I want to find the probability density function of $Y=\cos(X)$, where $X\sim N(\mu, \sigma^2)$. The answer is known when $X$ is uniformly distributed $U(-\pi, \pi)$ and it is an arcsin pdf, given by, $f_X(x) = {1\over{\pi\sqrt{(1-x^2)}}}$. But when $X$ is Gaussian, there are infinite roots of $Y=\cos(X)$ and the solution looks messy. Any help would be appreciated.

  • $\begingroup$ What do you mean by an "arccosine pdf"? $\endgroup$ – Avraham Jun 18 '14 at 3:36
  • $\begingroup$ It isn't going to have a nice form. Was the problem you were given to find the expected value of cos(X)? If so, thats a significantly easier problem -- look at the real part of the characteristic function evaluated at $1$. $\endgroup$ – Batman Jun 18 '14 at 4:08
  • $\begingroup$ Avraham: I have edited the question above with the details on the arcsin (not arccos - sorry) distribution. $\endgroup$ – ashaw2 Jun 18 '14 at 5:08
  • $\begingroup$ Batman: I have found one "messy" solution here, mathoverflow.net/questions/35260/… But I was told this becomes Wrapped Normal PDF which can be approximated by a uniform, which I don't see how. $\endgroup$ – ashaw2 Jun 18 '14 at 5:10
  • $\begingroup$ Well, see here en.wikipedia.org/wiki/Wrapped_normal_distribution , but this is not your distribution. You distribution has a normal wrapped onto a circle, then marginalized onto the $x$-axis, but having to transform from $\mathrm{d}x$s to $\mathrm{d}s$s, line elements along the circle. (Details are worked out in arc length from calculus.) $\endgroup$ – Eric Towers Jun 18 '14 at 8:56

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