$\operatorname{Var}(X+Y)=\operatorname {Var}(X)+\operatorname {Var} (Y)$ and $\operatorname{Var}(X-Y)=\operatorname {Var}(X)+\operatorname {Var} (Y)$

Question

$\operatorname{Var}(X+Y)=\operatorname {Var}(X)+\operatorname {Var} (Y)$ and $\operatorname{Var}(X-Y)=\operatorname {Var}(X)+\operatorname {Var} (Y)$ is also equal to $\operatorname {Var}(X)+\operatorname {Var} (Y)$, What is the main reason ? No need of proof. Thanks

Answer 1

Your statement is true only if X and Y are independent. In general,$$Var(X\pm Y)= Var(X) + Var (Y)\pm 2Cov(X,Y).$$If $X$ and $Y $ are independent $Cov(X,Y)=0$ and you will get $$Var(X\pm Y)= Var(X) + Var (Y).$$Since you don't want any proof, I'm skipping it.

Think of variance as being like “norm-squared” and independence as being the orthogonality.

Then you can see that the above mentioned formulas resemble the law of cosines and Pythagorean Theorem respectively.