<p>It's hard to tell without more context. But mostly sure it means &quot;<span class="math-container">$Z_t$</span> is integrated of order <span class="math-container">$g$</span>&quot;, i.e. <span class="math-container">$\Delta ^g Z_t=(1-d)^gZ_t$</span> (where <span class="math-container">$d$</span> is the lag operator) is stationary, in other words you have to difference <span class="math-container">$Z_t$</span> <span class="math-container">$g$</span> times to get an stationary time series. See <a href="https://en.wikipedia.org/wiki/Order_of_integration" rel="noreferrer">this</a>.</p>

CC BY-SA 4.0
2021-07-30T15:44:48.560Z
2021-07-30T15:44:48.560Z
manifold
<p>During my studies I came across the <span class="math-container">$I(g)$</span>-symbol, in the sense that <span class="math-container">$Z_t \sim I(g)$</span> for an integer <span class="math-container">$g$</span> and <span class="math-container">$Z_t$</span> is a times series. What exactly does this mean, and does it have to do with stationarity?</p>

<econometrics><time-series><definition>
$I(g)$-terminology
Hyperbolic PDE friend
# $I(g)$-terminology

(Asked by: username_0 on 2021-07-30)

During my studies I came across the $I(g)$-symbol, in the sense that $Z\_t \sim I(g)$ for an integer $g$ and $Z\_t$ is a times series. What exactly does this mean, and does it have to do with stationarity?

**Tags:** econometrics, time-series, definition

---

## 1 answers

### Answer by username_1 on 2021-07-30. Score: 7

It's hard to tell without more context. But mostly sure it means "$Z\_t$ is integrated of order $g$", i.e. $\Delta ^g Z\_t=(1-d)^gZ\_t$ (where $d$ is the lag operator) is stationary, in other words you have to difference $Z\_t$ $g$ times to get an stationary time series. See [this](https://en.wikipedia.org/wiki/Order_of_integration).


---

