Distribution of $R^2 = X^2 +Y^2$ where (X,Y) is a point on the unit circle

Question

So I have a point $(X,Y)$ chosen from the unit disk with uniform distribution. And I'm attempting to find the distribution of $R^2$, where $R$ is the distance from the point to the origin.

Now obviously the joint distribution:

$f_{X,Y}(x,y)=\frac{1}{\pi}, \forall (x,y)\in C$

where $C=\{(x,y)\in\mathbb{R}:x^2+y^2\le1\}$

Now, I attempted to take the marginal distributions, square them, and then add them back, which gave the following:

$F_X(x)=F_Y(x)=\frac{2}\pi\sqrt{1-x^2}, \forall x\in [-1,1]$

Applying the transformation, $g:[-1,1]\to\mathbb{R}, x\mapsto x^2$

\begin{align} f_{X^2}(x) & =f_{Y^2}(x)=f_X(\sqrt{x}) \left|\frac{1}{2}x^{-1/2}\right| + \space f_X(-\sqrt{x}) \left| -\frac{1}{2}x^{-1/2}\right| \\[10pt] & =\frac{2}\pi\sqrt{\frac{1-x^2}{x}}, \forall x\in[0,1] \end{align}

But then I realised I couldn't just add them as $X$ and $Y$ are dependent and as such so would $X^2$ and $Y^2$.

So I assume I have to apply the transformation, $h:[-1,1]\times[-1,1] \to \mathbb{R}^2, (x,y)\mapsto(x^2,y^2)$
Which would entail me finding the Jacobian and inverse, etc. which I am fully able to do.

However, $h$ is clearly not 1:1, and although I could get around that in the single variable case, I'm not sure what the anologue is for multy variables.

So if anyone has any ideas on where to go from here, any assistance would be appreciated.

Answer 1

Since $(X,Y)$ is uniformly distributed on the unit disk, for any reasonable subset of the unit disk we have $$ \mathbb{P}((X,Y)\in A)=\frac{|A|}{\pi} $$ where $|A|$ is the area of $A$.

Now consider the random variable $R=\sqrt{X^2+Y^2}$. For $0\leq r\leq 1$, the event $\{R\leq r\}$ is the same as the event that the point $(X,Y)$ lies in the disk centered at the origin of radius $r$. Therefore $$ \mathbb{P}(R\leq r)=\frac{\pi r^2}{\pi}=r^2 $$ for $0\leq r\leq 1$.

This can also be seen using the joint PDF you found and switching to polar coordinates: $$ \mathbb{P}(R\leq r)=\frac{1}{\pi}\int_{x^2+y^2\leq r^2}\;dxdy=\frac{1}{\pi}\int_0^{2\pi}\int_0^r\rho\;d\rho d\theta=\int_0^r2\rho\;d\rho=r^2$$

Finally, $\mathbb{P}(R^2\leq r)=\mathbb{P}(R\leq \sqrt{r})=r$ for $0\leq r\leq 1$.

Answer 2

For $0\le r\le 1$, we have $\Pr(R\le r)=\frac{\pi r^2}{\pi}$.

Let $W=R^2$. Then $$F_W(w)=\Pr(R^2\le w)=\Pr(0\le R\le \sqrt{w})=w.$$ So $W$ is uniformly distributed on $[0,1]$.

Answer 3

Hint: if the points are uniformly distributed on the unit disc then density measure for all points on a circle of radius ($r$) will be proportional to the circumference of that circle.

So $f_R(r) = cr\,\mathbf 1_{r\in[0;1]}$ and $\int_0^1 cr\operatorname d r=1$ and hence $c=2$

Hence $f_R(r)=2r \,\mathbf 1_{r\in[0;1]}$

Use a simple change of variables transformation to find $f_{R^2}$

$$f_{R^2}(\rho) = f_R(\surd\rho)\cdot\left\lvert\dfrac{\operatorname d \surd\rho}{\operatorname d \rho}\right\rvert$$