Variance of the sum of two random variables

Question

Don't really understand one of the questions on my passed test. :

Suppose X and Y are random variables such that E(XY) = 0. Suppose also each of X and Y has mean 1 and variance 3. Find the variance of X + Y.

I know the answer is 4, but why?

Answer 1

Compute $\operatorname{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=-1$ and hence $\operatorname{Var}(X+Y)=\operatorname{Var}X+\operatorname{Var}Y+2\operatorname{Cov}(X,Y)$ gives the value 4.

Answer 2

Because $$\operatorname{Var}(X+Y)=\operatorname{Var}(X) + \operatorname{Var}(Y) + 2 \operatorname{Cov}(X,Y)$$ and also $$\operatorname{Cov}(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y]$$ and since $ \mathbb{E}[XY] = 0$ then $\operatorname{Cov}(X,Y) = -1$. Then simply subsitute the values into the first equation, yielding \begin{align}\operatorname{Var}(X+Y)&=\operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)\\ &= 3 \ + 3 \ +2(-1)\\ &= 4. \end{align}