Why Variance Reveals Nonstationarity
A random walk can have a stable-looking sequence of one-step movements while its level changes substantially over time. The key distinction is between the random innovations added at each step and the accumulated level. Even when every innovation has the same distribution, the level has absorbed an increasing number of them. Its variance can therefore depend on the time index.
Weak stationarity provides a precise test for this behavior. A process is weakly stationary only if its mean is constant, its variance is finite and constant over time, and its covariance between two observations depends only on the lag between them. To establish that a process is not weakly stationary, it is enough to show that one of these requirements fails. For a nondegenerate random walk with the usual assumptions on its starting value and innovations, the variance alone is enough.
The Random Walk and Its Variance
Consider a random walk indexed by the nonnegative integers:
Here \(X_0\) is the starting value, and the innovations \( \varepsilon_t \) have mean zero and common finite variance \( \sigma^2 \). Assume that distinct innovations are uncorrelated. For the usual variance-growth result, also assume that \(X_0\) is fixed or has finite variance and is independent of all the innovations. Independence can be weakened to zero covariance between \(X_0\) and each innovation for this variance calculation. The innovations need not be normally distributed.
Repeated substitution expresses the level as the starting value plus all innovations accumulated so far:
This representation explains why the variance changes with time: later observations contain more random terms. The following derivation makes the role of the assumptions explicit.
\(X_t=X_0+\sum_{j=1}^{t}\varepsilon_j\). The variance of a sum includes the variances of its terms and twice the covariances between distinct terms.
The innovations are pairwise uncorrelated, and the starting value is fixed or uncorrelated with each innovation. Thus every covariance between distinct terms in the sum is zero.
\(\operatorname{Var}(X_t)=\operatorname{Var}(X_0)+\sum_{j=1}^{t}\operatorname{Var}(\varepsilon_j)=\operatorname{Var}(X_0)+t\sigma^2\). If \(X_0\) is fixed, its variance is zero.
The increase from one time to the next is \( \sigma^2 \) under these assumptions:
This is a statement about the level \(X_t\), not a claim that each realized observation must be larger than the previous one. An innovation can be negative, so a particular path can move down. The result concerns the variance across possible paths at a given time.
Worked Example: A Fixed Starting Value
Worked Example: A Fixed Starting Value
A hypothetical monthly index follows \(X_t=X_{t-1}+\varepsilon_t\), where \(X_0=30\) is fixed and the innovations are independent \(N(0,4)\). Compare the variance at months 2 and 7, then decide whether the level process is weakly stationary.
Method. Apply the random walk variance formula. The notation \(N(0,4)\) specifies a normal distribution with mean 0 and variance 4, so \( \sigma^2=4 \). The starting value is fixed, hence \( \operatorname{Var}(X_0)=0 \).
At month 2, two innovations have accumulated: $$ \operatorname{Var}(X_2)=0+2(4)=8. $$ At month 7, seven innovations have accumulated: $$ \operatorname{Var}(X_7)=0+7(4)=28. $$
The variances are 8 and 28, so they are not constant over time. The process is not weakly stationary. As a check, the increase over the five additional months is \(28-8=20\), which equals \(5(4)=20\).
Common mistake avoided: The variance of each innovation is 4, but the variance of the level at month 7 is not 4. The level contains seven uncorrelated innovations, so their variances add.
A Random Starting Value
A random starting value does not prevent the usual conclusion, provided it is independent of the innovations and has finite variance. Its variance contributes to every later level, while each new uncorrelated innovation contributes another \( \sigma^2 \). The starting value need not be zero or have zero variance.
Worked Example: Comparing Equal-Lag Covariances
A hypothetical reserve index has \(X_0\sim N(40,5)\), and its innovations are independent \(N(0,2)\), independent also of \(X_0\). Compare \( \operatorname{Cov}(X_2,X_3) \) with \( \operatorname{Cov}(X_5,X_6) \). Use the comparison to assess the covariance condition for weak stationarity.
Method. Expand each level into its starting value and accumulated innovations. Covariance includes only terms shared by both levels, because \(X_0\) and distinct innovations are mutually uncorrelated under the stated assumptions. The first argument of each normal distribution is the mean and the second is the variance.
The expansions are $$ X_2=X_0+\varepsilon_1+\varepsilon_2,\qquad X_3=X_0+\varepsilon_1+\varepsilon_2+\varepsilon_3. $$ The shared terms are \(X_0,\varepsilon_1,\varepsilon_2\). Therefore, $$ \operatorname{Cov}(X_2,X_3) =\operatorname{Var}(X_0)+\operatorname{Var}(\varepsilon_1)+\operatorname{Var}(\varepsilon_2) =5+2+2=9. $$
For the later pair, $$ X_5=X_0+\sum_{j=1}^{5}\varepsilon_j,\qquad X_6=X_0+\sum_{j=1}^{6}\varepsilon_j. $$ The shared terms are \(X_0\) and the first five innovations. Thus, $$ \operatorname{Cov}(X_5,X_6) =\operatorname{Var}(X_0)+5\operatorname{Var}(\varepsilon_1) =5+5(2)=15. $$
The two covariances are 9 and 15, respectively. Both pairs have lag 1, but their covariances differ, so the covariance condition for weak stationarity also fails. As a check, the level variances are \( \operatorname{Var}(X_2)=5+2(2)=9 \) and \( \operatorname{Var}(X_5)=5+5(2)=15 \), consistent with the shared-term calculations.
Common mistake avoided: For \( \operatorname{Cov}(X_2,X_3) \), there are two shared innovations, not one. For \( \operatorname{Cov}(X_5,X_6) \), there are five. Omitting a shared term understates the covariance.
What If the Starting Value Is Dependent?
The variance-growth conclusion must be tied to its assumptions. If the starting value is dependent on an innovation, a covariance term may offset some or all of that innovation’s variance. For mutually uncorrelated innovations, the general expansion is
Consequently, one cannot assert that every new innovation adds \( \sigma^2 \) to the level variance without controlling the covariances with \(X_0\). For example, let \(X_0=-\varepsilon_1\), where \(\varepsilon_1\sim N(0,1)\). Then \( \operatorname{Var}(X_0)=1 \), but \(X_1=X_0+\varepsilon_1=0\), so \( \operatorname{Var}(X_1)=0 \), not 2. This does not contradict the usual random walk result: the starting value in this example is deliberately dependent on the first innovation.
Worked Example: A Random Start Independent of the Innovations
A hypothetical portfolio indicator follows a random walk with \(X_0\sim N(12,5)\) and independent innovations \( \varepsilon_t\sim N(0,2) \), also independent of \(X_0\). Calculate the mean and variance at times 2 and 5, and determine which weak-stationarity conditions fail.
Method. There is no drift, so each innovation has mean zero. By linearity of expectation, $$ E[X_t]=E[X_0]+\sum_{j=1}^{t}E[\varepsilon_j]=12+t(0)=12. $$ Thus \(E[X_2]=12\) and \(E[X_5]=12\).
For the variances, \( \operatorname{Var}(X_0)=5 \) and \( \sigma^2=2 \): $$ \operatorname{Var}(X_2)=5+2(2)=9,\qquad \operatorname{Var}(X_5)=5+5(2)=15. $$
The mean is constant in this example, but the variance changes from 9 to 15. Therefore the process is not weakly stationary. The calculation also checks the distinction between the starting variance, which contributes 5 at both times, and the accumulated innovation variance, which is 4 at time 2 and 10 at time 5.
Common mistake avoided: A random starting value does not mean its variance is added only once and then ignored; its variance remains part of every later level. The accumulated innovation variance is added on top of it.
Drift and the Stationarity Test
A random walk may include a constant drift \(c\):
With a fixed starting value \(X_0\), zero-mean innovations, and the same uncorrelatedness assumptions, the mean is \(E[X_t]=X_0+tc\) and the variance is \(t\sigma^2\). If \(c\ne 0\), the mean changes with time; if \( \sigma^2>0 \), the variance changes with time as well. Either failure rules out weak stationarity. With no drift, the mean can remain constant, but the changing variance still rules it out when \( \sigma^2>0 \).
The normality of the innovations is not what causes nonstationarity. The variance argument uses only the finite common innovation variance and the relevant zero-covariance assumptions. Normality may make a distributional description convenient, but weak stationarity is assessed through means, variances, and covariances—not by requiring normal observations.
Common Exam Traps
- Confusing innovation variance with level variance. If each innovation has variance \( \sigma^2 \), that is not generally the variance of \(X_t\). Count the accumulated innovations and include the starting-value variance when it is random.
- Claiming variance must increase without stating assumptions. The formula \( \operatorname{Var}(X_t)=\operatorname{Var}(X_0)+t\sigma^2 \) requires the starting value to be fixed or uncorrelated with the innovations, as well as uncorrelated innovations. Dependence can change the result.
- Checking only the mean. A constant mean does not establish weak stationarity. The variance must also be constant, and the covariance must depend only on lag.
- Using a changing sample path as the proof. One path moving upward or downward does not establish nonstationarity. Compare population quantities such as \( \operatorname{Var}(X_t) \) at different times.
Check Your Understanding
For each problem, use the stated assumptions carefully. Variances in normal-distribution notation are the second parameters.
- A hypothetical sensor level has fixed \(X_0=8\) and independent innovations \( \varepsilon_t\sim N(0,3) \). Find \( \operatorname{Var}(X_4) \) and \( \operatorname{Var}(X_9) \), then state whether the level is weakly stationary. (Answer: 12 and 27; no, because the variance changes.)
- A hypothetical index has \(X_0\sim N(7,4)\), independent of uncorrelated innovations with mean 0 and variance 1.5. Find \( \operatorname{Var}(X_3) \) and \( \operatorname{Var}(X_8) \). (Answer: 8.5 and 16.)
- A hypothetical process has fixed \(X_0=10\) and follows \(X_t=X_{t-1}+0.2+\varepsilon_t\), with independent \( \varepsilon_t\sim N(0,0.25) \). Find its mean and variance at times 5 and 9, and identify the stationarity failures. (Answer: At time 5, mean 11 and variance 1.25; at time 9, mean 11.8 and variance 2.25; both mean and variance change.)
- A hypothetical random walk has \(X_0\sim N(0,6)\) independent of uncorrelated innovations with mean 0 and variance 3. Calculate \( \operatorname{Cov}(X_1,X_2) \) and \( \operatorname{Cov}(X_4,X_5) \), and explain the implication. (Answer: 9 and 18; both are lag-1 covariances but differ, violating the covariance condition.)
- Let \(X_0=-\varepsilon_1\), where \( \varepsilon_1\sim N(0,1) \), and let \(X_1=X_0+\varepsilon_1\). Calculate \( \operatorname{Var}(X_0) \) and \( \operatorname{Var}(X_1) \). Why does the usual variance-growth formula not apply? (Answer: 1 and 0; \(X_0\) is dependent on \(\varepsilon_1\), so the required zero-covariance assumption fails.)