Compute covariance matrix random walk 
Consider a random walk on the square lattice $\mathbb{Z}^2$ with diagonal jumps of size $2$, i.e. the jump probabilities are
$$P(X_1 = x) =
  \begin{cases}
    \frac{1}{4} & \quad \text{if } x \in \{(2,2), (-2,2), (2,-2), (-2,-2) \} \\
     0 & \quad \text{otherwise } \\
  \end{cases}$$
Compute the covariance matrix $(\operatorname{Cov}(X_1^i, X_1^j))_{i,j=1,2}$, where $X_1^{(i)}$ denotes the $i$-th component of $X_1$.

In my study material is written:
$$\operatorname{Cov}(X_1^{(i)}, X_1^{(j)}) = E[X_1^{(i)} X_1^{(j)}] = \sum_{x \in \mathbb{Z}^d} x^{(i)} x^{(j)} P(X_1 = x) = 
\begin{cases}
 \frac{1}{d} & \quad \text{if } i =j, \\
  0 & \quad \text{if } i \neq j, \\
\end{cases}$$
since $E[X_1]$ is the $d$-dimensional null-vector.
For $x \in \{(1,1), (1,2), (2,1)\}$ we have  $P(X_1 = x) =0$.
For $x = (2,2)$ we have then in the covariance matrix $A$: 
$$a_{22} = 2 \cdot 2 \cdot \frac{1}{4} = 1$$
This will give the covariance matrix
$$A = \begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix}$$
Is my way of computing correct?

Suppose that $i=1$ and $j=2$ and $d=2$.  Then we have
\begin{align}
& \sum_{x \in \mathbb{Z}^d} x^{(i)} x^{(j)} P(X_1 = x) = \sum_{x \in \mathbb{Z}^2} x^{(1)} x^{(2)} P(X_1 = x) \\[10pt]
= {} & \underbrace{2\cdot 2\cdot \frac 1 4}_{x \,=\, (2,2)} + \underbrace{2 \cdot(-2)\cdot \frac 1 4}_{x \,=\, (2,-2)} + \underbrace{(-2)\cdot2\cdot\frac 1 4}_{x \,=\, (-2,2)} + \underbrace{(-2)\cdot(-2)\cdot\frac 1 4}_{x \,=\, (-2,-2)} + \underbrace{\text{terms equal to 0}}_{x\,\in\,\mathbb Z^2 \,\setminus\, (\{\pm2\}\times\{\pm2\}) } \\[12pt]
= {} & 0.
\end{align}
By symmetry of covariance matrices, we get the same thing if $i=2$ and $j=1$.  Thus the matrix is
$$
\begin{bmatrix} \text{?} & 0 \\ 0 & \text{?} \end{bmatrix}.
$$
The two question marks would be filled in with the sum
$$
2^2\cdot \frac 1 4 + (-2)^2\cdot \frac 1 4 + (-2)^2 \cdot \frac 1 4 + 2^2 \cdot \frac 1 4 = 4,
$$
so we get
$$
\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}.
$$
