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Suppose that $X \sim \Gamma(\alpha_1,1)$, a random variable with gamma distribution with shape $\alpha$ and unit rate/scale.

Q: Can we found a reshaping function $f_{\alpha_1\rightarrow\alpha_2}$ such that $f_{\alpha_1\rightarrow\alpha_2}(X) \sim \Gamma(\alpha_2,1)$ ?

I do have a non-explicit evident solution : Suppose that $F_\alpha$ is the c.d.f. of a $\Gamma(\alpha,1)$ r.v. Then the function :

$$F_{\alpha_2}^{-1} \circ F_{\alpha_1}$$

works. This makes the link through a uniform r.v., but it is not explicit at all. What if we try to make the link by somewhere else ? Can we obtain a better expression for this function ?

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3 Answers 3

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If you want your "reshaping" transformation function to be monotonic, then the answer is no.

Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$.

Suppose that $X_2$ equals $f(X_1)$ in distribution for some continuous strictly increasing function $f$. Then for all real $x_2$ $$F_2(x_2)=P(X_2\le x_2)=P(f(X_1)\le x_2)=P(X_1\le f^{-1}(x_2)) =F_1(f^{-1}(x_2)),$$ so that $F_2=F_1\circ f^{-1}$ and hence necessarily $$f=F_2^{-1}\circ F_1,$$ just as you had it.

Similarly, if $X_2$ equals $f(X_1)$ in distribution for some continuous strictly decreasing function $f$, then $$f=F_2^{-1}\circ G_1,$$ where $G_1:=1-F_1$.


However, in the case when $a_1=a_2+1$, there is a curious somewhat related equidistribution phenomenon (that note can be read online for free).


I don't think that the expression $F_2^{-1}\circ F_1$ is hard to deal with. E.g., Mathematica produces the graph $\{(x,(F_2^{-1}\circ F_1)(x))\colon0<x<7\}$ for $a_1=1.2$ and $a_2=3.4$ in about $0.052$ sec; for a comparison, it produces the graph $\{(x,\ln x)\colon0<x<7\}$ in about $0.035$ sec. Also, I think it is not hard to deal with $F_2^{-1}\circ F_1$ analytically. Here is the graph production:

enter image description here

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  • $\begingroup$ Yes, i do need it to be monotonic. Thanks for the proof it's stunning now. The problem now becomes 'Is there a nice expression of $F_2^{-1} \circ F_1$' which i'm afraid has to be ansewred by a strict No... Too bad. $\endgroup$
    – lrnv
    Commented May 19, 2020 at 13:39
  • $\begingroup$ @lrnv : I have added a comment concerning the complexity of the expression $F_2^{-1}\circ F_1$. $\endgroup$ Commented May 19, 2020 at 14:40
  • $\begingroup$ Thanks about the comment regarding the complexity. However I do have more analytical need than computational : E.g, computing the c.g.f $K(t,s) = log\left(\mathbb{E}\left(e^{-tX - sf(X)}\right)\right)$ for later inversion purposes. Hence my goal to obtain a 'usable' form for $f$. $\endgroup$
    – lrnv
    Commented May 19, 2020 at 15:11
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Rejection sampling provides a general technique to convert from one gamma distribution to another, as described in arXiv:1304.3800. This is particularly effective if $\alpha_1=n$ is an integer, because then $X$ can be generated easily as a sum of $n$ exponential random variables.

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  • $\begingroup$ Thanks for the link, this is interesting; Although this is not exactly what i need, it's still interesting to read :) $\endgroup$
    – lrnv
    Commented May 19, 2020 at 11:44
  • $\begingroup$ Unfortunately, the rejection method will not work for $\text{gamma}(a_1,b_1)$ and $\text{gamma}(a_2,b_2)$ with $a_1\ne a_2$ and $b_1=b_2$, because then the ratio of the corresponding pdf's is unbounded. $\endgroup$ Commented May 19, 2020 at 12:58
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Actually, your reshaping function $F_{2}^{-1}\circ F_{1}$ is optimal in the sense it is the Monge's optimal transport (i.e., reshaping) map. Here is what this means: consider any two univariate probability measures $\mu,\nu$ with respective continuous c.d.f.s $F,G$. Consider all possible pushforward maps $T$ which reshape $\mu$ to $\nu$, i.e., $T_{\#}\mu = \nu$. Assume $c(x,y):=d(x-y)$ with $d$ strictly convex. Then

$$T^{\text{opt}} = \underset{T: T_{\#}\mu = \nu}{\arg\inf} \int c\left(x,T(x)\right)\:{\rm{d}}x$$

with $T^{\text{opt}} = G^{-1}\circ F$.

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  • $\begingroup$ Thanks for the head's up. Actualy, now the problem is to express $F_2^{-1}\circ F_1$ in a more usable form. $\endgroup$
    – lrnv
    Commented May 20, 2020 at 8:12

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