$X_n$ is a discrete random process. $P(X_n=2)=P(X_n=-2)=0.5$ for all $n$ Find autocovariance function of $X_n$.

Question

Find the autocovariance function of $X_n$ $$E[X_n]=E[X(n)]=0.5\cdot2+0.5\cdot (-2)=0$$ $$E[X_{n}^2]=E[[X(n)]^2]=0.5\cdot2^2+0.5(-2)^2=4$$ The autocorrelation function $$R_{XX}(n_1,n_2)=E[X(n_1)X(n_2)]=E[X(n_1)(X(n_1)+X(n_2)-X(n_1))]=E[X^2(n_1)]-E[X(n_1)]E[X(n_2)-X(n_1)]=E[X^2(n_1)]=4$$ assuming that $n_2>n_1$ What is next?

Answer 1

For $m>0$: $$\begin{align}E[X(n)X^*(n+m)]=(2)(2)(0.5)(0.5)+(-2)(2)(0.5)(0.5)+(2)(-2)(0.5)(0.5)+(-2)(-2)(0.5)(0.5)=0\end{align}$$ When $m=0$:

$E[X(n)X(n+m)]=E[|X(n)|^2]=0.5(-2)^2+0.5(2)^2=4$

Therefore, $$R_{XX}(n,n+m)=R_{XX}(m)=4\delta(m)$$ which means the process is uncorrelated.

Since $$E[X]=0.5(2)+0.5(-2)=0$$

The autocovariance function $$C_{XX}(m)=R_{XX}(m)-(E[X])^2=4\delta(m)$$