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Suppose we have densities $f_1, f_2$ from the random variables $W_1$ and $W_2$ where $W_i$ has known mean $\mu_i$ and variance $\sigma_i$. Consider the mixture of the two densities $$ f(x;\theta)=\theta f_1(x) + (1-\theta)f_2(x)$$ with $0< \theta < 1$ the unknown mixing proportion. A random sample $X_1, ..., X_n$ from $X \sim f(x;\theta)$ is available.

I calculated the method of moment estimator to be $$\hat{\theta}_n = \frac{\bar{X_n}-\mu_2}{\mu_1-\mu_2}.$$ I was also able to prove that this estimator is unbiased.

I'm now stuck on how to find the variance, MSE and the asymptotic normality result of this estimator.

$\textbf{EDIT : }$ I think I was able to calculate the variance and the MSE (which is equal to the variance since $\hat{\theta}_n$ is unbiased). $$Var(\hat{\theta}_n) = \left(\frac{1}{n(\mu_1 - \mu_2)^2}\right) \sum^n_{i=1}(\theta^2 \sigma^2_1 + (1-\theta)^2 \sigma^2_2).$$ But I'm not sure about this calculation.

I also got the Fisher Information $$I(\theta) = \left(\frac{1}{\theta} + \frac{1}{1 - \theta}\right)^2$$ which I'm also not too sure about. But this could easily give me the asymptotic normality result.

I also have a second question: Consider $\sigma=\sigma_1=\sigma_2$ and denote $\Delta = \mu_1 - \mu_2$ and assume $\Delta > 0$ How could one find an approximation for the probability that the method of moment estimator $\hat{\theta}_n$ differs more from $\theta$ than a given amount $b > 0$?

Geigercounter
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  • A simple way to look at it is to find the variance and MSE of $\bar{X}$, which should be simple as it's a sample mean. The actual estimator is of the form $a\bar{X}+c$, so its variance etc. will just be $a^2$ times the variance of $\bar{X}$. As for asymptotic Normality, is the sample mean asymptotically Normal? If so, then any linear transformation of it is too... – jbowman Jan 09 '22 at 18:13
  • @jbowman so are my calculations correct? – Geigercounter Jan 09 '22 at 18:30
  • https://stats.stackexchange.com/questions/447626/mean-and-variance-of-a-mixture-distribution – jbowman Jan 09 '22 at 18:50
  • @jbowman that link was helpful thank you very much! I'm still however stuck on the variance, since we're dealing with the sample mean here... Also, do you have any ideas for my second question? – Geigercounter Jan 09 '22 at 20:19

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