The original problem is: Given $Y_i = \beta X_i + \epsilon_i$, $i=1,2,...,n$, where $X \sim N(\mu, \tau^2)$ iid and $\epsilon \sim N(0, \sigma^2)$ iid, $X$ and $\epsilon$ are independent. What is the expectation and variance of $\frac{\sum Y_i}{\sum X_i}$
$\sum Y_i \sim N(n\beta \mu, n(\beta^2 \tau^2 + \sigma^2))$
$\sum X_i \sim N(n\mu, n\tau^2)$
$\sum \epsilon_i \sim N(0, n\sigma^2)$
I first found from the fact that $E(\epsilon_i) = 0$ that $$ E\left(\frac{\sum Y_i}{\sum X_i}\right) = E\left(\frac{\sum (\beta X_i + \epsilon_i)}{\sum X_i}\right) = E\left(\frac{\beta \sum X_i + \sum \epsilon_i}{\sum X_i}\right) = E\left(\beta + \frac{\sum \epsilon_i}{\sum X_i}\right) = \beta + E\left(\frac{\sum \epsilon_i}{\sum X_i}\right) = \beta + E(\Sigma \epsilon_i)\cdot E\left(\frac{1}{\Sigma X_i}\right) = \beta $$ And I tried a similar method for Variance to get $$ Var\left(\frac{\sum Y_i}{\sum X_i}\right) = Var\left(\beta + \frac{\sum \epsilon_i}{\sum X_i}\right) = Var\left(\frac{\sum \epsilon_i}{\sum X_i}\right) $$ but haven't been able to manipulate it further to find an answer. I tried using the Law of Total Variance to find $$ Var\left(\frac{\sum \epsilon_i}{\sum X_i}\right) = E\left(\frac{1}{\Sigma X_i}\right)^2Var\left(\Sigma \epsilon_i\right) + Var\left(\frac{1}{\Sigma X_i}\right)E\left(\Sigma \epsilon^2\right) $$ And then I tried to find the distribution for $1/X$ by letting $Z = \frac{1}{\sum X_i}$ and finding the transformation of $f_X$ but I couldn't find a way to integrate it. Does anyone have any advice for where I can go from here?
