For a special case of the gamma distribution:
$$f(x)=\frac{x}{\theta^2}e^{-x/ \theta}$$
$$E(x) = 2\theta\\ V(x) = 2\theta^2$$
I find MLE of $\hat{\theta}$ to be $\sum_{i=1}^n \frac{x_i}{2n}$
To show consistency, I'd like to show that:
- $\lim_{n\to \infty} E(\hat{\theta}) = \theta$
- $\lim_{n\to \infty} V(\hat{\theta}) = 0$
Focusing on $\lim_{n\to \infty} V(\hat{\theta}) = 0$:
$V(\sum_{i=1}^n \frac{x_i}{2n}) = \frac{1}{4n^2}V(\sum_{i=1}^n x_i)$
I'm somewhat stuck here, it seems obvious that this limit approaches, $0$, however, I'm not sure if I need to break the variance further (say $V(x) = E(x^2)-E(x)^2$) to prove this
I'd appreciate any help!
I assume that $x_1,\dots,x_n$ are independent, resulting in: \begin{align*} \lim_{n \to \infty} V(\hat{\theta}) = \lim_{n \to \infty} V(\sum_{i = 1}^n \frac{x_i}{2n}) = \lim_{n \to \infty} \frac{1}{4n^2}\sum_{i = 1}^n V(x_i) = \lim_{n \to \infty} \frac{1}{4n^2}\sum_{i = 1}^n 2\theta^2 = \lim_{n \to \infty} \frac{\theta^2}{2n} = 0 \end{align*}