Let $X$ and $Y$ two independent random variables for gamma distributions with common shape parameter $a$ and different rate parameter $b_{1}$ and $b_{2}.$ If $U=\min(X,X+Y),$ then what will be the density function for $U$?
It is well known that the density will be $f_{U}(z)=(1-F_{X}(z))f_{X+Y}(z)+(1-F_{X+Y}(z))f_{X}(z).$
Since, $X\sim G(a,b_{1}),$ so, $f_{X}(z)=\frac{b_{1}^{a}}{\Gamma(a)}z^{a-1}e^{b_{1}z},~z>0$
and
$1-F_{X}(z)=\frac{b_{1}^a}{\Gamma(a)}\int_{z}^{\infty}x^{a-1}e^{-bx}dx=\nu(a,z)$ (incomplete gamma function).
Then, I derived the joint density function for $X$ and $V=X+Y$ as $f_{X,X+Y}(x,v)=\frac{(b_{1}b_{2})^a}{(\Gamma(a))^2}x^{a-1}(v-x)^{a-1}e^{-(b_{1}x-b_{2}(v-x))},$ x>0,v>x.
From here I am interested to derive $f_{X+Y}(z)$ by taking integration on $f_{X,X+Y}(x,v)$ respect to $X$ and then from $f_{X+Y}(z)$ we may able to derive $F_{X+Y}(z).$
But, the derivation of $f_{X+Y}(z)$ seems complicated to me. Is there any other way or any technique to simplify derivation?