# Tag Info

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The expected value of a non-central $t$ with non-centrality $\delta$ and d.f. $\nu$ is $$E\left[t\right] = \delta \sqrt{\nu/2} \frac{\Gamma\left((\nu-1)/2\right)}{\Gamma\left(\nu/2\right)}.$$ Thus it is trivial to find an unbiased estimator of $\delta$: $$\hat{\delta} = t \sqrt{2/\nu} \frac{\Gamma\left(\nu/2\right)}{\Gamma\left((\nu-1)/2\right)}$$ In R: ...

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Likelihood integration (with respect to $M$) doesn't make sense because it doesn't even integrate/sum to $1$ if we iterate over all possible $M$'s. However, integration/sum of likelihood multiplied with prior makes sense because the integral is equal to the evidence. With respect to $x$, it's still a density/PMF and its integral is a CDF.

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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))

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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....

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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 $(((\phi + 180^{\circ}) \mod ... 4 The most obvious thing to do here would be to express the variable in polar coordinates and impose a prior on the angle and displacement. That is, you express your point$(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 on0 \leqslant \...

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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 ...

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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

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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 $$1.0105750026505362 \times \frac{\sqrt[4]{2 \pi }}{\sqrt[4]{1+ \mu^2/4}} \frac{(z-\sigma )^2 \sqrt{2 \pi }}{\sqrt{1+ \mu^2/4}} \exp\left(-\frac{(z-\... 2 Simplifications: replace$$\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\...

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Hint: the moment generating function is a natural tool for the study of exponential families.

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