I'm investigating model mismatch and have a wrapped Cauchy distribution of
f(x,p) = (1-p^2)/ (2*pi*(1+p^2-2p*cos(x)))
Is there a way to map this to a random uniform distribution, like
random.uniform(100, minval=-q, maxval=q)
i.e. If I have a value of q, what p does that equate to? Thanks
Edit:
I have decided to match the moments to map the parameter. The means would be 0, but the variance would give: $\frac{1}{12} (q--q)^2 = 1- e^\gamma$
Using the substitution for $\gamma$ to get to $p$, and rearranging, I get: $p = \sqrt{\frac{1 - \tanh (-log(1-q^2/3))}{1 + \tanh (-log(1-q^2/3))}}$ but this does not behave as expected. i.e. when $p$ is close to 1, the distribution should be tight and therefore I would expect $q \rightarrow 0$ and when $p \rightarrow 0$, $q$ should be high, but this is not what happens - does anyone spot a mistake?