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You may want t consider the von Mises distribution, aka Tikhonov distribution, and plays the role similar to the normal distribution in 1D statistics: $$p(\theta ; \alpha, \theta_0 ) = \frac{ e^{\alpha \cos (\theta -\theta_0)}} {2 \pi I_0(\alpha)}$$ For $\alpha=0$ it is uniform, for $\alpha >> 1$ the distribution is sharply peaked at $\theta_0$ C....
The most obvious thing to do here would be to express the variable in polar coordinates and impose a prior on the angle and displacement. That is, you express your point $(x,y)$ as a vector $(\theta,r)$ where: \begin{aligned} x &= r \cos \theta, \\[4pt] y &= r \sin \theta. \\[4pt] \end{aligned} You can then impose a prior on $0 \leqslant \... 2 You made a slight parentheses error, your pdf should look like this, with the variance in the denominator of the exponential: pdf = np.exp(-np.square(valores-mean)/(2*variance))/(np.sqrt(2*np.pi*variance)) 2 Simplifications: replace $$\frac{(z-\sigma )^2 \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi }}{1+0.25 \mu ^2}$$ by $$\frac{(z-\sigma )^2 \sqrt{2 \pi }}{\sqrt{1+0.25 \mu ^2}}$$ replace$$z-\sigma$$by$y$replace$\text{Erf}(x)\$ with $$\text{Erf}(x) = 2\Phi(x\sqrt{2}) - 1$$ This leads to consider instead $$\frac{2 y^2}{\pi^2\sqrt{2\pi}} \exp\left\{-\sqrt{2\... 2 Hint: the moment generating function is a natural tool for the study of exponential families. 2 First median is defined where F(x)=0.5. Right now you have a CDF that is defined in terms of p and x and it is possible to define the median in terms of p(I did not check if your CDF is correct or not since it is your task). You can think of p as a parameter and thus no need to worry if you have a definition that have p in it 1 If you have a prior for the angle, I'd use it as the reference. E.g. I'd rotate all data so that the prior is at 180^{\circ} and measure all angles on the scale [0^{\circ}, 360^{\circ}). I see no elegant solution to measuring distance between two angles, \phi and \psi. I'd calculate the differences (\phi - \psi) and (((\phi + 180^{\circ}) \mod ... 1 Since the function g is symmetric about the y-axis, and since this implies that g(0)=0, it is sufficient to find the function for values z>0. Thus, without loss of explanatory power, we will look at the transformation only over these values. Let F_* and Q_* denote the CDF and quantile function of the generalised error distribution. Using the ... 1 Not sure if this should be a comment or an answer but because this question was simultaneously posted on multiple forums, I placed an answer at Mathematica StackExchange. In short the pdf is approximately$$1.0105750026505362 \times \frac{\sqrt{2 \pi }}{\sqrt{1+ \mu^2/4}} \frac{(z-\sigma )^2 \sqrt{2 \pi }}{\sqrt{1+ \mu^2/4}} \exp\left(-\frac{(z-\...