1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Functions of Random Variables

  1. Oct 9, 2007 #1


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

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

    [tex] f_{X}(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{x^2}{2\sigma^2})} [/tex]

    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 [itex] Z \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) [/itex]

    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,

    [tex] |f_Z(z) dz| = |f_X(x) dx| [/tex]

    [tex] f_Z(z) = \left| \frac{dx}{dz} \right| f_X(x) [/tex]

    [tex] = \frac{d}{dz} (\tan z) f_X(x) [/tex]

    [tex] = (\sec^2(z)) f_X(x) [/tex]

    [tex] = (1+\tan^2(z)) \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{\tan^2(z)}{2\sigma^2})} [/tex] ​

    Am I doing this right?
    Last edited: Oct 9, 2007
  2. jcsd
  3. Oct 10, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    I don't see a problem; but you'll need to verify that it integrates to one between the bounds.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Functions of Random Variables