First Differences And The Difference Operator

Why First Differences Matter

A time series records observations in time order. Its levels describe what was observed at each time; its first differences describe how much the observed value changed from one time to the next. For example, a monthly account balance of 104 followed by 111 has increased by 7 units over that month. The difference focuses on the change, not the balance itself.

The earlier material on stationarity and random walks motivates why changes can be useful to examine. Here the goal is narrower: define adjacent changes precisely and express them with a difference operator. The calculations below are arithmetic transformations of hypothetical observations; they require no interest-rate, mortality, or probability-distribution assumptions.

Definition: For a time series \(y_1,y_2,\ldots,y_n\) observed at equally spaced times, the first difference at time \(t\) is the current observation minus the immediately preceding observation:

\Delta y_t = y_t-y_{t-1}, \qquad t=2,\ldots,n.

The symbol \(\Delta\) is the first-difference operator. Applying it to the series produces the difference sequence \(\Delta y_2,\Delta y_3,\ldots,\Delta y_n\), which contains \(n-1\) values.

Reading and Interpreting a Difference

The order of subtraction matters: \(\Delta y_t\) is the current value \(y_t\) minus the previous value \(y_{t-1}\). A positive difference means the series increased over that interval; a negative difference means it decreased; zero means the two adjacent values are equal. The difference has the same measurement units as the original series. If \(y_t\) is measured in thousands of dollars, then \(\Delta y_t\) is also measured in thousands of dollars.

A difference is a change over one observation interval, not automatically a change per unit of calendar time. If observations are quarterly, \(\Delta y_t\) is a change per quarter. If the times are not equally spaced, a difference still gives the change between the two values used, but those changes do not all cover the same duration. Dividing by elapsed time would give an average change per time unit, a distinct calculation.

The first observed value has no first difference within the observed series because there is no earlier observation in the series to subtract. Consequently, differencing a series of \(n\) observations gives \(n-1\) differences. It does not, by itself, supply a value for the unobserved change before the first observation.

Worked Example: Quarterly Changes in a Hypothetical Portfolio Measure

A hypothetical portfolio report records a quarterly measure, in thousands of dollars, of \(128, 135, 131, 144,\) and \(150\). Calculate the first differences, identify the largest increase and decrease, and check the total change. Use the difference-operator method; this avoids confusing a difference with a percentage change.

Step 1: Label the observations. Let \(y_1=128\), \(y_2=135\), \(y_3=131\), \(y_4=144\), and \(y_5=150\), all in thousands of dollars. Since the observations are quarterly, each difference is a change over one quarter.

Step 2: Subtract each previous observation from the current observation.

\begin{aligned} \Delta y_2 &= 135-128=7,\\ \Delta y_3 &= 131-135=-4,\\ \Delta y_4 &= 144-131=13,\\ \Delta y_5 &= 150-144=6. \end{aligned}

Thus the difference sequence is \(7,-4,13,6\), in thousands of dollars per quarter. The largest increase is \(13\) thousand dollars, from the third to the fourth observation. The largest decrease is \(4\) thousand dollars, from the second to the third observation.

Step 3: Check the total change. Add the differences: \(7+(-4)+13+6=22\). Compare with the change between the final and initial levels: \(150-128=22\). The calculations agree. Across the four quarters, the measure rose by \(22\) thousand dollars.

Why the Differences Add to the Endpoint Change

A useful check follows directly from the definition. The intermediate observations cancel when consecutive differences are added. This is called a telescoping sum:

\begin{aligned} \sum_{t=2}^{n}\Delta y_t &=\sum_{t=2}^{n}(y_t-y_{t-1})\\ &=(y_2-y_1)+(y_3-y_2)+\cdots+(y_n-y_{n-1})\\ &=y_n-y_1. \end{aligned}

Each interior value appears once with a positive sign and once with a negative sign, so only the final value \(y_n\) and initial value \(y_1\) remain. This identity is useful for checking a calculated difference sequence, but it does not say that every individual difference is equal or that the series changed at a constant rate.

The operator is also linear. For two series \(x_t\) and \(y_t\), and constants \(a\) and \(b\), subtracting consecutive values gives:

\Delta(ax_t+by_t) =a(x_t-x_{t-1})+b(y_t-y_{t-1}) =a\Delta x_t+b\Delta y_t.

A related consequence is that adding the same constant \(c\) to every observation does not change the first differences: \(\Delta(y_t+c)=(y_t+c)-(y_{t-1}+c)=y_t-y_{t-1}\). The differences describe changes, so a fixed shift in the level cancels out.

Differences Are Not Percentage Changes

A first difference measures an absolute change in the original units. A percentage change compares that difference with the previous value. For a nonzero previous value, the percentage change from \(t-1\) to \(t\) is:

100\frac{y_t-y_{t-1}}{y_{t-1}} =100\frac{\Delta y_t}{y_{t-1}}\%.

The denominator is the previous observation. Therefore a difference of 6 can represent different percentage changes depending on the value from which the series started. If \(y_{t-1}=0\), this percentage-change formula is undefined, even though the first difference is still defined.

Worked Example: Absolute and Percentage Changes in a Hypothetical Index

A hypothetical monthly index has values \(96,102,99,\) and \(108\). Find each first difference and each percentage change from the previous month, rounded to two decimal places. The method is to calculate the absolute difference first, then divide by the previous level; this avoids treating an absolute change as a percentage.

Step 1: Calculate the first differences. The three adjacent changes are \(102-96=6\), \(99-102=-3\), and \(108-99=9\). The difference sequence is \(6,-3,9\) index points.

Step 2: Use the previous value as the denominator. The percentage changes are:

\begin{aligned} 100\frac{6}{96} &= 6.25\%,\\ 100\frac{-3}{102} &\approx -2.94\%,\\ 100\frac{9}{99} &\approx 9.09\%. \end{aligned}

The absolute changes are in index points, whereas the percentage changes are relative to the respective starting values. As a check, the first difference is \(6\) points, not \(6.25\) points: \(6.25\%\) is the relative change from the starting value of 96. Percentages are rounded only at the final step to two decimal places.

Reconstructing Levels from Differences

If the initial level and a sequence of first differences are known, the later levels can be reconstructed one at a time. Rearranging the definition gives \(y_t=y_{t-1}+\Delta y_t\). Thus each new level is the previous level plus the change that leads to it. The starting level is essential: differences alone determine changes between levels but do not identify the absolute level.

Worked Example: Recovering Levels from a Starting Value

A hypothetical quarterly index starts at \(y_1=72\), followed by first differences \(5,-2,4,-1\). Recover the four later levels and verify the final net change. Use sequential reconstruction; this avoids assigning a difference to the wrong time index.

Step 1: Add each difference to the immediately preceding level.

\begin{aligned} y_2 &= y_1+\Delta y_2=72+5=77,\\ y_3 &= y_2+\Delta y_3=77+(-2)=75,\\ y_4 &= y_3+\Delta y_4=75+4=79,\\ y_5 &= y_4+\Delta y_5=79+(-1)=78. \end{aligned}

Step 2: Check by summing the changes. The differences sum to \(5-2+4-1=6\). The endpoint change is \(y_5-y_1=78-72=6\), as required. The reconstructed levels are \(72,77,75,79,78\); calculating their adjacent differences returns \(5,-2,4,-1\).

Common Traps and Practical Checks

The sign and time index in a first difference are frequent sources of avoidable errors. To find the change from one observation to the next, subtract the earlier value from the later value. Reversing the order gives the negative of the intended change. Also check that the observations truly correspond to consecutive time periods before describing a difference as a one-period change.

For example, if monthly observations are available for March and April, then \(y_{\text{Apr}}-y_{\text{Mar}}\) is a one-month change. If the next available observation is June, then \(y_{\text{Jun}}-y_{\text{Apr}}\) is the change over two months, not the observed one-month change for May or June separately. The missing value cannot be recovered from the two observed endpoints alone.

A final check is to add all the computed differences and compare their sum with the last observed level minus the first. A mismatch usually signals an arithmetic error, a sign reversal, or an omitted interval. The check confirms the total change; it does not validate the interpretation of irregularly spaced observations.

Key takeaway: The first-difference operator records adjacent change: \(\Delta y_t=y_t-y_{t-1}\). Keep the order, time interval, and units clear; use the endpoint identity \(\sum_{t=2}^{n}\Delta y_t=y_n-y_1\) as an arithmetic check.

Check Your Understanding

For each question, use the difference operator and keep the time interval and units in view.

A hypothetical quarterly count is \(41,38,44,43\). Give the first-difference sequence and identify the interval with the largest increase. (Answer: \(-3,6,-1\); the largest increase is 6, from the second to the third observation.)
A hypothetical monthly index is \(250,246,255,264\). Find the first differences and verify their sum using the endpoints. (Answer: \(-4,9,9\); their sum is 14, equal to \(264-250=14\).)
A hypothetical quarterly revenue series, in millions of dollars, is \(1.8,2.1,2.0,2.5\). Find the first differences and their average. (Answer: \(0.3,-0.1,0.5\) million dollars per quarter; average difference \(0.7/3\approx0.2333\) million dollars per quarter.)
A hypothetical series starts at 310 and has successive first differences \(-8,12,-5\). Reconstruct the later levels and state the total change. (Answer: \(302,314,309\); total change \(-1\), which equals \(309-310\).)
A hypothetical monthly series has observations 73 in March, 77 in April, 80 in June, and 78 in July; May is unobserved. Which observed adjacent changes can be reported, and can the April-to-June change be called a one-month first difference? (Answer: March-to-April is \(4\), April-to-June is \(3\) over two months, and June-to-July is \(-2\); the April-to-June change is not a one-month first difference, and the missing May value cannot be determined from these observations.)

What You'll Learn