We have 2 independent random variables $a \sim N(0,1)$ and $b \sim N(0,1)$. I want a proof (or even good intuition) to show that the chance of having the pair $(a_0,b_0)$ is a function of only the distance from $(0,0)$ , [if it actually the case, as I guess it is.] e.g: the chance of having $(1,1)$ is the same as the chance of having $(\sqrt{2}, 0)$, because the distance from $(0,0)$ is the same.
I guess convolution might help, but I'm not sure if and how we can apply some kind of convolution here (I'm not so familiar with convolutions).
here is density representation of a scatter plot, seem to suggest that indeed the shape is circle(i.e the geometrical distance is the parameter) and not rhombus as one might think:
eventually I think I have managed to solve that.
The key issue is to calculate the density function $f(x,y)$ of the 2 variables. which because the variables are independent we have: $$f(x,y) = \dfrac{e^{-x^2/2}}{\sqrt{2\pi}} * \dfrac{e^{-y^2/2}}{\sqrt{2\pi}} = \dfrac{e^{-(x^2+y^2)/2}}{2\pi}=g(x^2+y^2)$$