posterior distribution after having partial information on some linear combinations of unknown variables (Revised) $x_1$, $x_2$, and $x_3$ are i.i.d. normal random variables with distribution $N(0, \sigma_x^{2})$
$\epsilon_1$, $\epsilon_2$, and $\epsilon_3$ are i.i.d. normal random variables with distribution $N(0, \sigma_n^{2})$
lets define : 
$y_1 = x_1 + \epsilon_1$
$y_2 = x_2 + \epsilon_2$
$y_3 = x_3 + \epsilon_3$
given the values for :
$s_1 = y_1 + y_2$
$s_2 = y_1 + y_3$
what is the posterior distribution $f(x_1,x_2,x_3|s_1,s_2)$ ? 
what is the ML estimate for $x_1$, $x_2$, and $x_3$ ?
as an example, assume  $\sigma_x=2, \sigma_n=1, s_1=2, s_2=1$ , we are interested in an estimate for $x_1$, $x_2$, and $x_3$. 
Thanks,
MG
 A: (Edited: I seemed to have forgotten an $\epsilon$ in my earlier answer and would now say) 
$\hat{x}_1 = \frac{ 4 }{5}$, 
$\hat{x}_2 = \frac{ 4 }{5}$, and 
$\hat{x}_3 = 0 $, with 
$\hat{\epsilon}_1 = \frac{ 1 }{5}$, 
$\hat{\epsilon}_2 = \frac{ 1 }{5}$, and 
$\hat{\epsilon}_3 = 0 $.
Generalising further with some justification added as a response to a comment:
This stat.SE question and answer gives an explanation of how to derive this. We will take all means to be zero.  
The slightly harder part is estimating $Y_1$,$Y_2$ and $Y_3$.  They each have variance $\sigma_y^2=\sigma_x^2+\sigma_{\epsilon}^2 $.  So $S_1$ and $S_2$ each have variance $2\sigma_y^2$ and have covariance $\sigma_y^2$.  Let's define 
$$T_1 = 2Y_1$$ $$T_2=Y_2+Y_3$$ $$T_3 = S_1+S_2 = 2Y_1 +Y_2+Y_3=T_1+T_2 $$ $$T_4=S_1-S_2=Y_2-Y_3$$  
so $T_1$ has variance $4\sigma_y^2$ and $T_2$ has variance $2\sigma_y^2$  and they are uncorrelated.  This means $T_3$ has variance $6\sigma_y^2$ and $T_4$ has variance $4\sigma_y^2$ and they too are uncorrelated.  
The observation of $s_1$ and $s_2$ tells us now that $t_3 = s_1 + s_2$.  From this we can get maximum likelihood estimates of $t_1$ and $t_2$ in proportion to their variances so 
$$\hat{t}_1 = \frac{2(s_1 + s_2)}{3}$$ 
$$\hat{t}_2 = \frac{s_1 + s_2}{3}$$
but since $2Y_1 = T_1$ this gives   
$$\hat{y}_1 = \frac{s_1 + s_2}{3}$$
and we also have from the observation $t_4 = s_1 - s_2$, and from the definitions we have  $2Y_2=T_2+T_4$ and $2Y_3=T_2-T_4$, giving us  
$$\hat{y}_2 = \frac{2s_1-s_2}{3} $$
$$\hat{y}_3 = \frac{2s_2-s_1}{3} .$$
Getting from $\hat{y}_i$ to $\hat{x}_i$ is easier, since from the stat.SE question, we simply do this in proportion to variance, so  $\hat{x}_i = \hat{y}_i \frac{\sigma_x^2}{\sigma_x^2+\sigma_{\epsilon}^2}$ giving 
$$\hat{x}_1 = \frac{(s_1 + s_2)\sigma_x^2}{3(\sigma_x^2+\sigma_{\epsilon}^2)}$$
$$\hat{x}_2 = \frac{(2s_1-s_2)\sigma_x^2}{3(\sigma_x^2+\sigma_{\epsilon}^2)} $$
$$\hat{x}_3 = \frac{(2s_2-s_1)\sigma_x^2}{3(\sigma_x^2+\sigma_{\epsilon}^2)} $$
and 
$$\hat{\epsilon}_i = \hat{x}_i \frac{\sigma_{\epsilon}^2}{\sigma_x^2}.$$
