1
$\begingroup$

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 $k>2$.

enter image description here

This may help. Any special case is also useful.

Thanks in advance for any help you are able to provide.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber♦ Aug 24 '20 at 14:21
  • 1
    $\begingroup$ Also asked at math.stackexchange.com/q/3801641/321264. $\endgroup$ – StubbornAtom Aug 24 '20 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.