If $V_1$, $V_2$, $\dots$ , $V_n$ and $W_1$, $W_2$, $\dots$ , $W_n$ are independent random samples of size $n$ from normal populations with the means $\mu_1 = \alpha + \beta$ and $\mu_2 = \alpha - \beta$ and the common variance $\sigma^2 = 1$, find maximum likelihood estimators for $\alpha$ and $\beta$.
Is it enough to find the MLE for both $\mu _1$ and $\mu_2$ and add the solutions to obtain $2\alpha$? I'd like a hint on how to do this problem. Thanks.
On
In this situation, your suggested plan will work. Alternatively, you can just write the log likelihood and check that it does end up that way.
The log likelihood (up to some constant additive terms) is $$-\frac{1}{2}\sum_{i=1}^n (V_i - (\alpha + \beta))^2 - \frac{1}{2}\sum_{i=1}^n (W_i - (\alpha - \beta))^2.$$
Computing the partial derivatives with respect to $\alpha$ and $\beta$ and setting them to zero yields $$\alpha = \frac{1}{2n} \sum_{i=1}^n V_i + \frac{1}{2n} \sum_{i=1}^n W_i$$ $$\beta = \frac{1}{2n} \sum_{i=1}^n V_i - \frac{1}{2n} \sum_{i=1}^n W_i$$
As you can see, this is the same result as if you computed the MLEs of $\mu_1$ and $\mu_2$ and performed the necessary linear transformations to get estimates of $\alpha$ and $\beta$.
You can use the standard approach, just take alpha and beta as the parameters instead of $\mu _1$ and $\mu_2$. Also maximize the likelihood for the full data that is product of likelihood of Vs and Ws. This is because if you are given alpha and beta then pdf for both distribution of Vs as well as Ws are defined. My guess is the answer you get for alpha will be same as $\frac{\mu_1 + \mu_2}{2}$. However, more interesting case is when you have m samples for Vs and n samples for Ws. This time the answer will be some kind of weighted average.