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For $\alpha \geq 0$ the transformation $x \mapsto \log(x) - \alpha \log(1-x)$ maps the unit interval to the real line (in fact for $\alpha = 0$ the transformation is not surjective). For $\alpha=1$ this is the logit transformation, which has a well-known inverse. It is also not difficult to find an inverse for $\alpha = 0.5$ and I would guess along these lines for some more $\alpha = 1/N$, $N$ positive natural number. But I do not see a closed form for an arbitrary $\alpha > 0$.

Finding an inverse numerically is not too difficult because the function is strictly monotone, but this is at least aesthetically unpleasing ;)

Is there a way to show that for arbitrary $\alpha$ no such solution can exist, or is there maybe one?

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  • $\begingroup$ You need a minus sign rather than a plus in your formula. The logit transformation is $x \mapsto \log x - \log (1-x)$. $\endgroup$
    – user856
    Jan 27, 2011 at 23:00

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If $y = \log x - \alpha \log (1-x)$, then $e^y = x/(1-x)^\alpha$. If you pick $\alpha = 1/5$ or $\alpha = 5$, you will have to solve a quintic equation for $x$. Some of the coefficients will involve $e^y$ which is almost always transcendental, and (I think) this implies that the quintic will not in general be solvable.

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According to Wolfram Alpha, there is no analytic form for the inverse.

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  • $\begingroup$ How reliable are the answers from Wolfram Alpha? It probably just means Wolfram's algorithm can't find one... $\endgroup$
    – Philipp
    Jan 27, 2011 at 22:29

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