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I need to find some function $f:\mathbb{R} \rightarrow \mathbb{R}^+$ such that

If $\; \; x \sim \mathcal{N}(x; \mu, \sigma^2) \; \;$ then $\; \; f(x) \sim \mathcal{G}(y; \alpha, rate=\beta)$

Where $\mathcal{G}$ is the Gamma probability density function.

I think it will have to do with the derivation of the distribution function of $\sigma^2$ from a normal distribution, and $\alpha, \beta$ will have to be functions of $\mu, \sigma$. But I am having trouble from here.

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    $\begingroup$ The obvious one is to transform the normal to uniform $U=F(X)$ and the uniform to gamma $Y=G^{-1}(U)$ where F and G are the relevant cdfs. This composition of functions is discussed in many posts on site $\endgroup$
    – Glen_b
    Commented Oct 15, 2023 at 5:52
  • $\begingroup$ @Glen_b The idea of using an intermediary space then a composite function makes sense. However, I am a little confused on what you mean by the uniform though. Like a uniform distribution over [0, 1]? $\endgroup$ Commented Oct 15, 2023 at 6:08
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    $\begingroup$ If you transform any continuous random variable by its cdf you get a uniform on the unit interval, yes. en.wikipedia.org//wiki/Probability_integral_transform ... also see en.wikipedia.org/wiki/Inverse_transform_sampling to transform a uniform to some other desired distribution ... As indicated in my earlier comment, $Y= G^{-1}(F(X))$ has the desired distribution. $\endgroup$
    – Glen_b
    Commented Oct 15, 2023 at 7:44
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    $\begingroup$ There's some discussion here: stats.stackexchange.com/questions/243151/… but the same idea crops up in multiple places $\endgroup$
    – Glen_b
    Commented Oct 15, 2023 at 8:03
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    $\begingroup$ Please see our threads about the probability integral transform. The duplicate answers a generalization of this question. $\endgroup$
    – whuber
    Commented Oct 15, 2023 at 15:48

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