Correct formula for calculation covariances

Question

Let $X=(Y_1,Y_2),Y_1,Y_2 \,$ i.i.d $\sim U\{-1,0,1\}$.

I want to calculate $\operatorname{Cov}(Y_1,Y_2)$

$\operatorname E(Y_1)=\operatorname E(Y_2)=0$ and $\operatorname{Var}(Y_1)=\operatorname{Var}(Y_2)=\frac23$.

Then we obtain $$\operatorname{Cov}(Y_1,Y_2)=\operatorname E(Y_1Y_2)-\operatorname E(Y_1)\operatorname E(Y_2)=\frac13 \bigr((-1 \cdot 0)+(-1\cdot -1)+(-1\cdot1)\bigl )+ \frac13 \bigr((0 \cdot -1)+(0\cdot 0)+(0\cdot1)\bigl )+\frac13 \bigr((1 \cdot -1)+(1\cdot 0)+(1\cdot1)\bigl )-0=0$$

I am wondering if I used the formula to calculate this covariance correctly. Some help is welcome!

Answer 1

$$\text{Cov}(Y_1,Y_2) = \Bbb E [Y_1Y_2] - \Bbb E [Y_1] \Bbb E [Y_2] = \sum_{i=-1}^1 \sum_{k=-1}^1 ik \Bbb P (Y_1 = i, Y_2 = k) - 0 \\ = \sum_{i=-1}^1 i \sum_{k=-1}^1 k \underbrace{\Bbb P (Y_1 = i)}_{= \frac 1 3} \underbrace{\Bbb P( Y_2 = k)}_{= \frac 1 3} \quad \text{(because $Y_1$ and $Y_2$ are independent)} \\ = \frac 1 9 ((-1)\cdot \underbrace{(-1 + 0 +1)}_{=0} + \underbrace{0\cdot(-1 +0 +1)}_{=0} + 1\cdot \underbrace{(-1 +0 +1)}_{=0}) = 0$$

Answer 2

Since $Y_1$ and $Y_2$ are independent (this is the first "i" in "i.i.d."), their covariance is 0.