# Autocovariance for an explosive AR(1) process

I am struggling to understand how the below result of the autocovariance of an explosive AR(1) process is derived, taken from Time Series Analysis and its Applications (R. H. Shumway & D. S. Stoffer 2ed).

As I understand it, we can represent the covariance as

$$\mathbb{E}[(-\phi^{-1}w_{t+h+1}-\phi^{-2}w_{t+h+2}-...)(-\phi^{-1}w_{t+1}-...-\phi^{-h-1}w_{t+h+1}-\phi^{-h-2}w_{t+h+2}-...)]$$ Which then works out to $$\phi^{-1}\phi^{-h-1}\sigma_w^2+\phi^{-2}\phi^{-h-2}\sigma_w^2+...$$ And hence $$\sum\limits_{j = 1}^\infty {\phi^{-j}\phi^{-h-j}\sigma_w^2} = \sigma_w^2\phi^{-h}\sum\limits_{j = 1}^\infty {(\phi^{-2})^j}$$ Which we can treat as a geometric series, using

$$\sum\limits_{j=1}^\infty{a_jr^j}=\frac{a_1}{1-r}$$

However, I'm not sure we can do this because the limits aren't the same. Even if we did I keep getting $$\sigma_w^2\phi^{-h}\sum\limits_{j = 1}^\infty {(\phi^{-2})^j} = \frac{\sigma_w^2\phi^{-h}}{1-\phi^{-2}}$$ So I'm not sure where the additional $$\phi^{-2}$$ is coming from in the numerator of the textbook's result.

#### Forward-recursion uses a different variance than backwards-recursion

The essence of the issue here is that forward-recursion of the time-series uses a different variance than backwards-recursion. Your own working is quite sloppy, since you don't even use equation statements for your steps. In any case, the simplest way to proceed is to divide both sides of the original AR statement by $$\phi$$ to get the alternative form:

$$x_{t-1} = \phi^{-1} x_t - \phi^{-1} w_t.$$

Consequently, taking $${x}_t^* \equiv x_{-t}$$, $${\phi}^* \equiv \phi^{-1}$$ and $${w}_t^* \equiv - \phi^{-1} w_t \sim \text{IID N}(0, \sigma_w^2 \phi^{-2})$$ allows us to write the model in the stationary $$\text{AR}(1)$$ form:

$${x}_t^* = {\phi}^* {x}^*_{t-1} + {w}_t^* \quad \quad \text{with} \quad \quad |{\phi}^*|<1.$$

Using the standard result for the auto-correlation of a stationary $$\text{AR}(1)$$ process we then have:

\begin{align} \gamma_x(h) &= \mathbb{C}(x_t, x_{t+h}) \\[14pt] &= \mathbb{C}({x}_{-t}^*, {x}_{-t-h}^*) \\[14pt] &= \mathbb{C}({x}_{t+h}^*, {x}_{t}^*) \\[10pt] &= \frac{{\phi}^{*h}}{1 - {\phi}^{*2}} \cdot \mathbb{V}(w_t^*) \\[6pt] &= \frac{{\phi}^{-h}}{1 - {\phi}^{-2}} \cdot \sigma_w^2 \phi^{-2} \\[6pt] &= \frac{\sigma_w^2 {\phi}^{-2-h}}{1 - {\phi}^{-2}}. \\[6pt] \end{align}

As you can see, the additional term $$\phi^{-2}$$ in the numerator comes from the fact that the error term in the stationary form of the model has variance $$\sigma_w^2 \phi^{-2}$$ instead of $$\sigma_w^2$$. This reflects the different variance for the error term that applies when you change from backward to forward-recursion or vice versa.

• I've accepted your answer, thanks for your response. After some reflection, this makes sense. Feb 4, 2022 at 9:02