I don't understand how can I solve this. My only guess it's that it's related with the probability of the circle area of c.
The coordinates X and Y of a point are independent zero mean normal
random variables with common variance $\sigma^2$ (Given that the point is at a distance of at least c from the origin. Find the conditional joint PDF of X and Y).
Best Answer
You have $$f_X(x)={1\over\sigma\sqrt{2\pi}}\exp{-x^2\over 2\sigma^2}$$ and similarly for $f_Y(y)$. Since $X$ and $Y$ are assumed independent you then can write $$f_{(X,Y)}(x,y)=f_X(x)f_Y(y)={1\over2\pi\sigma^2}\exp{-(x^2+y^2)\over 2\sigma^2}\ .$$If outcomes $(x,y)$ with $x^2+y^2<c^2$ are discarded you need the probability $$P_{\geq c}:=\int_{x^2+y^2\geq c^2}f_{(X,Y)}(x,y)\>{\rm d}(x,y)={1\over2\pi\sigma^2}\int_c^\infty \exp{-r^2\over 2\sigma^2}\>2\pi r\>dr\ .$$ The conditional density is then given by $$f(x,y)=\cases{0\qquad&$(x^2+y^2<c^2),$ \cr {\displaystyle{1\over P_{\geq c}}f_{(X,Y)}(x,y)}\qquad &$(x^2+y^2\geq c^2).$ \cr}$$