Risk Modelling and Survival Analysis · Core concepts of time series models

Stationarity and Weak Stationarity of Time Series

Q: What is the difference between strict and weak stationarity?

Strict stationarity says the full joint distribution is unchanged by a time shift. Weak stationarity only needs a constant mean, finite variance and an autocovariance that depends on the lag alone. Strict stationarity with finite variance implies weak stationarity, but not the other way round in general.

Q: How do I check if a time series is stationary?

For a model, compute the mean, variance and autocovariance and see whether any depends on t. For data, plot the series and the sample ACF. A clear trend, seasonal pattern, growing variance or a very slowly decaying ACF suggests non-stationarity.

Q: What is the autocorrelation function?

The ACF at lag k is ρk = γk ÷ γ0, where γk is the autocovariance at lag k. It measures the linear dependence between values k periods apart. It always lies between −1 and 1 and equals 1 at lag 0.

Q: Is a random walk stationary?

No. Its variance grows with time, so it fails the conditions for weak stationarity. Taking first differences of a random walk gives white noise, which is stationary.

Updated 11 October 2026 · Fact-checked

A time series is stationary if its statistical behaviour does not change over time. Weak stationarity needs a constant mean and an autocovariance that depends only on the lag, not on time. To check it, find E[Xt] and Cov(Xt, Xt+k) and see whether either depends on t.

Understand Stationarity and Weak Stationarity of Time Series

A time series is a set of observations Xt recorded in time order, such as monthly claim counts or daily share prices. We model it as a stochastic process. To fit a model, we need some stable feature to estimate from a single observed path. Stationarity gives us that stable feature.

Strict stationarity is the strongest form. For any times t1, ..., tn and any shift k, the joint distribution of (Xt1, ..., Xtn) is the same as that of (Xt1+k, ..., Xtn+k). The whole distribution is unchanged by a shift in time. This is hard to verify in practice.

Weak stationarity (also called covariance or second-order stationarity) asks for less. You need three things: the mean E[Xt] is a constant μ; the variance is finite; and Cov(Xs, Xt) depends only on the lag t − s. If Xt is strictly stationary and has finite second moments, it is also weakly stationary. The reverse is not true in general. The exception is a Gaussian process, where weak stationarity does imply strict stationarity, because the normal distribution is fixed by its mean and covariances.

The autocovariance function is γk = Cov(Xt, Xt+k). At lag 0 it equals the variance, γ0 = Var(Xt). The autocorrelation function (ACF) is ρk = γk ÷ γ0. So ρ0 = 1 and |ρk| ≤ 1. For a real series, γ−k = γk, so the ACF is symmetric about lag 0.

To check stationarity of a model, compute the mean and the covariance and see whether they depend on t. Typical failures are a trend, a seasonal pattern, or a variance that grows with time. A random walk is the classic non-stationary example. For observed data, you plot the series and the sample ACF. A non-stationary series usually shows a visible trend, and a sample ACF that dies away very slowly.

Key rules to remember

Mean function: μt = E[Xt]; Weak stationarity requires μt = μ, the same for all t.
Autocovariance function: γk = Cov(Xt, Xt+k) = E[(Xt − μ)(Xt+k − μ)]; For a weakly stationary process this depends only on the lag k. γ0 = Var(Xt).
Autocorrelation function: ρk = γk ÷ γ0; ρ0 = 1, |ρk| ≤ 1 and ρ−k = ρk.
Weak stationarity conditions: E[Xt] = μ for all t; Var(Xt) < ∞; Cov(Xt, Xt+k) = γk for all t; All three must hold. Failing any one means the series is not weakly stationary.
Strict stationarity: (Xt1, ..., Xtn) has the same joint distribution as (Xt1+k, ..., Xtn+k) for all n, t1, ..., tn, k; Strict plus finite variance implies weak. Weak implies strict only for Gaussian processes.
Sample autocovariance and ACF: ck = (1 ÷ n) Σ (xt − x̄)(xt+k − x̄), sum over t = 1 to n − k; rk = ck ÷ c0; Used to estimate γk and ρk from data. Divide by n, not n − k, in the standard definition.

How to solve Stationarity and Weak Stationarity of Time Series questions

Use this method when a question asks you to decide whether a given process is stationary, or to find its autocovariance or autocorrelation.

1Write the process clearly, including the distribution of any noise term. For example, state whether the errors are white noise with mean 0 and variance σ².
2Compute E[Xt]. If it depends on t, stop: the process is not weakly stationary.
3Compute Var(Xt). It must be finite and not depend on t.
4Compute Cov(Xt, Xt+k) for a general lag k ≥ 0. Use the properties of covariance and the independence or uncorrelatedness of the noise terms.
5Check that the result depends only on k and not on t. If it does, the process is weakly stationary and this gives γk.
6Find ρk = γk ÷ γ0 if asked for the ACF.
7State the conclusion in words. Mention strict stationarity only if you have the full distribution, for example for a Gaussian process.

Quickest way: Mean and variance first

When to use it: Use this when you are short of time and the question only asks whether a process is stationary.

Take the expectation of Xt. A term like a·t, or a random walk's accumulated sum, makes the mean or variance depend on t.
Take the variance. If it grows with t, the process fails immediately.
If both are constant, compute Cov(Xt, Xt+1) and Cov(Xt, Xt+k) for general k.
If the covariance has t in it, the process fails. If it has only k, it passes.
Write the verdict with the one failing or passing line as your evidence.

Common mistakes in Stationarity and Weak Stationarity of Time Series

Checking only that the mean is constant and then declaring the process stationary.
The mean is the easiest condition to test, so students stop there.
Fix: Always test the variance and the autocovariance too. All three conditions must hold.
Saying weak stationarity implies strict stationarity.
The names suggest a hierarchy that runs both ways.
Fix: Strict plus finite variance implies weak. The reverse holds only in special cases such as Gaussian processes.
Writing the autocovariance as a function of both s and t, or leaving t in the answer, and still calling the series stationary.
Students compute the covariance but forget to check what it depends on.
Fix: After computing, ask: does t appear? If yes, it is not weakly stationary. If only |t − s| appears, it is.
Forgetting that ρ0 = 1 and that γ0 is the variance.
Students treat lag 0 as a special case they do not need.
Fix: Always compute γ0 first. It is the denominator of every autocorrelation.
Treating a series with a trend or seasonality as stationary because it looks stable over a short window.
A short sample can hide a trend.
Fix: Look at the whole series and the sample ACF. A slowly decaying ACF or a regular seasonal pattern points to non-stationarity.
Using γk = E[Xt Xt+k] without subtracting the mean.
Students remember the product but forget the centring.
Fix: Use Cov(Xt, Xt+k) = E[Xt Xt+k] − μ². Subtract μ² when the mean is not zero.

Worked examples

Example 1

Let et be white noise with mean 0 and variance σ². Define Xt = 5 + et + 0.6 et−1. Show that Xt is weakly stationary and find its ACF at lags 0, 1 and 2.

Show the solution

Mean: E[Xt] = 5 + 0 + 0.6 × 0 = 5. This is constant.
Variance: γ0 = Var(et) + 0.6² Var(et−1) = σ² + 0.36σ² = 1.36σ². This is finite and constant.
Lag 1: Cov(Xt, Xt+1) = Cov(et + 0.6 et−1, et+1 + 0.6 et). The only common term is et, so γ1 = 0.6 × Var(et) = 0.6σ².
Lag 2: Xt involves et and et−1, and Xt+2 involves et+2 and et+1. There is no common term, so γ2 = 0.
None of these depend on t, so the process is weakly stationary.
ρ0 = 1. ρ1 = 0.6σ² ÷ 1.36σ² = 0.6 ÷ 1.36 = 0.4412 (to 4 decimal places). ρ2 = 0.

Answer: Xt is weakly stationary with mean 5 and γ0 = 1.36σ², γ1 = 0.6σ², γ2 = 0. The ACF is ρ0 = 1, ρ1 ≈ 0.4412, ρ2 = 0.

Example 2

Let et be white noise with mean 0 and variance σ². Define Xt = Xt−1 + et for t ≥ 1 with X0 = 0. Determine whether Xt is weakly stationary.

Show the solution

Write Xt as a sum: Xt = e1 + e2 + ... + et.
Mean: E[Xt] = 0 for all t. This condition holds.
Variance: the et are uncorrelated, so Var(Xt) = tσ².
This depends on t and grows without limit, so the variance condition fails.
Check the covariance as well: for s < t, Cov(Xs, Xt) = Var(Xs) = sσ². This depends on s, not only on the lag t − s.
So the autocovariance is not a function of the lag alone.

Answer: Xt is a random walk. Its mean is constant but Var(Xt) = tσ² depends on t, so it is not weakly stationary (and so not strictly stationary either).

Exam tips

Start every stationarity answer by computing E[Xt]. It is quick and often settles the question.
Show the covariance working line by line. Marks are given for method, not only for the verdict.
State which noise assumptions you use, for example that et is white noise with mean 0 and variance σ².
If a question mentions a Gaussian process, say that weak stationarity then implies strict stationarity.
In computer-based papers, plot the series and use the acf() function in R to show the sample ACF, then explain what a slow decay means.

Practice questions from Core concepts of time series models