In much of the literature I'm studying it's one of those terms that occurs frequently yet without a rigorous definition to be found. Specifically, I am told:

For time-indexed random variables (RVs) $\{X_t\}$, the additive decomposition model is given as

$$X_t = {ll}(X_{t-1}, X_{t-2}, \ldots) + {fc}(X_{t-1}, X_{t-2}, \ldots, \varepsilon_t, \varepsilon_{t-1}, \ldots)$$


  • $ll$ is the long-term level, which is a stochastic process and can be visualised as a smoothed version of $\{X_t\}$, not to be confused with trends which are deterministic patterns
  • $fc$ is the fluctuation component which represents changes in local level, assumed stationary and with zero mean level
  • $\{\varepsilon_t\}$ are innovations, and are IID mean-zero RVs

But what is the difference in meaning between trend vs. long-term level vs. local level vs. mean level?

Additionally, aren't the fluctutation component and innovations modelling the same thing, which is the noise associated with each observation? So why complicate things by including both?


1 Answer 1


This has to do with the order of integration. A stochastic process $X_t$ is said to be integrated of order $0$, equivalently $X_t$~$I(0)$ if it is stationary. If $X_t$~$I(d)$ with $d>0, d \in \mathbb{N}$, the process is said to be integrated of order $d$ and is then nonstationary. The above decomposition attempts to filter out the stationary components (as fluctuation component and innovations) and the nonstationary stochastic trend component. A stochastic trend is different from a deterministic trend, and the usage of the word trend in the passage is sloppy.

Now, this makes it all sound more complicated than it is. Lets consider an example. Take $\varepsilon_t$~$(0,\sigma^2)$ as a white noise process and let $\varepsilon_t$ be $iid$. Define the following lag polynomial

\begin{align} C_1(L) &= 0.5L + 0.25L^2 -0.75L^3 -0.05 L^4 \\ \end{align}

The lag operator $L$ works on time-indexed random variables as $L^k\varepsilon:=\varepsilon_{t-k}$. Suppose now further that $X_t$ is generated as

\begin{align} X_t = X_{t-1} + C_1(L)\varepsilon_t + \varepsilon_t \end{align}

Then, using the terminology from your excerpt, the long term level would be defined by $X_{t-1}$, the seasonal/fluctuation component by $C_1(L)\varepsilon_t$ and the innovations by $\varepsilon_t$. As described in the excerpt, the fluctuation component and the innovations are stationary.

The reason why it is called that way is somewhat hard to see without making further remarks and relates back to the aforementioned order of integration. Usually, we don't encounter processes that are integrated of orders higher than $1$ or $2$, so lets consider the above example of integration order $1$.

First of, define $u_t := C_1(L)\varepsilon_t + \varepsilon_t$. $u_t$ is stationary, so $u_t$~$I(0)$. Now we can write \begin{align} X_t &= X_{t-1} + u_t \Longleftrightarrow\\ X_t - X_{t-1} &= (1-L)X_t = \Delta X_t = u_t \\ \end{align} this tells us that $X_t$~$I(1)$, because its first difference is integrated of order $0$. The meaning of this might be hard to grasp, until one realizes what $\Delta X_t = u_t$ actually means. It means that one can rewrite \begin{align} X_t &= \sum_{i=1}^{\infty} \Delta X_t = \sum_{i=1}^{\infty} u_t\\ \end{align} This might not look dramatic: $\mathbb{E}(X_t) = 0$, after all! However, the variance of this process is not finite and explodes to $\infty$. This is why we say the term defines a stochastic trend: while it is not deterministic (like for instance a linear trend), $X_t$ will only be stationary once we have filtered out the nonstationary component and substracted it from $X_{t}$. (In this case, as observed previously, $\Delta X_t = X_t - X_{t-1}=C_1(L)\varepsilon_t + \varepsilon_t$ would have filtered out the nonstationary component and would be stationary.) If you do not do this, your usual statistical inference procedures do not work anymore, since $X_{t}$ will converge to a brownian motion by the invariance principle/Functional Central Limit Theorem. These results replace standard CLR results for Autoregressions, Cointegration problems, and so forth.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.