Computing the covariance function

Question

I have an exercise, and the answer is given, but I can't figure out how they get to this answer. The question is: Let $e_0,e_1, . . .$ be a sequence of independent, identically distributed random variables with mean m and variance $\sigma^2$. Let $\{X_t , t = 1, \dots\}$ be the stochastic process defined by $X_t = 1.2e_t + 0.9e_{t−1}$. Compute the mean $m_X (t) = E[X(t)]$, and the covariance function $r_X (s, t) = C[X_s, X_t ]$. Show that $\{X_t\}$ is weakly stationary.

The part of the covariance function I just can't figure out. I'd think that I should compute it with this one: $\operatorname{Cov}(X_s, X_t)=E[X_s*X_t]−E[X_s]⋅E[X_t]$. This, however, won't give the answer I'm looking for.

This should be the solution

I hope someone can explain this to me!

Answer 1

We have $X_t=ae_t+be_{t-1}$ with $a=1.2$ and $b=0.9$. Let $s\leqslant t$. We will compute $c_{s,t}:=\operatorname{Cov}(X_sX_t)$. By linearity in each argument, $$ c_{s,t}=a^2\operatorname{Cov}(e_s,e_t)+ab\operatorname{Cov}(e_se_{t-1})+ab\operatorname{Cov}(e_{s-1}e_t)+b^2\operatorname{Cov}(e_{s-1},e_{t-1})=A+B+C+D $$

• If $s=t$, then $B=C=0$ (by independence) and $A=D=\sigma^2$.
• If $t=s+1$, then $A=D=C=0$ (by independence again) and $B=\sigma^2$.
• If $t\geqslant s+2$, then $A=B=C=D=0$.