Mapping Normal distribution to a new bounded distribution.
Solution 1
Every distribution on $\mathbb R$ (or on any interval of $\mathbb R$) is the distribution of a random variable $H(X)$, where the random variable $X$ is standard normal, for some nondecreasing function $H$.
Proof: Let $F$ denote the CDF of the desired distribution and $G$ the standard gaussian CDF. Then $G(X)$ is uniform on $(0,1)$ and, for every $U$ uniform on $(0,1)$, $F^{-1}(U)$ has CDF $F$. The inverse $F^{-1}$ must be defined with care when the CDF $F$ of the target distribution is not increasing but only nondecreasing, but this can be done, and it works. By composition $H=F^{-1}\circ G$ fits.
Solution 2
I'm not sure if the question makes much sense because of this:
Let $X$ be any continuous random variable (bounded or not) and $F_X$ its CDF. Then the random variable $U=F_X(X)$ has uniform distribution on $[0,1]$.
Now pick any other continuous random variable (bounded or not), say $Y$ and let $F_Y$ be the corresponding CDF. Then the random variable $F_Y^{-1}(U)$ has the same distribution as $Y$ has.
Concluding, if you pick $X$ to be normal, you can generate any arbitrary continuous distribution $F_Y$ by considering the random variable, $$F_Y^{-1}\left(F_X(X)\right).$$
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ashim
Updated on August 01, 2022Comments
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ashim over 1 year
My question is vague. I am looking for distributions that can be obtained from Normal distribution by mapping it to a bounded interval. "By mapping" I mean that a probability distribution function(PDF) is obtained from PDF for Normal distribution by applying some function. For example, log normal distribution is obtained from Normal distribution by using exponential function. What distribution do you know that obtained from Normal distribution by mapping to a bounded interval?
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ashim almost 10 yearsYour answer is awesome. I had to choose the answer by Did, because he answered earlier (and the essence is the same).
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Sergio Parreiras almost 10 yearsWhen I started typing he had not answered, if I knew someone was already working on it, I would pass. It was not my intention to replicate someone else answer.
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ashim almost 10 yearsYes, I know that ( I was in this situation :-) ) I just feel bad about not accepting two good answers.