I am relatively new to time series and studying this section - https://otexts.com/fpp3/stochastic-and-deterministic-trends.html.

From the above section, didn't quite understand the difference why one is called stochastic and the other is called deterministic. Went through several posts in stack overflow. But I don't see a relation to the answer in below posts to the above textbook. Can some one please explain this.

What is the difference between deterministic and stochastic model?

Explain what is meant by a deterministic and stochastic trend in relation to the following time series process?

I saw the youtube videos in the second link, and I understood the difference between deterministic and stochastic. But, I don't see any relation between the explanation in video vs. text books ( which talks about ARIMA process with D=1)

  • $\begingroup$ As a follow up to this question ..... What time series software distinguishes (AND HOW) between these two alternatives. I know one that does and like most data-driven solutions/algorithms it actually conducts a computer-based evaluation to determine the best strategy for any given time series. $\endgroup$
    – IrishStat
    Apr 17, 2020 at 19:29

1 Answer 1


Deterministic Trend

$$ y_t = \beta_0 + \beta_1 t + \epsilon_t $$ where $\{\epsilon_t\}$ is white noise, for simplicity. Same discussion applies to the case where $\{\epsilon_t\}$ is a covariance-stationary process (e.g. ARIMA with $d = 0$).

The process is random fluctuations around a deterministic linear trend $\beta_0 + \beta_1 t$. Hence the terminology "deterministic trend".

Such processes also called trend-stationary. If you remove the linear trend, you recover the stationary process $\{\epsilon_t\}$.

Stochastic Trend

$$ y_t = \beta_0 + \beta_1 t + \eta_t $$ where $\{\eta_t\}$ is a random walk, for simplicity. Same discussion applies to the case where $\{\eta_t\}$ is an $I(1)$ process (e.g. ARIMA with $d = 1$). Equivalently, $$ y_t = y_0 + \beta_0 + \beta_1 t + \sum_{s = 1}^{t} \epsilon_t $$ where $\{\epsilon_t\}$ is the white noise driving the random walk $\{\eta_t\}$. The "stochastic trend" terminology refers to $\eta_t$. The random walk is a highly persistent process, giving its sample path the appearance of a "trend".

Such processes are also called difference-stationary. If you take first-difference, you recover the stationary process $\{\epsilon_t\}$, i.e. $$ \Delta y_t = \beta_1 + \epsilon_t, $$ which is the same series (random walk with drift) from your second link.

Visual Similarity

You can observe via simulation that the sample paths from these two models can be visually similar---e.g. choose $\beta_1=1$ and $\epsilon_t \stackrel{i.i.d.}{\sim}(0,1)$.

This is because the linear trend $\beta_0 + \beta_1 t$ dominates. More precisely, for both models $$ \frac{y_t}{t} = \beta_1 + o_p(1). $$ Only the slope term $\beta_1$ is not negligible in the limit. For the deterministic trend case, it is clear that $\frac{\epsilon_t}{t} = o_p(1)$. For the stochastic trend case, $\frac{\eta_t}{t} = o_p(1)$ because $\frac{\eta_t}{\sqrt{t}}$ converges in distribution to a normal distribution (Central Limit Theorem).

Statistical Testing

The visual similarity of sample paths motivates the problem of statistically distinguishing these two models. This is the purpose of unit root tests---e.g. the (Augmented) Dickey-Fuller test, which is historically the first such test.

For the ADF test, you basically take the detrended series $\tilde{y}_t$ (residuals from regressing $y_t$ on $1$ and $t$), run the regression $$ \Delta \tilde{y}_t = \alpha \tilde{y}_{t-1} + \tilde{\epsilon}_t, $$ and consider the $t$-statistic for $\alpha = 0$. It the $t$-statistic is small, you reject the null of stochastic trend.

The empirical reasoning behind the ADF test is simple. Even though the sample paths themselves are similar, the detrended series would look quite different. Under trend-stationarity, the detrended series would appear stationary. On the other hand, if a difference-stationary model is mistakenly detrended, the detrended series would not appear stationary.

  • $\begingroup$ Thank you so much Michael. this is really helpful and cleared my confusion. Can you please let me know, which one to use(deterministic/stochastic) or how to choose if I want to model a linear trend ? Also how to model them - should I use simple linear regression for deterministic trend and 'ARIMA(010) with constant' for stochastic ? Also the text link says the error in deterministic trend is 'ARMA' process. does it mean it is ARMA (000) process - white noise and can this trend be modeled by simple linear regression? Also will unit root test gives that the stochastic trend has unit root ? $\endgroup$
    – tjt
    Apr 18, 2020 at 0:19
  • 1
    $\begingroup$ If you can reasonable assume that your data is realization of one model or the other, modeling is easy. For deterministic trend, the model is simple ARMA plus linear trend. For stochastic trend, model is ARIMA plus linear trend. The error term can be general ARMA---for simplicity I assumed ARMA(0,0), i.e. white noise, in the answer. $\endgroup$
    – Michael
    Apr 19, 2020 at 1:56
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    $\begingroup$ The tricky part is to pin down, to the extent possible, which one is the true model for your data. That's where unit root tests come in. The null hypothesis of most unit root tests is stochastic trend (i.e. unit root). There is also the KPSS test, whose null is deterministic trend. $\endgroup$
    – Michael
    Apr 19, 2020 at 1:58
  • $\begingroup$ Thank you very much Michael. pardon my ignorance. for deterministic and stochastic you gave the same models, except 'simple' in deterministic. Can you please elaborate on what is simple ARIMA/ARIMA. and by 'plus linear trend', does it mean above ARIMA models 'with a non seasonal difference - does it require a constant too'? $\endgroup$
    – tjt
    Apr 19, 2020 at 2:04
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    $\begingroup$ No, not the same. For deterministic trend---linear trend+ ARMA, i.e. linear trend + ARMA process. For stochastic trend, linear trend + ARIMA with d = 1. E.g ARMA(0,0) is white noise, and ARIMA(0,1,0,) is random walk. Yes, linear trends have intercept terms. $\endgroup$
    – Michael
    Apr 20, 2020 at 18:05

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