Question

. Let (X, Y ) be chosen uniformly from inside the circle of radius one centered at the origin. Let R denote the distance of the chosen point from the origin. Determine the density of R. From there, determine the density of the random variable R2 = x2 + y2

Answer 1

Denote by
`D` the unit disk centered at the origin,

`D=\{(x,y)\mid x^2+y^2\le1\}`

which has area
`\pi`. Then
`X,Y` have joint PDF

`f_(X,Y)(x,y)=\begin{cases}\frac1\pi&\text{for }(x,y)\in D\\0&\text{otherwise}\end{cases}`

Transform the pair of random variables
`X,Y` to a new pair
`R,\Theta` such that

`\begin{cases}X=R\cos\Theta\\Y=R\sin\Theta\end{cases}`

In this new coordinate system, the support is

`D=\{(r,\theta)\mid0\le r\le1,0\le\theta\le2\pi\}`

Using the method of transformations, the joint PDF of
`R,\Theta` is

`f_(R,\Theta)(r,\theta)=f_(X,Y)(r\cos\theta,r\sin\theta)|\det J|`

where
`J` is the Jacobian matrix for the transformation. We have

`J=\begin{bmatrix}(\partial x)/(\partial r)&(\partial x)/(\partial\theta)\\\\(\partial y)/(\partial r)&(\partial y)/(\partial\theta)\end{bmatrix}=\begin{bmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{bmatrix}\implies|\det J|=|r\cos^2\theta+r\sin^2\theta|=r`

Then

`f_(R,\Theta)(r,\theta)=\begin{cases}\frac r\pi&\text{for }0\le r\le1,0\le\theta\le2\pi\\0&\text{otherwise}\end{cases}`

Since the joint density doesn't depend on
`\theta` (it varies uniformly over the interval
`[0,2\pi]` with probability
`\frac1{2\pi}`), it follows that
`R` and
`\Theta` are independent of one another and

`f_R(r)=\begin{cases}2r&\text{for }0\le r\le1\\0&\text{otherwise}\end{cases}`

Then letting
`S=R^2`, by the method of transformations we have

`f_S(s)=f_R(\sqrt s)\left|(\mathrm dr)/(\mathrm ds)\right|=\begin{cases}1&\text{for }0\le s\le1\\0&\text{otherwise}\end{cases}`