1. ## MVUE help

Question:

Let $\bar{X_1}$ and $\bar{X_2}$ be the means of two independent random samples of size $n_1 > 1$ and $n_2 > 1$ from an infinite population that has mean $\mu$ and finite variances $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$ where $\sigma_{1}^{2}>0, \sigma_{2}^{2}>0, \sigma_{1}^{2} \neq \sigma_{2}^{2}$.

If we want $\hat\theta=w \bar{X_1}+(1-w) \bar{X_2}$ to be the minimum variance unbiased estimator of $\mu$, what must $w$ be?

My work:

$V \hat\theta = w^2 \frac{\sigma_{2}^{1}}{n_1}+(1-w)^2 \frac{\sigma_{2}^{2}}{n_2}$

Let $\frac{dV \hat\theta}{dw} = 0:$

$2w\frac{\sigma_{2}^{1}}{n_1}-2(1-w)\frac{\sigma_{2}^{2}}{n_2}=0$

$2w \left(\frac{\sigma_{2}^{1}}{n_1}+\frac{\sigma_{2}^ {2}}{n_2} \right)=2 \frac{\sigma_{2}^{2}}{n_2}$

$w=\frac{ \frac{ \sigma_{2}^{2} }{n_2} } {\left( \frac{\sigma_{2}^{1}}{n_1}+\frac{\sigma_{2}^{2}}{n _2} \right)}$

Look good?

2. you messed up on the sub/super scripts of $\sigma_1^2$

Otherwise, it looks ok.
But you should obtain the second derivative to prove that you have a min.