# Variance of sum of two uniform RV

Let $$X$$ and $$Y$$ be two independent random variables, each uniformly distributed on $$[-1,1],$$ then find $$\operatorname{Var}(X+Y).$$

My attempt : $$\operatorname{Var}(X+Y) =\operatorname{Var}(X) + \operatorname{Var}(Y) +2\operatorname{Cov}(X,Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) = 2\operatorname{Var}(X) = \frac{2}{3}.$$

But the answer given is $$2;$$ what am I missing?

• I don't see what you've done wrong here. – Don Thousand Mar 28 at 18:42
• I also get $2/3$. First, I calculated $E(X) = 0$ and $E(X^2) = 1/3$. Hence, $Var(X) = 1/3$. So, $Var(X+Y) = E((X+Y)^2)) = E(X^2 + 2XY + Y^2) = 2/3 + 2E(XY)$. And then I read that $E(XY) = 0$ since $X$ and $Y$ are independent. – amsmath Mar 28 at 19:07
• Your answer is correct, the "given" one is wrong. – Robert Israel Mar 28 at 19:55
• Ok thanks guys. – user601297 Mar 28 at 19:57