2
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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?

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    $\begingroup$ I don't see what you've done wrong here. $\endgroup$ – Don Thousand Mar 28 at 18:42
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    $\begingroup$ 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. $\endgroup$ – amsmath Mar 28 at 19:07
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    $\begingroup$ Your answer is correct, the "given" one is wrong. $\endgroup$ – Robert Israel Mar 28 at 19:55
  • $\begingroup$ Ok thanks guys. $\endgroup$ – user601297 Mar 28 at 19:57

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