# Functions of Random Variables

1. Oct 9, 2007

### cepheid

Staff Emeritus
1. The problem statement, all variables and given/known data

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 fZ(z) of Z = arctan(x).

3. 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?

Last edited: Oct 9, 2007
2. Oct 10, 2007

### EnumaElish

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