Expectation and Variance of Ratio Estimator Let $X$ and $Y$ be positive random variables such that 
$$E(Y\mid X)= aX $$
$$\operatorname{Var}(Y\mid X) = b^2X^2 $$
$$a,b > 0 \text{ are constants}.$$
Let $R = \dfrac{\bar{Y}}{\bar{X}}=\dfrac{\sum_{i=1}^nY_i}{\sum_{i=1}^nX_i}$.
Is there an easy way to solve for $E(R)$ and $\operatorname{Var}(R)$ using the info above, without having to take the Taylor series expansion of $\frac{\bar{Y}}{\bar{X}}$ around $\mu_X,\mu_Y$?
 A: There is an easy way to compute the expectation of $R$, provided one makes much more precise the stochastic structure of the samples... Namely, introducing $Z_i=(X_i,Y_i)$, one should assume that the sequence $(Z_i)_{1\leqslant i\leqslant n}$ is i.i.d. And the variance is another, more complicated, story.

Introduce the random variables $S_n=X_1+\cdots+X_n$ and $T_n=Y_1+\cdots+Y_n$, hence $R_n=T_n/S_n$. Let $\mathfrak X_n$ denote the sigma-algebra generated by $(X_i)_{1\leqslant i\leqslant n}$. 
To compute the expectation of $R_n$, note that, for every $i\leqslant n$, $E(Y_i\mid\mathfrak X)=E(Y_i\mid X_i)=aX_i$ hence $E(T_n\mid\mathfrak X_n)=aS_n$ and $E(R_n\mid\mathfrak X_n)=E(T_n/S_n\mid\mathfrak X_n)=E(T_n\mid\mathfrak X_n)/S_n=a$. In particular, 
$$E(R_n)=a.
$$
As regards the variance, one can start from
$$
T_n^2=\sum_iY_i^2+\sum_{i\ne j}Y_iY_j
$$
and note that, for every $i\leqslant n$, $E(Y_i^2\mid\mathfrak X_n)=E(Y_i^2\mid X_i)=(b^2+a^2)X_i^2$ and, for every $i\ne j$, 
$$E(Y_iY_j\mid\mathfrak X_n)=E(Y_iY_j\mid \sigma(X_i,X_j))=E(Y_i\mid X_i)E(Y_j\mid X_j)=a^2X_iX_j.
$$
This
implies
$$
E(T_n^2\mid\mathfrak X_n)=\sum_i(b^2+a^2)X_i^2+\sum_{i\ne j}a^2X_iX_j=b^2\sum_iX_i^2+a^2S_n^2.
$$ 
Dividing by $S_n^2$, taking the expectations and using the value of $E(R_n)$, one gets 
$$\text{Var}(R_n)=b^2E\left(\frac{U_n^2}{S_n^2}\right),\qquad
U_n^2=X_1^2+\cdots+X_n^2.
$$
This reduces $\text{Var}(R_n)$ to an expression involving (the value of $b$ and) the marginal distribution of the $X_i$s only, but there is no general formula for $\text{Var}(R_n)$ in terms of $n$, $a$ and $b$. The exception is the case $n=1$ since $U_1^2=X_1^2=S_1^2$ hence $\text{Var}(R_1)=b^2$.
Edit: 
(1) For every $n\geqslant1$, $\text{Var}(R_n)=nb^2E\left(\dfrac{X_1^2}{S_n^2}\right)\geqslant\dfrac{b^2}n$. 
(2) If the $X_i$s are uniform on an interval $[0,x]$, then, for every $n\geqslant1$, $\text{Var}(R_n)=\dfrac{2b^2}{n+1}$.
A: I'm doing the case for $n=1$ just for ease of notation. For the expectation, just condition on $X$:
$$E[R] = E[E[R|X]] = E[a] = a$$ 
For the variance, we have:
$$Var[R] = E[Var[R|X]] + Var[E[R|X]]$$$$E[Var[R|X]] = E[b^2X] = b^2E[X]$$
$$Var[E[R|X]] = Var[a] = 0$$
*For clarity: to move from 1 to $n$, now just condition on the vector valued $(X_1, X_2, X_3, \ldots X_n)$, where I'm assuming $X_i \sim X, Y_i \sim Y$. Everything works the same. 
