91Ó°ÊÓ

We have given the function

$f(x,y)=\frac{1}{\mathrm{π}}0≤x^2+y^2≤1$.

Considering the transformation $g:c→(0,1)×(0,2\mathrm{π}),$where $C$is a unit circle. Transformation $g$is defined by

$g(x,y)=(r^2,\mathrm{θ})=\left(x^2+y^2,{\tan }^{-1}\left(\frac{y}{x}\right)\right)$

We are interested in the distribution of random vector $(R^2,\mathrm{θ})=g(X,Y).$Using the theorem about the density of transformation of a random vector, we have that

$f{R}^2,\mathrm{θ}(r^2,\mathrm{θ})=fXY,\left(g^{-1}\right(r^2,\mathrm{θ}\left)\right)·\left|det∇g^{-1}(r^2,\mathrm{θ})\right|$

We have $fX,Y\left(g^{-1}\right(r^2,\mathrm{θ}\left)\right)=\frac{1}{\mathrm{π}}\mathrm{χ}{g}^{-1}(r^2,\mathrm{θ})∈C=\frac{1}{\mathrm{π}}\mathrm{χ}(r^2,\mathrm{θ})∈(0,1)×(0,2\mathrm{π})$and

$∇g(x,y)=\left[-\begin{array}{cc}2x& 2y\\ \frac{y}{x^2+y^2}& \frac{x}{x^2+y^2}\end{array}\right]$

which implies

$\left|det∇g(x,y)\right|=2⇒\left|det∇g^{-1}(r^2,\mathrm{θ})\right|=\frac{1}{2}$

Hence, we have obtained

$f{R}^2,\mathrm{θ}(r^2,\mathrm{θ})=\frac{1}{2\mathrm{π}}·\mathrm{χ}(r^2,\mathrm{θ})∈(0,1)×(0,2\mathrm{π})$

$fX,Y\left(g^{\_1}\right(r^2,\mathrm{θ}\left)\right)=\frac{1}{\mathrm{π}}\mathrm{χ}{g}^{-1}(r^2,\mathrm{θ})∈C=\frac{1}{\mathrm{π}}\mathrm{χ}(r^2,\mathrm{θ})∈(0,1)×(0,2\mathrm{π})$

We have seen that $R^2$and $\mathrm{θ}$are independent since their joint distribution can be factorized as

$f{R}^2,\mathrm{θ}(r^2,\mathrm{θ})=\mathrm{χ}{R}^2∈(0,1)·\frac{1}{2\mathrm{π}}\mathrm{χ}\mathrm{θ}∈(0,2\mathrm{π})$

and we also have that

$R^2~Unif(0,1),\mathrm{θ}~Unif(0,2\mathrm{π})$