This is almost the same as Suppose that X ∼ U ( $− π/2$ , $π/2$ ) . Find the pdf of Y = tan(X)., but making sure I am understanding the process:
Let $U \sim \textrm{Unif}(0, \pi/ 2)$. Find the PDF of $\sin(U)$.
$$\begin{align} F_Y(y) &= P(Y<y) = P(\sin(u)<y) = P(u < \sin^{-1}(y)) \\ &=F_U(\sin^{-1}(y)) = \int_0^{\sin^{-1}(y)}\frac{1}{\frac{\pi}{2}}du \\ &= \frac{1}{\frac{\pi}{2}}u |_0^{\sin^{-1}(y)} \\ &= \frac{2}{\pi}\sin^{-1}(y) \\ \end{align} $$
That is the CDF. To find the PDF, take the derivative with respect to $y$ to get:
$$ \frac{2}{\pi}\frac{1}{\sqrt{1-y^2}} $$
- Is the work & logic correct?
- Is the support of $Y$ the same as the support for U: $[0,pi/2]$?
Per Aditya Dua: The support for Y is [0,1]. Always good to confirm that the PDF integrates to 1 over [0,1]. In this case it does. You can compute the integral by substituting y=sinθ.