This is the question I'm answering:
$$ \text{Suppose that the method of maximum likelihood is used to fit the following model representing exponential growth,}\; Y_i = \beta_1 e^{x_{i}} + \epsilon_i \; \text{to the data} \; (x_i, y_i), \; \text{where} \; i = 1,...,n \; \text{and} \; \epsilon_i \sim^{inep}N(0, \sigma^2). \; \text{It follows that} \; Y_i \sim^{inep}N(\beta_1e^{x_{i}}, \sigma^2). \; \text{The maximum likelihood estimator of}\; \beta_1 \; \text{is}\\ \hat{\beta_1} = \frac{\sum_{i=1}^n Y_ie^{x_i}}{\sum_{i=1}^n e^{2x_i}}\\ \text{Derive} \; E[\hat{\beta_1}] \; \text{and}\; Var(\hat{\beta_1}) \; \text{and hence write down the distribution of}\; \hat{\beta_1} $$ So, the expectation was easy enough to derive, $E[\hat{\beta_1}] = \beta_1$. However, in deriving the variance I'm left with a sum in the denominator that I can't seem to simplify? $$\begin{align} Var(\hat{\beta_1}) &= Var\left(\frac{\sum_{i=1}^n Y_ie^{x_i}}{\sum_{i=1}^n e^{2x_i}}\right)\\ &= \frac{\sum_{i=1}^n e^{2x_i}}{\sum_{i=1}^n e^{4x_i}}Var(Y_i)\\ &= \frac{\sigma^2}{\sum_{i=1}^n e^{2x_i}} \end{align}$$ Is there any way to simplify more? I'm also not sure how to determine the distribution. Can I just treat the $e^{x_i}$ terms as constant and thus $\hat{\beta_1} \sim N\left(\beta_1, \frac{\sigma^2}{\sum_{i=1}^n e^{2x_i}}\right)$ ?
Variance: \begin{align} var\left(\frac{\sum Y_i e^{x_i}}{\sum e^{2x_i}}\right) = \frac{1}{ (\sum e^{2x_i})^2} var\left(\sum Y_i e^{x_i}\right) = \frac{1}{ (\sum e^{2x_i})^2} \sum e^{2x_i} var(Y_i) = \frac{ \sigma^2 }{ \sum e^{2x_i}} \end{align}
Distribution: \begin{align} Y_i &\sim N(\beta e^{x_i}, \sigma^2)\\ e^{x_i}Y_i &\sim N(\beta e^{2x_i}, e^{2x_i}\sigma^2)\\ \sum e^{x_i}Y_i &\sim N(\beta \sum e^{2x_i}, \sum e^{2x_i}\sigma^2)\\ \frac{\sum e^{x_i}Y_i}{\sum e^{2x_i}} &\sim N(\beta \sum e^{2x_i}/\sum e^{2x_i}, \sum e^{2x_i}\sigma^2/\sum e^{2x_i})\\ \hat \beta & \sim N(\beta, \sigma ^ 2/ \sum e^{2x_i}) \end{align} that is basically just a linear combination of a normal random variables $Y_i$, $i=1,...,n$.