In my exercise, I'm given two independent discrete stochastic variables, with the probability function;
$$ p_{X}\left(x\right)=\left\{ \begin{array}{cc} \frac{1}{4} & \text{if } x = -1 \\ \frac{1}{4} & \text{if } x = 0 \\ \frac{1}{2} & \text{if } x = 1 \\ 0 & \text{otherwise} \end{array} \right. $$
I'm given the task to find several properties, for instance the mean ($EX = \frac{1}{4}$), and variance ($VarX=\frac{11}{16}$). I was able to do this with any issues.
Now I'm given the task of calculating, the variance; $Var(X-5Y)$, based upon the fact that independent variables have a covariance of 0, I've found this to be; $Var(X-5Y)=\frac{143}{8}$.
However when I'm trying to confirm that the covariance is 0, I get in trouble; $$ Cov(X,Y) = E(XY) - EX \cdot EY $$ With; $$E(XY) = \left(\frac{1}{4} \cdot -1 \cdot -1 \right) + \left(\frac{1}{4} \cdot 0 \cdot 0 \right) + \left(\frac{1}{2} \cdot 1 \cdot 1 \right) = \frac{3}{4}$$ I'm getting; $$ Cov(X,Y) = \frac{3}{4} - \frac{1}{4} \cdot \frac{1}{4} = \frac{11}{16} $$ Clearly I'm doing something wrong, but what?
The expected value of the product of two random variables is obtained by summing over all possible pairs of outcomes,
$$E(XY) = \sum_{y \in \Omega_Y}\sum_{x\in\Omega_X} P(X=x,Y=y)\cdot x\cdot y.$$
In the particular case of two independent random variables, the joint probability of a pair is the product of the probabilities of the components,
$$P(X=x,Y=y) = P(X=x) \cdot P(Y=y),$$
and then the sum splits
$$\begin{align}E(XY) &= \sum_{y \in \Omega_Y}\sum_{x\in\Omega_X} P(X=x)P(Y=y)\cdot x\cdot y\\ &= \sum_{x\in\Omega_X} P(X=x)\cdot x \cdot \sum_{y\in\Omega_Y} P(Y=y)\cdot y\\ &= E(X)\cdot E(Y). \end{align}$$
For two inependent identically distributed variables with the given distribution, that is
$$\begin{align}E(XY) &= P(X=1,Y=1)\cdot1 + P(X=-1,Y=-1)\cdot 1\\ &\qquad + P(X=1,Y=-1)\cdot(-1) + P(X=-1,Y=1)\cdot (-1)\\ &= \left(\frac{1}{2^2} + \frac{1}{4^2}\right)\cdot 1 + \left(\frac12\frac14 + \frac14\frac12\right)\cdot (-1)\\ &= \frac{1}{16} \end{align}$$
by the explicit summation (ignoring $0$ terms).