Assume I have a sample of $n \in \mathbb{N}$ data points $x_1,\ldots, x_n$, which are asummed to come from iid drawings of a Gamma-distribution. I may assume the shape paramter $k\in\mathbb{R}_+$ and the sample mean $\bar{x}$, but not the sample variance, are known.
To avoid confusion the Gamma(k, $\theta$) density function used here is $f(x) = \frac{1}{\Gamma(k) \theta^k} x^{k-1} e^{-\frac{x}{\theta}}.$
I am trying to calculate the Maximum-Likelihood Estimator $\hat{\theta}$ and then find an asymptotic 95% Confidence Interval for that estimate.
The MLE was no problem but I'm not sure about the asymptotic CI and would be very grateful, for corrections if I made a mistake there.
The likelihood function has the form $$\mathcal{L} = \prod_{i=1}^{n} f(x_i, \theta) = \left(\frac{1}{\Gamma(k)\theta^k}\right)^n \prod_{i=1}^{n} x_i^{k-1} e^{-\frac{x_i}{\theta}}.$$ From there we get the log-likelihood $$ \ell = \ln(\mathcal{L}) = (k-1) \sum_{i=1}^{n} \ln(x_i) - \sum_{i=1}^{n} \frac{x_i}{\theta} - n\cdot \ln(\Gamma(k)) - nk\cdot\ln(\theta).$$ Differentiating with respect to $\theta$ and then solving for it yieds the Maximum-Likelihood Estimator $$\hat{\theta} = \frac{\bar{x}}{k}.$$
To find now an asymptotic 95%-Confidence Interval for $\hat{\theta}$ I first calculated the mean and variance of the Gamma(k,$\theta$) distribution, which are $$\mathbb{E}[X] = k\theta, \quad\quad \text{Var}(X)= k^2 \theta.$$ Given that the sample variance is unknown, I first calculated the variance of my estimate and then an estiamte of that variance: $\begin{align*} \text{Var}(\hat{\theta}) =& Var(\frac{1}{k} \frac{1}{n} \sum_{i=1}^{n}X_i) \\ =& \left(\frac{1}{kn} \right)^2 \sum_{i=1}^{n} \text{Var}(X_i) \\ \overset{iid}{=}& \left(\frac{1}{kn} \right)^2 \cdot n \cdot k\theta^2 \\ =& \frac{\theta^2}{kn} \; =: \sigma^2 \\ \hat{\sigma^2} =& \frac{\hat{\theta^2}}{kn} \\ =& \frac{(\frac{\bar{x}}{k})^2}{kn} \\ =& \frac{1}{kn} \cdot \frac{1}{k^2} \frac{1}{n^2} \left(\sum_{i=1}^{n}x_i\right)^2 \\ =& \frac{1}{k^3} \cdot \frac{1}{n^3} \left(\sum_{i=1}^{n}x_i\right)^2 \\ =& \frac{1}{k^3n} \bar{x}^2 \end{align*}$
From that I have constructed the asymptotic Confidence Interval
$\begin{array}{lrcccl} &z_{0.025} &\leq& \frac{\hat{\theta} - \mathbb{E}\left[\hat{\theta}\right]}{ \sqrt{ \frac{\hat{\sigma^2}}{n} } } &\leq& z_{0.975} \\ \Leftrightarrow & \hat{\theta} - z_{0.975} \sqrt{ \frac{\hat{\sigma^2}}{n} } &\leq& \theta &\leq& \hat{\theta} + z_{0.975} \sqrt{ \frac{\hat{\sigma^2}}{n} } \\ \Leftrightarrow & \frac{\bar{x}}{k} - 1.96 \sqrt{ \frac{\bar{x}^2}{k^3n^2}} &\leq& \theta &\leq& \frac{\bar{x}}{k} - 1.96 \sqrt{ \frac{\bar{x}^2}{k^3n^2}} \end{array}$
Although I can't find any mistakes (though admittedly I'm also not an expert in statistics), the results from calculating $\hat{\sigma^2}$ on, seem a bit strange to me. As I said, I would be very glad if someone could check whether I indeed made a mistake or if this can be solved somewhat more elegantly.