The ACF Of A Random Walk

Why Random-Walk Levels Are Correlated

A random walk accumulates shocks over time. Two observations taken close together therefore contain many of the same shocks, even when each new shock is uncorrelated with the previous ones. This shared history is why the levels can have substantial autocorrelation.

For a stationary series, theoretical autocorrelation is ordinarily described as a function of lag alone. A random walk is nonstationary, so its level correlation must instead be specified using both the earlier observation time and the lag. Confusing these two descriptions can lead to treating the random walk as if it had one fixed ACF.

Definition: Let $X_t=X_0+\sum_{j=1}^{t}\varepsilon_j$, where $X_0$ is fixed and the innovations have mean $0$, common variance $\sigma^2>0$, and zero covariance across distinct times. For an origin $t>0$ and lag $h\geq 0$, the random-walk level autocorrelation is $$ \operatorname{Corr}(X_t,X_{t+h}) = \frac{\operatorname{Cov}(X_t,X_{t+h})} {\sqrt{\operatorname{Var}(X_t)\operatorname{Var}(X_{t+h})}}. $$

The zero-covariance condition is sufficient for the covariance calculation; independence and normality are not required. A drift can also be included, as in $X_t=X_0+ct+\sum_{j=1}^{t}\varepsilon_j$, without changing the correlation, because adding deterministic constants changes means but not covariances or variances. When examples specify a normal innovation distribution, it is an additional assumption for that example, not a requirement for the formula.

Deriving the Random-Walk Correlation

At time $t$, the level includes the first $t$ innovations. At time $t+h$, it includes those same $t$ innovations plus $h$ later ones. With a fixed starting value, write the two observations as

X_t=X_0+\varepsilon_1+\cdots+\varepsilon_t,\qquad X_{t+h}=X_0+\varepsilon_1+\cdots+\varepsilon_t+\varepsilon_{t+1}+\cdots+\varepsilon_{t+h}.

Each shared innovation contributes its variance to the covariance. Cross-covariances between distinct innovations are zero by assumption. Thus the shared $t$ innovations contribute $t\sigma^2$; the later $h$ innovations contribute nothing to the covariance with $X_t$. The variances count all innovations in each level.

Calculate each variance.
$\operatorname{Var}(X_t)=t\sigma^2$ and $\operatorname{Var}(X_{t+h})=(t+h)\sigma^2$, because each level sums that many uncorrelated innovations.

Count the shared innovations.
$\operatorname{Cov}(X_t,X_{t+h})=t\sigma^2$, since only the first $t$ innovations appear in both observations.

Divide covariance by the product of standard deviations.
Substituting the covariance and variances gives the correlation formula below.

\operatorname{Corr}(X_t,X_{t+h}) = \frac{t\sigma^2}{\sqrt{(t\sigma^2)((t+h)\sigma^2)}} = \sqrt{\frac{t}{t+h}},\qquad t>0,\ h\geq 0.

This derivation explains the shared-history effect directly. The scale $\sigma^2$ cancels, so the correlation does not depend on the innovation variance. The formula does depend on the origin $t$. At lag zero, it equals $1$. At a positive lag, it is less than $1$, because the later observation contains additional innovations that were not present in the earlier level.

Key result: For a random walk with fixed initial value and uncorrelated, equal-variance innovations, the correlation between levels $h$ time steps apart is $\sqrt{t/(t+h)}$ when the earlier observation is at time $t>0$. It is not a function of $h$ alone, so it is not a stationary ACF.

Worked Correlation Calculations

Worked Example: Correlation Three Steps Apart

Suppose $X_0=0$ and $X_t=\sum_{j=1}^{t}\varepsilon_j$, where the innovations are independent and $N(0,9)$. Find the correlation between $X_4$ and $X_7$. State the origin and lag, and check the result.

Method. The earlier time is $t=4$ and the lag is $h=7-4=3$. The innovation variance is $\sigma^2=9$. The observations share $\varepsilon_1,\ldots,\varepsilon_4$.

The variances and covariance are

\operatorname{Var}(X_4)=4(9)=36,\qquad \operatorname{Var}(X_7)=7(9)=63,\qquad \operatorname{Cov}(X_4,X_7)=4(9)=36.

Therefore,

\operatorname{Corr}(X_4,X_7) = \frac{36}{\sqrt{36\cdot63}} = \frac{36}{\sqrt{2268}} = \sqrt{\frac{4}{7}} \approx 0.7559.

Rounded to four decimal places, the correlation is $0.7559$. The check is that it is positive and less than $1$, and it agrees with $\sqrt{t/(t+h)}=\sqrt{4/7}$. A common mistake is to use the lag $3$ as if the correlation were $\sqrt{1/(1+3)}=0.5000$. That calculation silently sets the origin to $t=1$; the specified earlier time is $t=4$, not $1$.

Worked Example: Drift Does Not Change the Correlation

A hypothetical index follows $X_t=100+2t+\sum_{j=1}^{t}\varepsilon_j$, where the innovations are independent and $N(0,4)$. Find the correlation between $X_5$ and $X_{10}$. Show why the drift does not enter the calculation.

Method. The earlier time is $t=5$, the lag is $h=5$, and $\sigma^2=4$. The deterministic parts of the levels are $100+2(5)=110$ and $100+2(10)=120$. Deterministic constants do not affect covariance or variance.

The two variances are $5(4)=20$ and $10(4)=40$. The five shared innovations give covariance $5(4)=20$. Hence

\operatorname{Corr}(X_5,X_{10}) = \frac{20}{\sqrt{20\cdot40}} = \frac{20}{\sqrt{800}} = \sqrt{\frac{5}{10}} \approx 0.7071.

Rounded to four decimal places, the answer is $0.7071$. As a check, the drift changes the means but does not appear in the variance or covariance calculations; the result is also the formula $\sqrt{t/(t+h)}$. The trap is to include $110$ or $120$ in the covariance calculation. Means are not additional shared random shocks.

Worked Example: Same Lag, Different Origins

Let $X_0=0$ and let the independent innovations have distribution $N(0,2)$. Compare the correlation at lag $1$ when the earlier time is $t=4$ with the correlation at lag $1$ when it is $t=20$. Also calculate the correlation between $X_4$ and $X_8$. Report each to four decimal places.

Method. For the first pair, the observations are $X_4$ and $X_5$. Their variances are $4(2)=8$ and $5(2)=10$, and their covariance is $4(2)=8$. Thus

\operatorname{Corr}(X_4,X_5) = \frac{8}{\sqrt{8\cdot10}} = \sqrt{\frac{4}{5}} \approx 0.8944.

For the second pair, the observations are $X_{20}$ and $X_{21}$. Their variances are $20(2)=40$ and $21(2)=42$, and their covariance is $20(2)=40$. Therefore

\operatorname{Corr}(X_{20},X_{21}) = \frac{40}{\sqrt{40\cdot42}} = \sqrt{\frac{20}{21}} \approx 0.9759.

For $X_4$ and $X_8$, the lag is $4$. Their variances are $8$ and $8(2)=16$, and their covariance is $4(2)=8$. Hence

\operatorname{Corr}(X_4,X_8) = \frac{8}{\sqrt{8\cdot16}} = \sqrt{\frac{4}{8}} \approx 0.7071.

The checked results are $0.8944$, $0.9759$, and $0.7071$, respectively. At lag $1$, the later-origin pair is more highly correlated because its two levels share a larger accumulated history relative to the one new innovation. A common mistake is to report one value for “the lag-1 ACF” without giving the origin. These calculations show that the origin matters.

Interpreting the Formula and Avoiding Common Traps

For fixed $t>0$, increasing $h$ lowers $\sqrt{t/(t+h)}$: the later level contains more innovations that the earlier level does not share. For fixed positive $h$, increasing $t$ raises the correlation toward $1$. The same number of new shocks makes up a smaller fraction of a level that has accumulated a longer history. These comparisons describe correlations of levels; they do not imply that the innovations themselves are correlated.

In particular, the fact that nonzero-lag correlations between levels are positive does not contradict the zero covariance of distinct innovations. The levels overlap in their sums. For $X_t$ and $X_{t+h}$, the first $t$ shocks are present in both, while only the later $h$ shocks are new at time $t+h$.

Do not use lag alone. State the pair of times, or give both $t$ and $h$, before calculating the level correlation.
Do not confuse levels with increments. The level correlation comes from shared accumulated shocks; it is not the autocorrelation of the differenced series.
Do not add drift to the covariance. A deterministic trend changes the mean, not the covariance or variance.
Check the time index. The formula assumes a fixed starting value and $t>0$. If $X_0$ is fixed, $X_0$ itself has zero variance; at $t=0$, the correlation is undefined because $X_0$ is nonrandom.
Do not call this a stationary ACF. Weak stationarity requires the covariance between two observations to depend only on their separation. Here the correlation changes with the earlier time.

A random walk can therefore show strong level persistence even though its innovations have zero covariance at distinct times. For analysis, the key question is whether a reported correlation concerns the accumulated levels or the step-to-step changes. The formula derived here applies to the former, under its stated assumptions, and must be indexed by the observation origin as well as by lag. This distinction is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each problem, use the earlier time as $t$, the separation as $h$, and the random-walk level correlation formula. Unless stated otherwise, the initial value is fixed and the innovations are independent normal random variables with the given mean and variance.

$X_0=0$, with innovations $N(0,16)$. Find $\operatorname{Corr}(X_3,X_8)$. (Answer: $0.6124$)
A random walk has deterministic drift $2$ per time step and innovations $N(0,9)$. Find $\operatorname{Corr}(X_6,X_9)$, explaining whether the drift changes the result. (Answer: $0.8165$; the drift does not change it.)
For innovations $N(0,5)$, find $\operatorname{Corr}(X_{10},X_{12})$. (Answer: $0.9129$)
For innovations $N(0,1)$, compare lag-2 correlations at origins $t=2$ and $t=18$. (Answer: $0.7071$ at $t=2$, and $0.9487$ at $t=18$)
For innovations $N(0,5)$, calculate $\operatorname{Corr}(X_5,X_8)$ by first finding the covariance and both variances. (Answer: covariance $25$, variances $25$ and $40$, correlation $0.7906$)

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