# Suppose that X ∼ U ( $− π/2$ , $π/2$ ) . Find the pdf of Y = tan(X).

Suppose that X ∼ U $(−π/2,π/2$) . Find the pdf of Y = tan(X). Make sure to define the support of the density function.

My work so far: $F_Y(y) = P(Y < y) = P (tan(x) < y) = P (x < tan^{-1}(y)) = F_X(tan^{-1}(y)) = \int_{-\pi/2}^{tan^{-1}(y)} 1/\pi = tan^{-1}(y)/\pi + 1/2 -> derivative -> 1/(\pi + \pi y^{2})$

This is the cauchy distribution. Thanks for the help in the comments!

• Double check the solution to your integral (keep in mind what to integrate with respect to). Also, you computed the CDF. Try taking a derivative and then I think you'll recognize the pdf (after you fix the solution to your integral) – Brenton Mar 15 '16 at 1:46
• Did you mean $\mathcal U(-\pi^2, \pi^2)$ or $\mathcal U(-\pi/2, \pi/2)$ ? – Graham Kemp Mar 15 '16 at 2:04
• Oops, I messed up in the title there. Editting it into the question – Sam N. Mar 15 '16 at 2:12

So you have:

$$F_X(tan^{-1}(y)) = \int_{-\pi/2}^{tan^{-1}(y)} \frac{1}{\pi} dx = \frac{1}{\pi}\left(tan^{-1}(y) + \frac{\pi}{2}\right) = \frac{1}{\pi}tan^{-1}(y) + \frac{1}{2}$$.

Take a derivative with respect to $y$ and you'll end up with a very familiar distribution