# Characteristic function of comonotone gammas

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 $$i \in \{1,...,n\}$$.

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:

• 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?
– lrnv
May 5, 2020 at 14:54
• @lrnv : I think everything is correct as written. $-x$ in the exponent comes from $dF(x)=e^{-x}\,dx$. May 5, 2020 at 15:00
• Ho yeah you are right. Here comes down all my hopes ;( Thanks anyway !
– lrnv
May 5, 2020 at 15:08
• @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 $x$. May 5, 2020 at 15:13
• Hahaha it got worse :)
– lrnv
May 5, 2020 at 15:37