I've got following scenario:
I have two observations with $\mu$ unknown:
$$ X_a \sim N(\mu, \sigma_a), X_b \sim N(\mu, \sigma_b) $$
Both are observation for the same quantity $X$, just measured with different systems. I want to estimate the parameters of $X \sim N(\mu, \sigma)$ based on the two observation.
For that I took a basic MLE to estimate $\mu$:
$$ L = \frac{1}{\sqrt{2 \pi \sigma^2_a}} exp(\frac{-(x_a - \mu)^2}{2 \sigma_a^2}) \frac{1}{\sqrt{2 \pi \sigma^2_b}} exp(\frac{-(x_b - \mu)^2}{2 \sigma_b^2}) $$
This basically yields the equation of a weighted mean: $$\mu = \frac{\sigma^2_a x_b + \sigma^2_b x_a}{\sigma^2_a + \sigma^2_b}$$
Now I want to calculate the standard deviation of $X$, but somehow I am not sure how to approach this. Can I just assume (based on $\operatorname{VAR}(Z) = \operatorname{VAR}(aY) = a^2 \operatorname{VAR}(Y)$), that: $$ \sigma^2 = var = (\frac{\sigma^2_a}{\sigma^2_a + \sigma^2_b})^2 \sigma^2_b + (\frac{\sigma^2_b}{\sigma^2_a + \sigma^2_b})^2 \sigma^2_a $$
Or am I mistaken about that?
Thanking you in advance!
Let $(X_1,X_2,...,X_n)$ be independent Gaussian rvs s.t. $X_k\sim \mathcal{N}(\mu,\sigma_k^2)$.
(1) We calculate the MLE of $\mu$: we set the first order condition and we obtain the estimator $$\frac{\partial}{\partial \mu}\ln(L(x_1,...,x_n;\mu))=\sum_{1\leq k \leq n}\frac{x_k-\mu}{\sigma_k^2}=0$$ $$\implies\hat{\mu}_{\textrm{MLE}}(X_1,...,X_n)=\sum_{1\leq k \leq n}\frac{X_k}{\sigma_k^2}\bigg/\sum_{1\leq k \leq n}\frac{1}{\sigma_k^2}$$ (2) Denote the denominator with $\gamma:=\sum_{1\leq k \leq n}\frac{1}{\sigma_k^2}$. We get, by independence $$V[\hat{\mu}_{\textrm{MLE}}(X_1,...,X_n)]=\gamma^{-2}V\bigg[\sum_{1\leq k \leq n}\frac{X_k}{\sigma_k^2}\bigg]=\gamma^{-2}\sum_{1\leq k \leq n}\frac{\sigma_k^2}{\sigma_k^4}=\gamma^{-2}\gamma=\gamma^{-1}$$ (3) By using your case: $$\frac{\sigma_a^4\sigma_b^2+\sigma_b^4\sigma_a^2}{(\sigma_a^2+\sigma_b^2)^2}=\frac{\sigma_b^2+\sigma_b^4/\sigma_a^2}{(1+\sigma_b^2/\sigma_a^2)^2}=\frac{1/\sigma_b^2+1/\sigma_a^2}{(1/\sigma_b^2+1/\sigma_a^2)^2}=\frac{1}{1/\sigma_a^2+1/\sigma_b^2}$$ as expected.