Simplifying the expectation of a ratio of exponential random variables. Suppose there are i.i.d. exponential random variables $x_1,\dots,x_n, y_1,\dots,y_m$. Can we write the following expectation
$$E\left(\cfrac{a_1x_1+\dots+a_nx_n}{b_1y_1+\dots+b_my_m}\right),$$ as the sum of some functions of the coefficients i.e., of the form $$\cfrac{f_1(a_1)+\dots+f_n(a_n)}{g_1(b_1)+\dots+g_m(b_m)},$$
where $f_i$ and $g_j$ are functions and $a_i$ and $b_j$ are constants. I think the numerator can be written as the sum of expectations $a_1Ex_1+\dots+a_nEx_n$ and of course $Ex_i$ are all equal by i.i.d. I am not sure how to simplify the denominator.
 A: Assume that $m\geqslant2$, that every $b_k$ is positive, and, without loss of generality that every $y_k$ is standard exponential. Apply to the denominator the identity, valid for every $z\gt0$,
$$
z^{-1}=\int_0^\infty\mathrm e^{-zs}\mathrm ds.
$$
For $z=\sum\limits_kb_ky_k$, this yields
$$
E[z^{-1}]=\int_0^\infty \prod_kE[\mathrm e^{-b_ky_ks}]\mathrm ds.
$$
For every standard exponential $y$ and every positive $s$, $E[\mathrm e^{-sy}]=1/(1+s)$, hence
$$
E[z^{-1}]=\int_0^\infty Q(s)\mathrm ds,\qquad Q(s)=\prod_k\frac1{1+b_ks}.
$$
Assuming the real numbers $b_k$ are distinct, the decomposition of $Q$ reads
$$
Q(s)=\sum_kc_k\frac{b_k}{1+b_ks},
$$
where one can note that $\sum\limits_kc_k=0$ since $sQ(s)\to0$ when $s\to\infty$. Thus,
$$
E[z^{-1}]=\sum_k\left.c_k\log(1+b_ks)\right|_0^\infty=\sum_kc_k\log b_k.
$$
To be complete, note that
$$
c_k=b_k^{m-2}\prod_{j\ne k}\frac1{b_k-b_j},
$$
hence 
$$
E[z^{-1}]=\sum_k\left.c_k\log(1+b_ks)\right|_0^\infty=\sum_kb_k^{m-2}\log b_k\prod_{j\ne k}\frac1{b_k-b_j}.
$$
This rules out the possibility of a formula such as the one you have in mind.
