I am struggling to find the PDF of the difference of two Beta Prime distribution.
DefinitionA random variable is said to have a Beta Prime distribution $\text{B}'(\alpha, \beta)$ with $\alpha, \beta>0$ if its density $f$ is defined as:$$ \forall x>0,~f(x):=\frac{x^{\alpha-1}}{\text{B}(\alpha, \beta)(1+x)^{\alpha+\beta}}$$where $\text{B}(\alpha, \beta)$ is the beta function.
PDF of the sum of two Beta Prime
In this paper, the authors derive an expression of the PDF of the sum of two Beta Prime r.v.. More precisely, if $X_{a,b}\sim \text{B}'(a, b)$ and $Y_{c,d}\sim \text{B}'(c, d)$ two independent variables, proposition 2 states that:
$$f_{X+Y}(x) = \frac{C_{a,b,c,d}x^{a+c-1}}{(1+x)^{c+d}}F_1(a,a+b,c+d, a+c;-x, x/(x+1)).$$
They get the result by changing the variable from $\mathbb{R}_+$ to $[0,1]$ and use integral repsentation of hypergeometric series. ($C$ is a constant)
PDF of the difference of two Beta Prime, my attempt
What I am looking for is an expression of the PDF of $X-Y$ with special functions. To proceed, what I have done is, for $z>0$, use the convolution:
$$f_{X-Y}(z) = \int_z^\infty f_X(t)f_Y(t-z)dt.$$
This might be rewritten as:$$f_{X-Y}(z) = \frac{1}{\text{B}(a, b)\text{B}(c, d)}\int_z^\infty \frac{t^{a-1}(t-z)^{c-1}}{(1+t)^{a+b}(1+t-z)^{c+d}}dt.$$
In order to get on $[0,1]$ (hope that might work(*)), I use the change of variable $u=(t-z)/z$, it comes then:
$$f_{X-Y}(z) = \frac{1}{\text{B}(a, b)\text{B}(c, d)}z^{a+c-1}\int_0^1 \frac{(1-u)^{b+d}u^{c-1}}{(1-u+z)^{a+b}(1-u(1-z))^{c+d}}du.$$
From here, I struggle to find a integral repsentation of a special function (this one is close to hypergeometric).
Any correction of my computations, hints, or solutions will be highly appreciated!
Thank you!
(*) The intuition is that this types of integral are close to hypergeometric ones (see here).