# Why can this $(1- \phi L)^{-1}* u_{t}$ be written as $(1+ \phi L+\phi^{2}L + \cdots)* u_{t}$?

I was looking at AR processes in Chris Brooks (2008) and at one point, when deriving the variance of an AR(1) process he writes this part $$(1- \phi L)^{-1}* u_{t}$$ as $$(1+ \phi L+\phi^{2}L + \cdots)* u_{t}$$ but I do not understand why. It might be a rather basic question but I would really appreciate if you could at least point me in the right direction.

Thank you!

Equation 5.86 page 220 if you want to have a look.

Lukasz's answer is correct. Just to elaborate:

# Quick review of geometric series

### Finite geometric series ($g_k = 1 + r^2 + r^3 + \ldots + r^k$):

• Let $r$ be some scalar.
• Let $g_k$ be a geometric series of $r$, that is, $g_k = \sum_{j=0}^k r^j = 1 + r + r^2 + \ldots + r^k$

Observe that $g_{k+1} = g_k + r^{k+1}$ and that $g_{k+1} = 1 + r g_k$. Equating the two sides:

$$g_k + r^{k+1} = 1 + rg_k$$

Then some simple algebra gives us:

$$g_k = \frac{1 - r^{k+1}}{1 - r}$$

### Infinite geometric series ($k=\infty$):

If $|r| < 1$ then the limit of $g_k$ as $k\rightarrow \infty$ exists:

$$\lim_{k \rightarrow \infty} \frac{1 - r^{k+1}}{1 - r} = \frac{1}{1-r}$$

To recap, if $|r| < 1$ then: $$1 + r + r^2 + r^3 + \ldots = (1-r)^{-1}$$

# Time series, lag polynomials, and geometric series

If we have some time series process:

$$y_t = \phi y_{t-1} + u_t$$

We can write this with the lag operator as: $$\left( 1 - \phi L \right) y_t = u_t$$

Some manipulation (yes this is allowed, if the lag polynomial is invertible): $$y_t = \left(1 - \phi L \right)^{-1} u_t$$

Our previous results of geometric series turns out to hold here as well. For $|\phi| < 1$ we have:

$$\left(1 - \phi L \right)^{-1} = 1 + \phi L + \phi^2 L^2 + \phi^3 L^3 + \ldots$$

Hence: \begin{align*} y_t &= \left(1 + \phi L + \phi^2 L^2 + \phi^3 L^3 + \ldots \right) u_t \\ &= u_t + \phi u_{t-1} + \phi^2 u_{t-2} + \phi^3 u_{t-3} + \ldots \end{align*}

• Exceptional answer! – Alex Feb 9 '17 at 15:26

Because the sum telescopes

$$1 = \frac{(1 - xL)}{(1 - xL)} = (1 + xL + x^2L^2 + \dots)(1 - xL) = 1 -xL + xL - x^2L^2 + x^2L^2 \dots = 1$$

Note: we need $x < 1$