CFA Level II Exam · Time-Series Analysis

Linear and Log-Linear Trend Models in Time-Series Analysis

Updated 7 October 2026 · Fact-checked

A trend model regresses a time series on time. A linear trend, y = b0 + b1t + e, assumes a constant change per period. A log-linear trend, ln y = b0 + b1t + e, assumes a constant growth rate. Pick the one whose residuals look random. Use Durbin-Watson to test for serial correlation.

Understand Trend Models: Linear and Log-Linear Trends

A trend model explains a variable using only time. You number the periods t = 1, 2, 3, and so on, and regress the variable on t. It is the simplest time-series model.

In a linear trend model, the value changes by a constant amount each period. The model is y(t) = b0 + b1t + e(t). The slope b1 is the expected change in y per period. A plot of y against time looks like a straight line.

In a log-linear trend model, the natural log of the value is linear in time: ln y(t) = b0 + b1t + e(t). Here y grows at a constant rate. The slope b1 is the continuously compounded growth rate per period. Exponentiating gives y = e^(b0 + b1t), a curve that bends upward when b1 is positive. Use it for series such as sales or prices that grow exponentially. It fits only when all values are positive, since you cannot take the log of zero or a negative number.

How do you choose? Plot the data and plot the residuals. If a linear model on a growing series leaves residuals that are mostly positive, then mostly negative, then positive again, the fit is poor. The errors are correlated over time, which signals that the model form is wrong. Switching to log-linear often fixes this when growth is exponential.

Trend models have a key limitation. The error in one period is often related to the error in the previous one. This is serial correlation. When it is present, the model's coefficient standard errors are unreliable and the t-tests cannot be trusted. The Durbin-Watson test checks for it. If a trend model shows serial correlation, move to an autoregressive (AR) model.

Key formulas to remember

Linear trend model: y(t) = b0 + b1t + e(t); b1 is the constant change in y per period. t = 1, 2, ..., T.
Log-linear trend model: ln y(t) = b0 + b1t + e(t); b1 is the constant continuously compounded growth rate per period. y must be positive.
Forecast from log-linear model: ŷ(t) = exp(b0 + b1t); Compute the ln forecast first, then take the exponential. Do not forget this last step.
Durbin-Watson statistic: DW ≈ 2(1 − r); r is the correlation between residuals and lagged residuals. DW near 2 means no serial correlation. DW below 2 suggests positive serial correlation. DW above 2 suggests negative.
Durbin-Watson decision rule: Reject H0 (no positive serial correlation) if DW < dl; inconclusive if dl ≤ DW ≤ du; do not reject if DW > du; dl and du come from a table given in the exam. They depend on sample size and number of independent variables.
Trend model limitation: Serial correlation → biased standard errors; t-tests unreliable; The Durbin-Watson test is not valid for autoregressive models with a lagged dependent variable. For those, test residual autocorrelations with a t-test.

How to solve Trend Models: Linear and Log-Linear Trends questions

Use this method for any item-set question on trend models.

1Find the model in the vignette or exhibit. Check whether the dependent variable is y or ln y. This decides whether the slope is an amount or a growth rate.
2Read the slope correctly. For linear, b1 is the change in y per period. For log-linear, b1 is the growth rate per period, as a decimal.
3If asked to forecast, set t to the right period number. Count carefully from the start of the sample.
4For a log-linear forecast, compute b0 + b1t, then take exp of it. Only then do you have a forecast in original units.
5To choose between models, look for the one with randomly scattered residuals, a better fit for the data pattern, and no sign of serial correlation. Exponential growth points to log-linear.
6For the Durbin-Watson test, compare DW with dl and du from the table in the exhibit. Below dl means reject the null of no positive serial correlation.
7If serial correlation is found, state the conclusion: the trend model is inadequate and standard errors are unreliable, so an AR model is the next step.

Quickest way: Three-check shortcut for trend questions

When to use it: Use when time is short and the vignette gives a fitted equation, a residual comment, or a DW statistic.

Look at the left side of the equation. If it says ln, the model is log-linear and the slope is a growth rate.
For a forecast, plug in t, and exponentiate if the model is log.
For DW, use the approximation 2(1 − r). Values far below 2 mean positive serial correlation. Then compare with dl and du only if a table is provided.
If a question says the residuals show a pattern over time, choose the answer that says serial correlation exists and a different model is needed.

Common mistakes in Trend Models: Linear and Log-Linear Trends

Reading the log-linear slope as a change in the value of y.
The linear model trains you to read the slope as units per period.
Fix: If the dependent variable is ln y, b1 is a growth rate. Multiply by 100 for a percentage.
Forgetting to exponentiate a log-linear forecast.
You stop once the regression equation gives a number.
Fix: That number is ln y. Take e raised to it to get y.
Using the wrong value of t.
The first observation is t = 1, but students count from 0 or confuse the year with the period.
Fix: Write down which date is t = 1 and count forward. A forecast five periods after a sample of 40 uses t = 45.
Treating a high R-squared as proof the trend model is good.
Trending series fit well on R-squared almost by construction.
Fix: Check residuals and the Durbin-Watson statistic. A high R-squared with serial correlation still gives unreliable t-tests.
Reversing the Durbin-Watson decision rule.
Students mix up dl and du, or think a larger DW means more serial correlation.
Fix: Below dl is evidence of positive serial correlation. Above du is no evidence. Between is inconclusive. DW near 2 is the good outcome.
Applying the Durbin-Watson test to an AR model.
The test was learned for trend regressions and is applied everywhere.
Fix: With a lagged dependent variable as a regressor, DW is not valid. Use the t-test on residual autocorrelations.

Worked examples

Example 1

An analyst fits a log-linear trend to a company's quarterly revenue over 40 quarters: ln(Revenue) = 4.20 + 0.025t, where revenue is in ₹ crore and t = 1 is the first quarter. Q1: What does the slope mean? Q2: What is the forecast revenue for t = 44? Q3: The Durbin-Watson statistic is 0.85, with dl = 1.44 and du = 1.54. What is the conclusion?

Show the solution

Q1: The dependent variable is ln(Revenue), so the slope 0.025 is the continuously compounded growth rate per quarter, which is 2.5% per quarter.
Q2: Compute ln forecast = 4.20 + 0.025 × 44 = 4.20 + 1.10 = 5.30.
Then revenue = e^5.30. Since e^5 = 148.41 and e^0.30 = 1.3499, revenue ≈ 148.41 × 1.3499 = 200.3, so about ₹200.3 crore.
Q3: DW = 0.85 is below dl = 1.44. Reject the null of no positive serial correlation.
This means the residuals are positively correlated, so the standard errors and t-tests are unreliable. The trend model is not adequate.

Answer: Q1: 2.5% continuously compounded growth per quarter. Q2: about ₹200.3 crore. Q3: Positive serial correlation exists; the model is inadequate and an AR model should be considered.

Example 2

An analyst models annual sales of a firm over 20 years. Linear trend: Sales = 50 + 8t (₹ crore). Log-linear trend: ln(Sales) = 3.90 + 0.06t. A residual plot of the linear model shows long runs of positive then negative errors, and the log-linear residuals look random. Q1: Which model is preferred and why? Q2: What does the log-linear model forecast for t = 21?

Show the solution

Q1: Long runs of same-sign residuals in the linear model signal serial correlation and a poor functional form. The log-linear residuals look random, so the log-linear model fits better. Sales are growing at a roughly constant rate, which suits the log-linear form.
Q2: ln(Sales) = 3.90 + 0.06 × 21 = 3.90 + 1.26 = 5.16.
Sales = e^5.16. Since e^5 = 148.41 and e^0.16 = 1.1735, Sales ≈ 148.41 × 1.1735 = 174.2.
So forecast sales are about ₹174.2 crore.

Answer: Q1: The log-linear model, because its residuals are random and the data show constant growth. Q2: about ₹174.2 crore.

Exam tips

Always read the left side of the equation first. ln y versus y changes the whole interpretation.
Questions often give a Durbin-Watson statistic and dl and du. Practise the three-zone rule until it is automatic.
When a vignette says residuals show a pattern, the answer is almost always serial correlation and a need for a different model, not a higher R-squared.
For log-linear forecasts, calculate carefully with the exponential. Options are usually spaced so that forgetting to exponentiate gives an obviously wrong answer, but check.
Remember that DW is for trend and standard regressions. For AR models, the exam expects the residual autocorrelation t-test.

Trend Models: Linear and Log-Linear Trends in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

FRM Part IDeterministic Trends and Trend Models

Trend Models: Linear and Log-Linear Trends: frequently asked questions

How do I choose between a linear and log-linear trend model?

Plot the data and the residuals. If the series grows by a roughly constant amount, use linear. If it grows by a roughly constant percentage, use log-linear. Prefer the model with randomly scattered residuals and no serial correlation.

What does the slope mean in a log-linear trend model?

It is the continuously compounded growth rate per period. A slope of 0.03 means about 3% growth per period. It is not a change in units.

What does a Durbin-Watson statistic near 2 mean?

It means there is no evidence of serial correlation in the residuals. Values well below 2 suggest positive serial correlation. Values well above 2 suggest negative serial correlation.

What should I do if a trend model has serial correlation?

The coefficient standard errors are unreliable, so you cannot trust the t-tests. The usual next step is to use an autoregressive model that captures the dependence on past values.