Explanation for Additive Property of Variance? I'm wondering why variance has additive property, as opposed to why this property doesn't extend to standard deviation? Additive property is defined as:
Var(A+B) = Var(A) + Var(B)
I imagine this as adding two distribution together which makes sense. But in that case SD should have similar property as well. Why does variance possess this magical property?
 A: It doesn't!
In general:
Var(A+B) = Var(A) + Var(B) + Cov(A, B)

The additive property only holds if the two random variables have no covariation.  This is almost a circular statement, since a legitimate definition of the covariation could be:
Cov(A, B) = Var(A) + Var(B) - Var(A + B)

This means that the covariance measures the failure of the additive property of variance.
This leads to the true heart of the matter, the covariance is bi-linear:
Cov(A_1 + A_2, B) = Cov(A_1, B) + Cov(A_2, B)
Cov(A, B_1 + B_2) = Cov(A, B_1) + Cov(A, B_2)

For an intuitive understanding of this, I'll link to the wonderful: How would you explain covariance to someone who understands only the mean?.  In particular, see @whuber's answer.
A: 
The first thing to notice is that Var(A+B) equals VarA + Var B only when Cov(A,B)=0.
To gain some intuition behind the relationship between sd(A+B) and sd(A)+sd(B), notice that in order to complete the square in this expression  
Cov(A,B) would have to equal sd(A)*sd(B). The next question is whether that ever happens? Indeed it does. The Cauchy Schwartz inequality gives us the inequality below:
. 
Whenever the following equality holds 
, we can complete the square and obtain that sd(A+B)=sd(A)+sd(B). However, in all other cases, sd(A+B) will not equal sd(A)+sd(B).
