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Consider the following Beta distribution for $x$ $$P(x |n,k) = \lambda x^k (1-x)^{n-k}, \quad \quad \lambda=\frac{(n+1)!}{(n-k)!k!}$$ Now transform this to a distribution over $y$ where $$x = \frac{1}{2} \left( \mathrm{erf}(y)+1 \right)$$ Using the fact that $\mathrm{d}x = \pi^{-\frac{1}{2}} \exp{\left[-y^2\right]} \mathrm{d}y$ we may write: $$P(y |n,k) = \frac{\lambda}{2^n\sqrt{\pi}} \left( 1+\mathrm{erf}(y) \right)^k \left(1 - \mathrm{erf}(y) \right)^{n-k} \exp{\left[-y^2\right]}$$ I'm interested in the mean, variance and mode of $y$, but I haven't been able to get at them after some playing around in Mathematica.

Are there any tricks we can play here? Perhaps as the moments of the original Beta distribution are known?

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    So, there is a r.v. $X$ (Beta distributed.) Do you really have to do the following transformation: $$Y=\operatorname{erf}^{-1}(2X-1)?$$ – 2017-02-23
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    @zoli It doesn't have to be that transformation specifically, but I'm looking to change the support of the distribution from $[0,1]$ to $[-\infty, \infty]$. The error function could therefore be replaced by any sigmoid-type function, and I'm sure there are other possibilities as well. It would just be convenient if after the transformation some properties of the distribution could be obtained analytically. – 2017-02-23
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    What about $Y=\frac{2X-1}{X(1-X^2)}$? – 2017-02-23
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    @zoli Would we need a closed expression for $X = f(Y)$ in order to write down $P(Y|n, k)$? Does one exist for this choice of $Y$? – 2017-02-24

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