# Find $\mathbb P(\sqrt{V} \cos(\pi U)\leq c)$, $\mathbb P(\sqrt{V} \sin(\pi U)\leq c)$

Find $$IC=\mathbb P(\sqrt{V} \cos(\pi U)\leq c),$$ $$IS=\mathbb P(\sqrt{V} \sin(\pi U)\leq c),$$ and $$ICS=\mathbb P(\sqrt{V} \cos(\pi U)\leq c_1, \sqrt{V} \sin(\pi U)\leq c_2),$$ where $$V\sim \chi^2_{k}$$ and $$U\sim Beta(a,b)$$, that is, $$f_V(v)=\frac{1}{\Gamma(\alpha) 2^{\alpha/2}} v^{\alpha/2-1} e^{-\frac{v}{2}}1_{v>0}$$ and $$f_U(u)=\frac{1}{Beta(a,b)}u^{a-1}(1-u)^{b-1}1_{(0,1)}(u)$$.

For a special case $$a=b=1$$, $$\alpha=2$$ , the distribution of $$\sqrt{V} \cos (\pi U)$$ is standard normal, the distribution of $$\sqrt{V} \sin (\pi U)$$ is standard normal too. The Box-Mueller transformation is special case of this transformation(for $$k=2$$ and $$a=b=1$$). The random variable $$Z=\sqrt{V} \cos(\pi U)$$, has a good property. By a simple simulation, it can be symmetric for $$a=b$$, and otherwise is asymmetric. It is also be bimodal for $$k>2$$.

This may help. Any special case is also useful.

• Although you can readily simplify these two questions to a single question about the distribution of $V\cos(\pi U/2),$ at that point the problem looks analytically intractable except when $a=b=1,$ where the solution is simple to obtain. – whuber Aug 24 '20 at 14:21