Get standard deviation of a distribution made by subtracting two distributions with known standard deviations I have some data 
D1 (N=4)    D2 (N=4)
Unfortunately, I don't have the data points. I only have the statistics:
(mean d1,mean d2) = (2.5,8)
(SD1,SD2 (standard deviation))= (1.290994449,1.414213562)
I can compute:
   (SE1,    SE2 (standard error)) = (0.322748612,0.353553391)
My problem is that I need to find the mean and standard deviation of D1-D2 = d11-d21,d12-d22,d13-d23,d14-d24.
The mean is easy: meand1-meand2.
But how to find the standard deviation of the new distribution??? 
Now, 
D1
3
2
1
4
D2
8
7
7
10
Diff
5
5
6
6
meanDiff
5.5
SDDiff
0.577350269
seDiff
0.144337567
but 
sqrt(1.290994449* 1.290994449 + 1.414213562* 1.414213562) = 1.914854216 != 0.577350269
 A: I think this is not possible without original data. Imagine that we know the original data and we change the order of data in the first group. After that mean and standard deviation of this group do not change but differences $d_{11}-d_{21},d_{12}-d_{22},d_{13}-d_{23},d_{14}-d_{24}$ change (because of different order  ).
A: If the two random variables are independent with means $\mu_{D_1}$ and $\mu_{D_2}$ and standard deviations  $\sigma_{D_1}$ and $\sigma_{D_2}$ then the mean of the difference is the difference of the means, as you have said, so   $\mu_{D_1-D_2}=\mu_{D_1}-\mu_{D_2}$, while the variance of the difference is the sum of the variances $\sigma^2_{D_1-D_2}=\sigma^2_{D_1}+\sigma^2_{D_2}$, so the standard deviation of the difference is the square-root of this $$\sigma_{D_1-D_2}=\sqrt{\sigma^2_{D_1}+\sigma^2_{D_2}}.$$
If they are not independent (if you are using sample data then the sample data is unlikely to be uncorrelated) than you cannot say this, though you can if you think appropriate assume it about the underlying distributions. 
