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Can one consider that a general probability distribution for some random variable X over a Euclidean space is actually the probability distribution of a uniform random variable over a curved 1D-space ?

(I was thinking about a map of the type $\frac{1}{\sqrt{1+f'^2}}$ between the two distributions)

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    $\begingroup$ You might want to explain your idea of a curved 1D-space. Extrinsic curvature of a curve is allright (given an embedding) but you would need intrinsic curvature, which does not exist... Or am I mistaken? $\endgroup$
    – Did
    Jul 12, 2011 at 21:19
  • $\begingroup$ @Didier Piau : Suppose you have a function y=f(x) which is not necessarily invertible. Suppose now that s is the curvilinear abscissa along this curve (The map I was talking about then gives the metric of the 1D-space) If X is a random variable with uniform probability distribution over s, does it not induce an apparent, non-uniform, probability distribution over x ? Sorry if this sounds fuzzy, I'm not very good at formalizing intuitions... $\endgroup$
    – AlexPof
    Jul 13, 2011 at 5:24

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If $X$ has a cumulative distribution function $F(x)$ and $F$ is invertible, and if $U$ is a random variable with a uniform distribution on $(0,1)$, then you can generate $X$ using $F^{-1}(U)$.

Now all you have to do is translate $F^{-1}$ into whatever you mean by a curved 1D-space.

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  • $\begingroup$ I suppose you're referring to the probability integral transform, however that's not exactly what I was thinking about, more like a uniform distribution over the curvilinear path of a curve. $\endgroup$
    – AlexPof
    Jul 13, 2011 at 8:18

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