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A die is rolled $twice$. Let $X_i$ denote the outcome of the i-th roll, and put $S = X_1 + X_2$ and $D = |X_1 − X_2|$.

(a) Show that $E(SD) = E(S)E(D)$.

(b) Are $S$ and $D$ independent?

$Cov(S,D)=Cov(X_1+X_2,|X_2-X_1|)=$ I tried splitting to cases to remove the absolute value

$X_1X_2 ------- X_1=X_2 $

$Cov(X_1+X_2,X_2-X_1)+Cov(X_1+X_2,X_1-X_2)+Cov(X_1+X_2,0) = $

$Var(X_1)-Cov(X_1,X_2) + Cov(X_2,X_1) - Var (X_2) + Cov(X_1,X_2) - Var(X_1) + Var(X_2) - Cov(X_2,X_1) + 0 = 0 $

Now as $X_1,X_2$ are independent $Cov(X_1,X_2)=0$ so finally I get $Cov(S,D)=0$

$Cov(S,D)=E(SD)-E(S)E(D) = 0$

$E(SD)=E(S)E(D)$

Is this is the way to approach this question? How do I proceed?

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    For the second: if $D=0$ then $S$ has to be even. – 2017-01-13
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    So my solution to (a) is correct? Thanks alot! – 2017-01-13
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    I did not check the argument for $a$, but it should be a routine sum. Not hard to do it by simply summing over the $36$ pairs (which, I think, is what your calculation does). – 2017-01-13

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