Probability of choosing a point $(X,Y)$ on $\mathbb{R}^2$

Question

Picking a point $(X,Y)$ of $\mathbb{R}^2$.

We first randomly generate a radius $R$ by a continuous distribution and density function $f_R$. Then, choose a point $(X,Y)$ uniformly at random inside the disc centred at $(0,0)$ with radius $R$.

Find a density function $f_R$, so that $(X,Y)$ has standard gaussian distribution of $\mathbb{R}^2$.

I am not sure where to start? My only thoughts are to consider the form $(X,Y)$ should be in, in order for it to have a standard gaussian distribution and work backwards?

Answer 1

Conditionally on $R$, the density of $(X,Y)$ is $$ f_{X,Y\mid R}(x,y\mid r)=\frac{1}{\pi r^2}1\!\left\{x^2+y^2\le r^2\right\}. $$ Thus, the unconditional density is $$ f_{X,Y}(x,y)=\int_{\sqrt{x^2+y^2}}^\infty \frac{1}{\pi r^2}f_R(r)\, dr. $$ It is clear that in order to get the standard Gaussian density, $f_R(r)$ should be proportional to $r^3\cdot \exp(-r^2/2)$. You may check that $R\sim \chi_4$ in this case.