Stationarity, ACF and PACF

Idea: An inventory that never settles

Daily fills: mean $0$, variance $9$, independent. Inventory adds them up, so its variance keeps growing.

• $4$ days: $4\times 9=36$; $25$ days: $25\times 9=225$
• No fixed variance: the series is not stationary

Idea: Weak stationarity

A series is stationary when $3$ things hold. Every estimate from history leans on them.

• The mean does not depend on $t$
• The variance is finite and fixed
• Covariance depends only on the lag $k$

Formula: $\mathrm{AR}(1)$ variance

$V=\frac{{\sigma }^2}{1-{\varphi }^2}$

The level a stationary $\mathrm{AR}(1)$ settles at. At φ = 1 it breaks: a unit root.

Quick check: A desk that unloads

Inventory follows $I_t=0.8I_{t-1}+N_t$ with fill variance $9$. What is its long-run variance?

Formula: $\mathrm{ACF}$ of $\mathrm{AR}(1)$

$\rho (k)={\varphi }^k$

A slow geometric fade: $0.8$, $0.64$, $0.512$ for $\varphi =0.8$; signs alternate when φ < 0.

Idea: $\mathrm{ACF}$ of $\mathrm{MA}(1)$

Only neighbours share a shock: $\rho (1)=\theta /(1+{\theta }^2)$, and zero past lag $1$.

• $\theta =0.5$: $\frac{0.5}{1+0.25}=0.4$
• In size $\rho (1)$ never exceeds $\frac{1}{2}$
• $\mathrm{MA}(q)$ cuts off after lag $q$

Formula: Partial autocorrelation

$\mathrm{PACF}(2)=\frac{\rho (2)-\rho (1{)}^2}{1-\rho (1{)}^2}$

The correlation at lag 2 beyond what lag 1 alone passes on.

Example: A lag 2 partial

A series has $\rho (1)=0.5$ and $\rho (2)=0.4$. Find the partial autocorrelation at lag $2$.

1. $0.5\times 0.5=0.25$
2. $\frac{0.4-0.25}{1-0.25}=0.2$

Idea: Read the plots

Match the cut-offs to the model.

• $\mathrm{AR}(p)$: $\mathrm{ACF}$ fades, $\mathrm{PACF}$ cuts after $p$
• $\mathrm{MA}(q)$: $\mathrm{ACF}$ cuts after $q$, $\mathrm{PACF}$ fades
• Noise band $\frac{2}{\sqrt{n}}$: $\frac{2}{20}=0.1$ for $400$ points

Common mistake: Differencing on sight

A slow $\mathrm{ACF}$ is not a unit root: $\mathrm{AR}(1)$ with $\varphi =0.95$ is stationary, yet $0.95^{10}\approx 0.6$.

Instead Run a unit root test first. Needless differencing makes $\rho (1)=-\frac{1}{2}$ appear from nothing.