Finding the PDF of Z = arctan(x) from a Gaussian Distribution

[cepheid]

Homework Statement

If X is represented by the Gaussian distribution, that is,

$$f_{X}(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{x^2}{2\sigma^2})}$$
​

find an expression for the pdf f_Z(z) of Z = arctan(x).

The Attempt at a Solution

If Z =g(X), then g(X) is multivalued unless the range of the function is restricted to $Z \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Under this condition, the function is one to one, and so the probability that z will be in some interval (z, z + dz) is equal to the probability that x will be in the corresponding interval (x, x + dx). In other words,

$$|f_Z(z) dz| = |f_X(x) dx|$$

$$f_Z(z) = \left| \frac{dx}{dz} \right| f_X(x)$$

$$= \frac{d}{dz} (\tan z) f_X(x)$$

$$= (\sec^2(z)) f_X(x)$$

$$= (1+\tan^2(z)) \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{\tan^2(z)}{2\sigma^2})}$$​

Am I doing this right?

 

[EnumaElish]

I don't see a problem; but you'll need to verify that it integrates to one between the bounds.