# Show that if $x_{t+l}=Ax_t$, then $\rho(l)=1$ if $A>0$, and $\rho(l)=−1$ if $A<0$.

Given a single stationary series $$x_t$$ with zero mean and autocovariance function $$\gamma(h)$$ and autocorrelation function $$\rho(h)$$

Show that if $$x_{t+l}=Ax_t$$, then $$\rho(l)=1$$ if $$A>0$$, and $$\rho(l)=−1$$ if $$A<0$$.

What I have done is

$$\text{By definition we know that}\quad \rho(l) = \frac{\gamma(l)}{\gamma(0)}$$

So for $$x_{t+l}=Ax_t$$ we have

$$\gamma(l) = cov(x_t, x_{t+l}) = cov(x_t, Ax_t) = A\gamma(0)$$

and it follows that

$$\rho(l) = A\frac{\gamma(0)}{\gamma(0)} = A$$

which is not what I want. Where did I go wrong?

EDIT: Can I say that because of stationarity that

$$\frac{x_{t+l}}{x_t} = \pm 1$$

Because of stationarity

$$Var(x_t) = Var(x_{t+l}) = Var(Ax_t) = A^2Var(x_t)$$

which implies that $$A^2 = 1$$, and thus

$$\rho(l) = \sqrt{1} = \pm 1$$

• It's fine so far, but you're not finished. More explicitly, you should use the assumption of "stationary" to show $A^2=1$. Mar 11, 2021 at 17:13
• To your last question (in the edit): Yes, but it's not automatic. Try evaluating $\operatorname{Var}(x_{t+l})$ in two ways. Mar 11, 2021 at 17:15
• Thanks for the help. Mar 11, 2021 at 18:12

$$\text{By definition we know that}\quad \rho(l) = \frac{\gamma(l)}{\gamma(0)}$$

So for $$x_{t+l}=Ax_t$$ we have

$$\gamma(l) = cov(x_t, x_{t+l}) = cov(x_t, Ax_t) = A\gamma(0)$$

and it follows that

$$\rho(l) = A\frac{\gamma(0)}{\gamma(0)} = A$$

Because of stationarity

$$Var(x_t) = Var(x_{t+l}) = Var(Ax_t) = A^2Var(x_t)$$

which implies that $$A^2 = 1$$, and thus

$$\rho(l) = \sqrt{1} = \pm 1$$