I'm reading a book about time series analysis where an example for stationary is given:
$X_n$ independent with $X_i \sim N(0,1)$ if i is odd and $P(X = 1) = P(X = -1) = \frac{1}{2}$, if i is even.
$E[X_i] = 0$ for all i, which is clear.
What I don't understand is that $Cov(X_t, X_{t+h}) = 0$ for $h \neq 0$ and $= 1$ for $h = 0$.
How do I get this solution? Thanks!
For both cases, you can get the results by using the definition of the covariance, $\operatorname{Cov}(X,Y) = E[XY] - E[X] E[Y]$.
For the case of $h=0$ you might notice that $\operatorname{Cov}(X,X) = \operatorname{Var}(X)$ from the above. For the $h\ne 0$ case, use the independence. When two random variables $X$ and $Y$ are independent, $E[XY] = E[X]E[Y]$.