Expectation of the ratio of sum (XY) and sum(X) I want to know (mathematically) how the following expression changes as $M$ increases but still have no clue after thinking about it for a while. Any suggestions or comments will be much appreciated.
$$E\left[ {\frac{{\sum\limits_{i = 1}^M {\frac{{\sigma _{0,i}^2}}{{1 - \varphi _i^2}} \cdot \left( {5 - 4\varphi _i^2} \right)} }}{{\sum\limits_{i = 1}^M {\frac{{\sigma _{0,i}^2}}{{1 - \varphi _i^2}}} }}} \right]$$
where $\sigma _{0,i}^2 \sim U\left( {a,b} \right),b > a > 0$ and $\varphi _i^2 \sim U\left( {0,1} \right)$.
This expression is the expectation of the ratio of the variance of the total output of a system and the total variance of $M$ different input stationary autoregressive (AR(1)) sequences.
According to @Sextus Empiricus's answer, if we denote $\alpha_{i}=1-\varphi_{i}^{2} \sim U(0,1)$ and $\beta_{i}=\sigma_{0, i}^{2} \sim U(a, b)$, the expression can be simplified as
$$E\left[\frac{\sum_{i=1}^{M} \frac{\beta_{i}}{\alpha_{1}} \cdot\left(1+4 \alpha_{i}\right)}{\sum_{i=1}^{M} \frac{\beta_{i}}{\alpha_{i}}}\right]=E\left[\frac{\sum_{i=1}^{M} (\frac{\beta_{i}}{\alpha_{i}}+4 \beta_{i})}{\sum_{i=1}^{M} \frac{\beta_{i}}{\alpha_{i}}}\right]=1+4E\left[\frac{\sum_{i=1}^{M} \beta_{i}}{\sum_{i=1}^{M} \frac{\beta_{i}}{\alpha_{i}}}\right]$$

I will assume $a=0$ and $b=1$ in the following.
Here is a simulation experiment to look at the variability of the expectation in $M$:
e=rep(0,N)
f=matrix(0,T,N)
for(t in 1:T){
  phi2=runif(N)
  we=runif(N)/(1-phi2)
  f[t,]=cumsum(we*(5-4*phi2))/cumsum(we)
  e=e+f[t,]}

with the plot of the averaged e (in red) against the f[t,]'s (in gray):

The expectation thus appears to decrease with $M$. Due to the lack of expectation of the individual weights $\sigma^2_{0,i}/(1-\varphi_i^2)$, it is unclear that the average within the expectation enjoys a finite variance, as indicated by the repeated jumps in the individual gray curves.
Note that an equivalent expression for the expectation is
$$1+4\mathbb E^{U,W}\left[\sum_{i=1}^M U_i\Big/\sum_{i=1}^M W_i \right]$$
with$$(U_i,W_i)\sim \frac{u}{w^2}\mathbb I_{(0,1)}(u)\mathbb I_{w>u}$$
by a change of variables and that it is approximately
$$1+2\mathbb E^{W}[\overline W_M^{-1}]\tag{1}$$for $M$ large. (Or asymptotically equivalent by Slutsky's theorem.) Note also that the marginal distribution of the $W_i$'s is
$$W\sim \frac{(w\wedge 1)^2}{2w^2}=\frac{\mathbb I_{0<w<1}}{2}+
\frac{\mathbb I_{w>1}}{2w^2}$$
This means that an asymptotic equivalent to (1) is
$$1+4\mathbb E^{W}[1\big/(\overline S_{M/2}+\overline R_{M/2})]$$
where$$\overline S_{n}=\frac{1}{n}\sum_{i=1}^n U_i\qquad \overline R_n=\frac{1}{n}\sum_{i=1}^n V_i^{-1}\qquad U_i,V_i\sim\mathcal{U}(0,1)$$hence
$$1+8\mathbb E^{W}[1\big/(1+2\overline R_{M/2})]$$
Comparing the distribution of$$1+4\sum_{i=1}^M U_i\Big/\sum_{i=1}^M W_i$$ with the distribution of$$1+8\big/(1+2\overline R_{M})$$does not exhibit any significant difference:


The distribution of a sum of Pareto variates is particularly intricate. However, the limiting distribution of the centred average is a stable distribution. Namely,
$$\frac{\overline{R}_M-\log(M)-C}{\pi/2}\approx F_{1,1}$$
where $C\equiv 0.8744...$ and $F_{1,1}$ is the stable distribution for $\alpha=\beta=1$. With cdf
$$F_{1,1}(x)=2\left(1-\Phi(2/\sqrt\pi \exp\{-1/2-\pi x\sqrt2/4)\}\right)$$
