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I have the following problem that I wanna solve: I need something like a normal distribution that is not as usual defined on the real line, but instead on the interval $(-\pi/2, \pi/2)$, which means that the density has to decay faster than it would in a normal distribution. My idea was to transform the interval $(-\pi/2, \pi/2)$ to the real line using the map $tan(x)$ and insert this into the Gaussian density, yielding $\frac 1{\sqrt{2\pi\sigma^2}}exp\left(-\left(\frac{tan(x)-\mu}\sigma\right)^2\right)$. Now I think this lacks some kind of normalisation term affiliated to the derivative of the tangent function. Can anybody tell me where I have to place the term, and what exactly it is, so that the distribution becomes normalised?

Any help will be appreciated!

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    as you have said: the density would have to decay faster. But then it wouldn't be a normal distribution anymore. What I would do instead is to cut off the distribution: so if $X \sim N(\mu, \sigma^2)$ you work with $Y:= X 1_{X \in (-\pi/2, \pi/2)}$. The normailsation constant in this case would be $\Bbb P (X \in (-\pi/2, \pi/2))$. – 2017-01-12
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    Thanks, I will try that instead! – 2017-01-12

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