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I have a random variable which is an angle $\Theta$ that follows a wrapped Normal distribution. The angle $\Theta$ has a relatively small variance, so despite having a range from $(-\pi,\pi)$, practically all the PDF is contained within the range $(-\pi/2,\pi/2)$. (Sorry, I don't know how to word it better). What I want to calculate is the PDF of $Y=\sin(\Theta)$. I know how to calculate the PDF of a function of a random variable but I don't know how to do this with the wrapped Normal, the PDF seems much more untractable. I am okay making simplifications and approximations. Could just assuming that $\Theta$ is a Normal distribution (not wrapped) work?

EDIT: Thank you for your reply @whuber. I made some progress, could someone indicate me whether I'm on the right track? I still have doubts.

I am not sure about the true distribution of $\Theta$. I know some things about it: it's symmetric and centered at 0, mostly bounded between $(-\pi/6,\pi/6)$ and I need it to be function of only one parameter, so that's why I'm starting with a Normal distribution. But perhaps doing this with a distribution that are supported on bounded intervals makes more sense?

I actually needed the PDF of the function $Y = k \frac{1}{\cos(\Theta)}$, where $k$ is a constant. (I asked for the sine just to simplify the question).

Using the CDF technique to find the PDF of $Y$,

$$ F(y) = P(Y\le y) = P\left(k \frac{1}{\cos(\Theta)} \le y\right) = 1-P(\Theta \le \arccos k/y)$$

$$ F(y) = 1 - \int_{\theta = -\infty}^{\theta = \arccos k/y} f_N(\theta;\mu,\sigma^2) d\theta$$

And then we could get the PDF by differentiating. My doubts come from the bounds of this integration. Would this be correct?

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    $\begingroup$ By definition, when $Z$ has a Normal distribution then $\theta = Z\mod (2\pi)$ has a wrapped Normal distribution. But then $\sin\theta=\sin Z,$ so you are asking about the distribution of the sine of a Normal variable, which is something you state you know how to calculate! $\endgroup$
    – whuber ♦
    Commented Jun 28, 2023 at 20:27
  • $\begingroup$ Thank you very much for your help @whuber. I've updated my question after progressing with your input. $\endgroup$
    – James Craft
    Commented Jun 28, 2023 at 21:29
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    $\begingroup$ +1 It's a good question. As in many similar cases, my recommendation is to draw a picture. That's almost de rigueur when the inverse of your function is not one-to-one. You can use the approach I illustrate at stats.stackexchange.com/a/138922/919: it extends from the finite domain considered there to the infinite domain in your problem (so there is an infinite sum involved rather than a finite one: see the last formula). That sum has no closed form expression, but you can readily approximate it with a tiny partial sum (in your case, having three or fewer terms!). $\endgroup$
    – whuber ♦
    Commented Jun 28, 2023 at 22:11
  • $\begingroup$ Thank you @whuber for your help! I'll give it a try. $\endgroup$
    – James Craft
    Commented Jun 30, 2023 at 2:21
  • $\begingroup$ I wouldn't, if only because your question has completely changed. Could you back up and explain how you know this information about the distribution of $\theta$? And why you are trying to model it so explicitly? Moreover, the information you provide is extremely limited: all you have really told us is you think the distribution of $\theta$ is symmetric around $0.$ $\endgroup$
    – whuber ♦
    Commented Jun 30, 2023 at 14:08

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