Differencing with and Without Drift

Why the Difference Between Models Matters

A series that rises over time may be rising because each new observation builds on the previous one, or because it fluctuates around a deterministic line that slopes upward. Those mechanisms can produce similar-looking plots, but first differencing affects them differently. The resulting mean and dependence structure help explain what has changed and how forecasts should be constructed.

Let \(Y_t\) denote the observed value at integer time \(t\). Its first difference is the change from one time point to the next. Differencing is a transformation, not a model-selection decision by itself: the model assumed for \(Y_t\) determines what the transformed series means.

Definition: The first difference of a series is \(\Delta Y_t=Y_t-Y_{t-1}\), for \(t\geq 1\). A random walk with drift \(c\) has \(Y_t=Y_{t-1}+c+\varepsilon_t\), where the innovations \(\varepsilon_t\) have mean zero and are independent over time, with finite variance. A linear-trend model has \(Y_t=\alpha+\beta t+u_t\), where \(\alpha\) and \(\beta\) are fixed parameters and \(u_t\) is a mean-zero stationary error process.

For the random walk, the level at time \(t\) depends directly on the preceding level. The increment \(c+\varepsilon_t\) is the new change. In the trend model, the mean level follows the line \(\alpha+\beta t\), while \(u_t\) describes deviations from that line. These definitions make different assumptions about how persistence in the levels arises.

Differencing a Random Walk

Suppose \(Y_t=Y_{t-1}+c+\varepsilon_t\), with a fixed drift \(c\) and zero-mean innovations. Subtracting \(Y_{t-1}\) from both sides gives the first difference directly. When \(c=0\), this is a random walk without drift; when \(c\ne 0\), the average change per time step is \(c\).

\Delta Y_t=Y_t-Y_{t-1}=c+\varepsilon_t,\qquad E[\Delta Y_t]=c.

If the innovations are independent with common variance \(\sigma^2\), the differenced series has variance \(\sigma^2\) and zero covariance between distinct time points. Thus, under these assumptions, differencing removes the dependence created by the random walk’s accumulated level. It does not remove the drift: the differenced series has constant mean \(c\). With no drift, its mean is zero.

The result follows from cancellation: the previous level appears in both \(Y_t\) and the subtraction \(Y_t-Y_{t-1}\), leaving only the new drift and innovation. It is this cancellation, rather than a general property of differencing, that makes first differences of a random walk equal to its one-step innovations plus drift.

Worked Example: Differencing a Random Walk With Known Drift

A hypothetical monthly portfolio index follows \(Y_t=Y_{t-1}+4+\varepsilon_t\), where the drift is 4 index points per month. The observed innovations for four months are \(-2,1,0,3\), and \(Y_0=120\). Find the four observed first differences, the resulting levels, and the three-month-ahead forecast from month 4 using the stated model.

Method. Apply the recurrence one month at a time. Each difference is \(4+\varepsilon_t\):

Month 1: \(\Delta Y_1=4+(-2)=2\), so \(Y_1=120+2=122\).
Month 2: \(\Delta Y_2=4+1=5\), so \(Y_2=122+5=127\).
Month 3: \(\Delta Y_3=4+0=4\), so \(Y_3=127+4=131\).
Month 4: \(\Delta Y_4=4+3=7\), so \(Y_4=131+7=138\).

For a forecast, future innovations have conditional mean zero. Therefore, each future expected increment is the known drift of 4. Over three months, the forecast is \(138+3(4)=150\) index points.

Check. The four differences sum to \(2+5+4+7=18\), and \(Y_4-Y_0=138-120=18\), as required. The method avoids confusing the observed increments, which include innovations, with the model’s mean increment of 4.

Differencing a Linear-Trend Series

Now suppose \(Y_t=\alpha+\beta t+u_t\), where \(u_t\) is stationary with mean zero. Subtract the expression at time \(t-1\) from the expression at time \(t\). The intercept cancels, and the deterministic part changes by \(\beta\):

\begin{aligned} \Delta Y_t &=(\alpha+\beta t+u_t)-(\alpha+\beta(t-1)+u_{t-1})\\ &=\beta+(u_t-u_{t-1}). \end{aligned}

Since both \(u_t\) and \(u_{t-1}\) have mean zero, the differenced series has mean \(\beta\). This mean is the slope of the deterministic trend: it is the average change per time step. But the remaining term is a difference of successive errors, not a new independent innovation in general.

A useful special case is white-noise trend error: the \(u_t\) are independent with mean zero and variance \(\sigma^2\). Then \(w_t=u_t-u_{t-1}\) has variance \(2\sigma^2\). Its lag-one covariance is negative:

\begin{aligned} \operatorname{Cov}(w_t,w_{t-1}) &=\operatorname{Cov}(u_t-u_{t-1},u_{t-1}-u_{t-2})\\ &=-\operatorname{Var}(u_{t-1})=-\sigma^2. \end{aligned}

The other covariance terms in the expansion are zero because distinct \(u_t\)'s are independent. At lags greater than one, the two differences share no error term, so their covariance is zero. Consequently, differencing a line plus independent level noise produces a series with mean \(\beta\), variance \(2\sigma^2\), and negative lag-one covariance. This differs from the random walk with independent innovations, whose differences have zero covariance at every nonzero lag.

Worked Example: Differencing a Series Around a Straight-Line Trend

A hypothetical monthly claim-processing index follows \(Y_t=50+3t+u_t\), where the errors are \(u_0=1\), \(u_1=-2\), \(u_2=0\), \(u_3=2\), and \(u_4=-1\). Calculate the levels and first differences from \(t=0\) through \(t=4\), and explain why the observed average difference need not equal the slope exactly.

Method. Substitute each time and error value into the level equation:

\(Y_0=50+3(0)+1=51\).
\(Y_1=50+3(1)-2=51\).
\(Y_2=50+3(2)+0=56\).
\(Y_3=50+3(3)+2=61\).
\(Y_4=50+3(4)-1=61\).

Subtract consecutive levels: \(\Delta Y_1=51-51=0\), \(\Delta Y_2=56-51=5\), \(\Delta Y_3=61-56=5\), and \(\Delta Y_4=61-61=0\). The same values follow from \(\Delta Y_t=3+u_t-u_{t-1}\): respectively \(3-2-1=0\), \(3+0-(-2)=5\), \(3+2-0=5\), and \(3-1-2=0\).

The observed average difference is \((0+5+5+0)/4=2.5\), whereas the model’s mean difference is \(\beta=3\). There is no contradiction: over this finite stretch, the error changes do not average to zero. In fact, the error-change terms telescope to \(u_4-u_0=-2\), so the four differences sum to \(4(3)-2=10\), giving average \(10/4=2.5\).

Check. The difference sum is \(Y_4-Y_0=61-51=10\), consistent with the calculated average. The method avoids treating a finite-sample average as though it must equal the model’s expected change.

What the Mean Says About Forecasts

The mean of the differenced series is useful, but it must be interpreted in the context of the level model. For a random walk with drift, a future level is the current level plus the sum of future increments. If the drift \(c\) is known and each future innovation has conditional mean zero, the \(h\)-step-ahead conditional mean is:

E[Y_{T+h}\mid Y_0,\ldots,Y_T]=Y_T+hc.

The forecast therefore extends from the latest observed level, adding the mean increment \(c\) at each step. When there is no drift, the forecast is \(Y_T\) at every horizon. This is a statement about the conditional mean, not a claim that the future level will equal that value.

For the linear-trend model, the mean level at time \(T+h\) is \(\alpha+\beta(T+h)\). If the errors are independent over time with mean zero, a future error has conditional expectation zero given the observations through \(T\), so:

E[Y_{T+h}\mid Y_0,\ldots,Y_T]=\alpha+\beta(T+h).

Here the forecast follows the fitted line, rather than extending the most recent level by a drift from the last observed change. If the errors are stationary but dependent, their future conditional means need not be zero given the observed history; the forecast must then account for that dependence. The displayed simple forecast formula applies to independent mean-zero errors.

Worked Example: Comparing Forecast Paths After Differencing

At time \(T=20\), a hypothetical index has value \(Y_{20}=210\). Compare its four-step-ahead conditional mean under two alternative models: (A) a random walk with known drift \(c=2.5\), and (B) a linear-trend model with \(\alpha=100\), \(\beta=2\), and independent mean-zero errors. For model B, assume the observed value is consistent with a current error of zero.

Model A: random walk with drift. Add four expected increments of 2.5 to the current level: \(E[Y_{24}\mid Y_{20}]=210+4(2.5)=210+10=220\).

Model B: linear trend. The time-24 trend level is \(100+2(24)=100+48=148\). Since future errors have conditional mean zero under the stated independence assumption, the four-step conditional mean is 148.

These values differ substantially because the model B trend line does not pass through the stated current level: at time 20 its mean is \(100+2(20)=140\), while the observed value is 210. The assumed current error is therefore not zero; it is \(210-140=70\). The problem’s conditions are inconsistent as written, which must be noticed rather than silently ignored.

To make the comparison coherent, use a current level of 140 for both models. Then model A gives \(140+4(2.5)=150\), while model B gives \(100+2(24)=148\). These different conditional means reflect different assumptions: the random walk carries its current level forward, while the trend model returns to its deterministic line under independent future errors.

Check. The trend model’s current fitted mean is 140, so changing the common current observation to 140 resolves the inconsistency. The method avoids combining incompatible level information and model parameters.

Common Exam Traps

Assuming any constant mean difference proves a random walk. A linear-trend model also has mean first difference \(\beta\). The dependence pattern of the differences and the assumptions about levels matter as well.
Forgetting the error difference in a trend model. The correct expression is \(\Delta Y_t=\beta+u_t-u_{t-1}\), not simply \(\beta+u_t\). Successive differences share a level error, which can create dependence.
Confusing a model mean with a realized sample average. A finite collection of observed differences need not average exactly to \(c\) or \(\beta\). Random deviations and endpoint error changes affect the observed average.
Forecasting from a differenced mean without returning to the level. An expected increment is a change, not a level. For a random walk, add the expected increments to the current level; for a trend model, evaluate the trend at the forecast time under the stated error assumptions.
Ignoring the time step. If \(t\) indexes months, \(c\) and \(\beta\) are changes per month. A four-step forecast adds four monthly expected changes, not one annual change.

Key takeaway: A random walk with drift differences to a constant mean plus its new innovation; a linear trend with stationary errors differences to its slope plus the change in its level error. Both can have a nonzero differenced mean, but their dependence assumptions and level forecasts are not interchangeable.

Check Your Understanding

Work each problem from the stated model and conditions; the final answer is provided for checking.

A monthly balance follows \(Y_t=Y_{t-1}-1+\varepsilon_t\), with mean-zero independent innovations. What is the mean of \(\Delta Y_t\), and what is the four-month-ahead conditional mean if \(Y_T=80\)? (Answer: Mean difference \(-1\); forecast \(76\).)
A series is \(Y_t=12+0.8t+u_t\), where \(u_t\) is stationary with mean zero. Write \(\Delta Y_t\) and its mean. (Answer: \(\Delta Y_t=0.8+u_t-u_{t-1}\); mean \(0.8\).)
For a trend model with independent errors of variance \(9\), what are the variance and lag-one covariance of the differenced error \(u_t-u_{t-1}\)? (Answer: Variance \(18\); lag-one covariance \(-9\).)
A random walk has known drift \(3\) per quarter and current level \(Y_T=45\). Find the six-quarter-ahead conditional mean. (Answer: \(45+6(3)=63\).)
A series has level model \(Y_t=40+1.5t+u_t\), with independent mean-zero errors. At \(T=10\), calculate the three-step-ahead conditional mean. Explain why the answer is not obtained by adding three times the last observed change. (Answer: \(40+1.5(13)=59.5\); the forecast is the trend level at time 13 under independent future errors, not an extrapolation of a realized last change.)

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