Suppose that $X_1,\ldots,X_n$ are normal with mean $\mu_1$; $Y_1,\ldots,Y_n$ are normal with mean $\mu_2$; and $W_1,\ldots,W_n$ are normal with mean $\mu_1+\mu_2$. Assuming that all $3n$ random variables are independent, with a common variance, find the maximum likelihood estimators of $\mu_1$ and $\mu_2$.
Solving for $\mu_1$ using $X_1,\ldots,X_n$ I got $\mu_1$ = the sample mean of $X$. Similarly, solving for $\mu_2$ using $Y_1,\ldots,Y_2$ I got $\mu_2$ = the sample mean of $Y$. However, I'm not sure if that's what this question meant, especially since I don't understand what the purpose of giving the mean of $W_1,\ldots,W_n$ is.
Best Answer
The likelihood function based on $X_1,\ldots,X_n,Y_1,\ldots,Y_n,W_1,\ldots,W_n$ is given by $$ L(\mu_1,\mu_2,\sigma^2)=\left[\prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(x_i-\mu_1)^2\right)\right]\times\left[\prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(y_i-\mu_2)^2\right)\right]\times \left[\prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(w_i-(\mu_1+\mu_2))^2\right)\right] $$
Then the log-likelihood function is $$ l(\mu_1,\mu_2,\sigma^2)=-\frac{3n}{2}\log(2\pi\sigma^2)-\frac{1}{2\sigma^2}\left(\sum_{i=1}^n(x_i-\mu_1)^2+\sum_{i=1}^n(y_i-\mu_2)^2+\sum_{i=1}^n(w_i-\mu_1-\mu_2)^2\right) $$ and so $$ \frac{\partial l}{\partial \sigma^2}(\mu_1,\mu_2,\sigma^2)=-\frac{3n}{2\sigma^2}+\frac{1}{2(\sigma^2)^2}\left(\sum_{i=1}^n(x_i-\mu_1)^2+\sum_{i=1}^n(y_i-\mu_2)^2+\sum_{i=1}^n(w_i-\mu_1-\mu_2)^2\right). $$ This is equal to $0$ if and only if $$ \sigma^2=\hat{\sigma}^2=\frac{1}{3n}\left(\sum_{i=1}^n(x_i-\mu_1)^2+\sum_{i=1}^n(y_i-\mu_2)^2+\sum_{i=1}^n(w_i-\mu_1-\mu_2)^2\right). $$ Thus $$ l(\mu_1,\mu_2,\sigma^2)\leq l(\mu_1,\mu_2,\hat{\sigma}^2)=-\frac{3n}{2}\log(2\pi\hat{\sigma}^2)-\frac{3n}{2}. $$ Now maximize this expression with respect to $\mu_1$ and $\mu_2$.