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Suppose we have two random variables $X$ and $Y$ both distributed as gamma random variables with parameters $\alpha_X,\beta_{X},\alpha_{Y},\beta_Y$, with characteristic functions given by:

$$\phi_{X}(t) = \left(1 - \frac{it}{\beta_X}\right)^{-\alpha_X} \;\;\text{ and }\;\;\phi_{Y}(t) = \left(1 - \frac{it}{\beta_Y}\right)^{-\alpha_Y}$$

Furthermore, suppose that $X$ and $Y$ are comonotone : the joint c.d.f is then given, via the upper frechet-hoffding bound, by :

$$F_{X,Y}(x,y) = \min(F_X(x),F_Y(y))$$

Note that this random vector has no density.

Can we obtain a formulation for the joint characteristic function ? Same question for more than $2$ comonotone gammas, say $X_i$ with parameters $\alpha_i,\beta_i$ for $i \in \{1,...,n\}$.

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If you are asking for a closed-form expression, this is quite impossible, at least with the current state of the art. Indeed, for any cdf $F$ and its generalized inverse defined by $$F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\} =\min\{x\in\mathbb R\colon F(x)\ge u\}$$ for $u\in(0,1)$, we have $$x\ge F^{-1}(u)\iff F(x)\ge u$$ for all $x\in\mathbb R$ and $u\in(0,1)$.

Letting now $F:=F_X$ and $G:=F_Y$, we see that the joint distribution of comonotone $X$ and $Y$ with cdf's $F$ and $G$ is the same as the joint distribution of $F^{-1}(U)$ and $G^{-1}(U)$, where $U$ is a random variable uniformly distributed on the interval $(0,1)$.

Therefore, the joint characteristic function (c.f.) $f$ of $X$ and $Y$ is given by $$f(s,t)=\int_0^1\exp\{i(s F^{-1}(u)+t G^{-1}(u))\}\,du$$ for real $s,t$. This integral cannot be taken in closed form when $X$ and $Y$ have arbitrary gamma distributions.

This cannot be done even in the simplest case when $\alpha_X=2$, $\beta_X=1$, $\alpha_Y=1$, $\beta_Y=1$. Indeed, then $F(x)=1-(x+1)e^{-x}$ and $G(x)=1-e^{-x}$ for $x>0$, whence for real $s,t$ $$f(s,t)=\int_0^\infty\exp\{i(s x+t G^{-1}(F(x)))\}\,dF(x) \\ =\int_0^\infty \exp\{i(s+t)x-x\}\,\frac{x\,dx}{(x+1)^{it}}.$$ Mathematica cannot do anything with this integral, returning just a trivially identical expression:

enter image description here

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  • $\begingroup$ Thanks for all the work; Except you made a mistake at the end : the $i$ is in factor of everything, not just (s+t)x, so you obtain an exponent of $i(s+t-1)x$ if i'm right. Can mathematica solve it with this correction? $\endgroup$
    – lrnv
    Commented May 5, 2020 at 14:54
  • $\begingroup$ @lrnv : I think everything is correct as written. $-x$ in the exponent comes from $dF(x)=e^{-x}\,dx$. $\endgroup$ Commented May 5, 2020 at 15:00
  • $\begingroup$ Ho yeah you are right. Here comes down all my hopes ;( Thanks anyway ! $\endgroup$
    – lrnv
    Commented May 5, 2020 at 15:08
  • $\begingroup$ @lrnv : Oops! I did confuse $F$ and $G$ at some point ($G$ must be the simpler one, as compared to $F$, to make $G^{-1}$ explicit). This is now fixed, and the integral has become only slightly more complicated, with the integrand now containing the extra factor $x$. $\endgroup$ Commented May 5, 2020 at 15:13
  • $\begingroup$ Hahaha it got worse :) $\endgroup$
    – lrnv
    Commented May 5, 2020 at 15:37

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