I normalized my data using max-min normalization as follows $$X_{normed}=\frac{X-\min(X)}{\max(X)-\min(X)}$$

How can I find the distribution of $X$ given the distribution of $X_{normed}$, if

  1. $X_{normed} \sim Gamma(\alpha,\beta)$ where $\alpha,\beta$ repectively denote the shape and rate parameter.

  2. $X_{normed} \sim Beta(\alpha,\beta)$ where $\alpha,\beta$ denote the shape parameters.

  • 1
    $\begingroup$ Your renormalised data is over $(0,1)$ while the Gamma distribution is over $(0,\infty)$... $\endgroup$
    – Xi'an
    Sep 6 at 15:49
  • 2
    $\begingroup$ Given that you normed variable is bounded to the unit interval, wouldn't a beta distribution make more sense? (That would also make reversing it quite straightforward.) $\endgroup$
    – Glen_b
    Sep 6 at 15:49
  • $\begingroup$ @Glen_b: That makes sense, so I edited my question. But I still don't get how reversing is "straightforward" in the case of Beta distribution? Also, one more issue: what if I have new data, saying, $X_i>\max(X)$. If I normalize it using the formula in the post, $X_{normed}>1$ so it can't be fitted into the Beta distribution... $\endgroup$
    – soma
    Sep 7 at 1:16
  • $\begingroup$ If $X\sim B(\alpha,\beta)$, then $1-X\sim B(\beta , \alpha)$. The other issue is a problem with you choice of norm. I believe a couple of questions on site touch on that issue. Note that the problem exists on both sides, not just to the right $\endgroup$
    – Glen_b
    Sep 7 at 1:29
  • $\begingroup$ I'm really sorry, but I still can't relate how $1-X_{normed} \sim B(\beta,\alpha)$ is useful for the derivation of the distribution of $X$ here? It just implies $\frac{max(X)-X}{max(X)-min(X)} \sim B(\beta,\alpha)$. $\endgroup$
    – soma
    Sep 7 at 1:55

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