I see @Xi'an just posted the algebra in his answer.)
– whuber
Jan 17, 2018 at 21:13
• Re the edit: I also obtain several different expressions and (using FullSimplify) they simplify to different formulas in Mathematica. However, they are equivalent. This is easily shown by plotting their difference. Apparently Mathematica doesn't know that $\tan ^{-1}\left(\sqrt{z^2-1}\right)=\sec ^{-1}(z)$ when $1\lt z \lt \sqrt{2}$.
– whuber
Jan 18, 2018 at 15:55
• The edge of the surface, $\sqrt{r^2-x^2}$, in your last picture should be a (semi-)circle with center (0,0). Thus concave instead of (your currently drawn) convex. Jan 21, 2018 at 14:35

That the pdf is correct can be checked by a simple simulation

samps=sqrt(runif(1e5)^2+runif(1e5)^2)
hist(samps,prob=TRUE,nclass=143,col="wheat")
df=function(x){pi*x/2-2*x*(x>1)*acos(1/(x+(1-x)*(x<1)))} Finding the cdf without the polar change of variables goes through \begin{align*} \mathrm{Pr}(\sqrt{X^2+Y^2}\le z) &= \mathrm{Pr}(X^2+Y^2\le z^2)\\ &= \mathrm{Pr}(Y^2\le z^2-X^2)\\ &=\mathrm{Pr}(Y\le \sqrt{z^2-X^2}\,,X\le z)\\ &=\mathbb{E}^X[\sqrt{z^2-X^2}\mathbb{I}_{[0,\min(1,z)]}(X)]\\ &=\int_0^{\min(1,z)} \sqrt{z^2-x^2}\,\text{d}x\\ &=z^2\int_0^{\min(1,z^{-1})} \sqrt{1-y^2}\,\text{d}y\qquad [x=yz\,,\ \text{d}x=z\text{d}y]\\ &=z^2\int_0^{\min(\pi/2,\cos^{-1} z^{-1})} \sin^2{\theta} \,\text{d}\theta\qquad [y=\cos(\theta)\,,\ \text{d}y=\sin(\theta)\text{d}\theta]\\ &=\frac{z^2}{2}\left[ \min(\pi/2,\cos^{-1} z^{-1}) - \sin\{\min(\pi/2,\cos^{-1} z^{-1})\}\cos\{\min(\pi/2,\cos^{-1} z^{-1}\}\right]\\ &=\frac{z^2}{2}\begin{cases} \pi/2 &\text{ if }z<1\\ \cos^{-1} z^{-1}-\sin\{\cos^{-1} z^{-1})\}z^{-1}&\text{ if }z\ge 1\\ \end{cases}\\ &=\frac{z^2}{2}\begin{cases} \pi/2 &\text{ if }z<1\\ \cos^{-1} z^{-1}-\sqrt{1-z^{-2}}z^{-1}&\text{ if }z\ge 1\\ \end{cases} \end{align*} which ends up with the same complexity! (Plus potential mistakes of mine along the way!)

• The case $1\le z<\sqrt 2$ is where it gets a bit fuzzy. I guess I don't end up with the correct pdf differentiating the expression for $z\ge 1$. Jan 18, 2018 at 12:40

## $f_z(z)$ :

So, for $1\le z<\sqrt 2$, we have $\cos^{-1}\left(\frac{1}{z}\right)\le\theta\le\sin^{-1}\left(\frac{1}{z}\right)$

You can simplify your expressions when you use symmetry and evaluate the expressions for $\theta_{min} < \theta < \frac{\pi}{4}$. Thus, for half of the space and then double the result.

Then you get:

$$P(Z \leq r) = 2 \int_0^r z \left(\int_{\theta_{min}}^{\frac{\pi}{4}}d\theta\right) dz = \int_0^r z \left(\frac{\pi}{2}-2\theta_{min}\right) dz$$

and your $f_z(z)$ is

$$f_z(z) = z \left(\frac{\pi}{2}-2\theta_{min}\right) = \begin{cases} z\left(\frac{\pi}{2}\right) & \text{ if } 0 \leq z \leq 1 \\ z \left(\frac{\pi}{2} - 2 \cos^{-1}\left(\frac{1}{z}\right)\right) & \text{ if } 1 < z \leq \sqrt{2} \end{cases}$$

## $F_z(z)$ :

You can use the indefinite integral:

$$\int z \cos^{-1}\left(\frac{1}{z}\right) = \frac{1}{2} z \left( z \cos^{-1}\left(\frac{1}{z}\right) - \sqrt{1-\frac{1}{z^2}} \right) + C$$

note $\frac{d}{du} \cos^{-1}(u) = - (1-u^2)^{-0.5}$

This leads straightforward to something similar as Xi'ans expression for $Pr(Z \leq z)$ namely

if $1 \leq z \leq \sqrt{2}$ then:

$$F_z(z) = {z^2} \left(\frac{\pi}{4}-\cos^{-1}\left(\frac{1}{z}\right) + z^{-1}\sqrt{1-\frac{1}{z^2}} \right)$$

The relation with your expression is seen when we split up the $cos^{-1}$ into two $cos^{-1}$ expressions, and then converted to different $sin^{-1}$ expressions.

for $z>1$ we have

$$\cos^{-1}\left(\frac{1}{z}\right) = \sin^{-1}\left(\sqrt{1-\frac{1}{z^2}}\right) = \sin^{-1}\left(\frac{\sqrt{z^2-1}}{z}\right)$$

and

$$\cos^{-1}\left(\frac{1}{z}\right) = \frac{\pi}{2} -\sin^{-1}\left(\frac{1}{z}\right)$$

so

$$\begin{array}\\ \cos^{-1}\left(\frac{1}{z}\right) & = 0.5 \cos^{-1}\left(\frac{1}{z}\right) + 0.5 \cos^{-1}\left(\frac{1}{z}\right) \\ & = \frac{\pi}{4} - 0.5 \sin^{-1}\left(\frac{1}{z}\right) + 0.5 \sin^{-1}\left(\frac{\sqrt{z^2-1}}{z}\right) \end{array}$$

which results in your expression when you plug this into the before mentioned $F_z(z)$ for $1<z<\sqrt{2}$

For $$0 \leq z \leq 1$$, $$P\left(\sqrt{X^2+Y^2} \leq z\right)$$ is just the area of the quarter-circle of radius $$z$$ which is $$\frac 14 \pi z^2$$. That is, $$\text{For }0 \leq z \leq 1, ~\text{area of quarter-circle} = \frac{\pi z^2}{4} = P\left(\sqrt{X^2+Y^2} \leq z\right).$$

For $$1 < z \leq \sqrt{2}$$, the region over which we need to integrate to find $$P\left(\sqrt{X^2+Y^2} \leq z\right)$$can be divided into two right triangles $$\big($$one of them has vertices $$(0,0), (0,1)$$ and $$(\sqrt{z^2-1}, 1)$$ while the other has vertices $$(0,0), (1,0)$$ and $$(1, \sqrt{z^2-1})$$ $$\big)$$ together with a sector of a circle of radius $$z$$ and included angle $$\frac{\pi}{2}-2\arccos\left(\frac{1}{z}\right)$$. The area of this region (and hence the value of $$\left( P(\sqrt{X^2+Y^2} \leq z\right)$$) is easily found. We have that for $$1 < z \leq \sqrt{2}$$, \begin{align}\text{area of region} &= \text{area of two triangles plus area of sector}\\ &=\sqrt{z^2-1} + \frac 12 z^2\left( \frac{\pi}{2}-2\arccos \left(\frac{1}{z}\right)\right)\\ &= \frac{\pi z^2}{4} + \sqrt{z^2-1} - z^2\arccos \frac{1}{z}\\ &= \left( P(\sqrt{X^2+Y^2} \leq z\right)\end{align} which is the result in Martijn Wetering's answer.