Deriving a property regarding variance. I am studying for the P-exam for actuaries and I've encountered a property that said,

$Var(x\pm y)=Var(x)+Var(y)$

I come from a math major and it has been years since I was taught statistics or probability, so I wanted to prove this using another property that I am comfortable using

$E[X^2]-(E[X])^2=Var(X)$

Using this definition, I actually got

$Var(x \pm y)= Var(x)+Var(y)\pm 2(E[XY]-E[X]E[Y])$

However, I was not able to confirm that the last two terms are equilavent to each other. I was thinking that independence in the random variables $X$ and $Y$ were the key, but my book is more of a problem set book, so it did not give me enough info to show it.
I also tried to look around online, but now the formula was too general and it was a bit above my head.
Can someone help me out ?
 A: The variance of a sum (or difference) is not necessarily the sum of the variances. But the result does hold if the two random variables are independent, or, more weakly, uncorrelated. 
To prove the result (say for sums) in the case $X$ and $Y$ are independent, we can proceed as you did. If $X$ and $Y$ are independent, then $E(XY)=E(X)E(Y)$, which finishes things. 
A: It is $Var(X\pm Y)=Var(X)+ Var(Y)\pm 2Cov(X,Y)$ .
If two variables are uncorellated then $Cov(X,Y)=0$ (Note that if two variables are independent then they are uncorellated but not the opposite)
For the covariance you can find more information here
A: Variance is a special case of covariance. The covariance operator is multilinear and symmetric, so you can write more generally that
$$\text{Var}(aX+bY) = \text{Cov}(aX+bY,aX+bY)=a\text{Cov}(X,aX+bY)+b\text{Cov}(Y,aX+bY)$$
$$=a^2\text{Cov}(X,X)+ab\text{Cov}(X,Y)+ab\text{Cov}(Y,X)+b^2\text{Cov}(Y,Y)$$
$$=a^2\text{Var}(X) + b^2\text{Var}(Y) + 2ab\text{Cov}(X,Y)$$
That is,
$$\boxed{\text{Var}(aX+bY) = a^2\text{Var}(X) + b^2\text{Var}(Y) + 2ab\text{Cov}(X,Y)}$$
where $\text{Cov}(X,Y) \stackrel{\text{def}}{=} \text{E}(XY) - \text{E}(X)\text{E}(Y)$ and $\text{Var}(X)\stackrel{\text{def}}{=}\text{Cov}(X,X)$.
Take $a=1$ and $b=\pm 1$ to get the result you want.
