EngWiPy said:
we can average over each ##G_k## as if the random variables are independent with no ordering, but what about the limits? I mean, ##G_3=\frac{X_{(3)}}{Y_3}## has the limits ##0## to ##\eta##, but ##G_2## contains ##X_{(2)}## which is less than or equal ##X_{(3)}## in ##G_3##. Or this doesn't affect the evaluation?
You are right to point out that the new ##G_k##s are not mutually independent, because they are based on order statistics. So we have to change the integrand in your triple integral, because it is based on the Zs being independent.
Instead we'll have to do (take a deep breath!):
\begin{align*}
\text{Pr}&\left[\sum_{i=1}^3G_k\leq \eta\right]
=
\text{Pr}\left[\frac{X_{(1)}}{Y_1}+\frac{X_{(2)}}{Y_2}+\frac{X_{(3)}}{Y_3}\leq \eta\right]
\\
&=
\int_{y_1=0}^\infty \int_{x_1=0}^{y_1\eta}
\int_{y_2=0}^\infty \int_{x_2=0}^{y_2(\eta - x_1/y_1)}
\int_{y_3=0}^\infty \int_{x_3=0}^{y_3(\eta - x_1/y_1-x_2/y_2)}
\\&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
f_Y(y_1) f_Y(y_2) f_Y(y_3)
f_{X_{(1)},X_{(2)},X_{(3)}}(x_1,x_2,x_3)
\,dx_3\,dy_3\,dx_2\,dy_2\,dx_1\,dy_1
\\
&=
\int_{y_1=0}^\infty f_Y(y_1) \int_{x_1=0}^{y_1\eta}
\int_{y_2=0}^\infty f_Y(y_2) \int_{x_2=0}^{y_2(\eta - x_1/y_1)}
\int_{y_3=0}^\infty f_Y(y_3) \int_{x_3=0}^{y_3(\eta - x_1/y_1-x_2/y_2)}
\\&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
f_{X_{(1)},X_{(2)},X_{(3)}}(x_1,x_2,x_3)
\,dx_3\,dy_3\,dx_2\,dy_2\,dx_1\,dy_1
\end{align*}
It looks horrible, but it should simplify down if all your distributions are exponentials.
You can get the three-way joint probability function for the Xs from the wiki section I linked above. The formula is on the last line in the section, with ##n=3##.
EDIT: Actually the formulas in that section don't give exactly what we want, which is the joint density for the three smallest order stats, but we can infer a formula using the same principles. It should be:
$$\frac{K!}{(K-3)!}f_X(x_1)f_X(x_2)f_X(x_3) \left[1-F_X(x_3)\right]^{K-3}$$
where ##x_1\le x_2\le x_3##, otherwise zero. To enforce that inequality condition we need to change the limits for the X integrals, to get:
\begin{align*}
\text{Pr}&\left[\sum_{i=1}^3G_k\leq \eta\right]
=
\int_{y_1=0}^\infty \int_{x_1=0}^{y_1\eta}
\int_{y_2=0}^\infty \int_{x_2=x_1}^{\max(x_1,y_2(\eta - x_1/y_1))}
\int_{y_3=0}^\infty \int_{x_3=x_2}^{\max(x_2,y_3(\eta - x_1/y_1-x_2/y_2))}
\\&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
\frac{K!}{(K-3)!}
f_Y(y_1)f_Y(y_2)f_Y(y_3)
f_X(x_1)f_X(x_2)f_X(x_3) \left[1-F_X(x_3)\right]^{K-3}
\,dx_3\,dy_3\,dx_2\,dy_2\,dx_1\,dy_1
\\&=
\frac{K!}{(K-3)!}
\int_{y_1=0}^\infty f_Y(y_1) \int_{x_1=0}^{y_1\eta} f_X(x_1)
\int_{y_2=0}^\infty f_Y(y_2) \int_{x_2=x_1}^{\max(x_1,y_2(\eta - x_1/y_1))} f_X(x_2)
\int_{y_3=0}^\infty f_Y(y_3) \int_{x_3=x_2}^{\max(x_2,y_3(\eta - x_1/y_1-x_2/y_2))}
\\&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
f_X(x_3) \left[1-F_X(x_3)\right]^{K-3}
\,dx_3\,dy_3\,dx_2\,dy_2\,dx_1\,dy_1
\end{align*}
