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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:

$$ \frac{2}{\pi}\frac{1}{\sqrt{1-y^2}} $$

  1. Is the work & logic correct?
  2. Is the support of $Y$ the same as the support for U: $[0,pi/2]$?
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    $\begingroup$ 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 $y = \sin \theta$. $\endgroup$ – Aditya Dua Dec 8 '18 at 22:21
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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.

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