Linear and Log Linear Trend Models

Quantitative Methods 7 min readModule 1 of 5Save

What you'll learn

LOS 2.a
Calculate and evaluate the predicted trend value for a time series, modeled as either a linear trend or a log linear trend, given the estimated trend coefficients.
LOS 2.b
Describe factors that determine whether a linear or a log linear trend should be used with a particular time series and evaluate limitations of trend models.

A time series is a set of observations for a variable over successive periods of time (e.g., monthly stock market returns for the past 10 years). The series has a trend if a consistent pattern can be seen by plotting the data (i.e., the individual observations) on a graph. For example, a seasonal trend in sales data is easily detected by plotting the data and noting the significant jump in sales during the same month(s) each year.

Linear Trend Model

A linear trend is a time series pattern that can be graphed using a straight line. A downward sloping line indicates a negative trend, while an upward sloping line indicates a positive trend.

The simplest form of a linear trend is represented by the following linear trend model:

y_{t} = b_{0} + b_{1} t + ε_{t}

Ordinary least squares (OLS) regression is used to estimate the coefficient in the trend line, which provides the following prediction equation:

y^_{t} = b^_{0} + b^_{1} t

Don’t let this model confuse you. It’s very similar to the simple linear regression model we covered previously; only here, (t) takes on the value of the time period. For example, in Period 2, the equation becomes:

y^_{2} = b^_{0} + b^_{1} (2)

And, likewise, in Period 3:

y^_{3} = b^_{0} + b^_{1} (3)

This means $y^$ increases by the value of $b^_{1}$ each period.

Worked example: Using a linear trend model

Suppose you are given a linear trend model with $b^_{0} = 1.70$ and $b^_{1} = 3.0$

Calculate $y^_{t}$ for t = 1 and t = 2.

Solution

y^_{1} = 1.70 + 3.0 (1) = 4.70

y^_{2} = 1.70 + 3.0 (2) = 7.70

Note that the difference between $y^_{1}$ and $y^_{2}$ is 3.0, or the value of the trend coefficient $b_{1}$ .

Worked example: Trend analysis

Consider hypothetical time series data for manufacturing capacity utilization.

Manufacturing Capacity Utilization

Applying the OLS methodology to fit the linear trend model to the data produces the results shown here.

Time Series Regression Results for Manufacturing Capacity Utilization

Based on this information, predict the projected capacity utilization for the time period involved in the study (i.e., in sample estimates).

Solution

As shown in the regression output, the estimated intercept and slope parameters for our manufacturing capacity utilization model are $b^_{0} = 82.137$ and $b^_{1} = - 0.223$ , respectively. This means that the prediction equation for capacity utilization can be expressed as:

y^_{t} = 82.137 - 0.223 t

With this equation, we can generate estimated values for capacity utilization, $y^_{t}$ , for each of the 14 quarters in the time series. For example, using the model capacity utilization for the first quarter of 2020 is estimated at 81.914:

y^_{1} = 82.137 - 0.223 (1) = 81.914

Note that the estimated value of capacity utilization in that quarter (using the model) is not exactly the same as the actual, measured capacity utilization for that quarter (82.4). The difference between the two is the error or residual term associated with that observation:

residual = actual value - predicted value \approx 82.4 - 81.914 = 0.486

Note that since the actual, measured value is greater than the predicted value of y for 2020.1, the error term is positive. Had the actual, measured value been less than the predicted value, the error term would have been negative.

The projections (i.e., values generated by the model) for all quarters are compared to the actual values here.

Projected Versus Actual Capacity Utilization

The following graph shows visually how the predicted values compare to the actual values, which were used to generate the regression equation. The residuals, or error terms, are represented by the distance between the predicted (straight) regression line and the actual data plotted in blue. For example, the residual for t = 10 is 81.9 − 79.907 = 1.993.

Predicted vs. Actual Capacity Utilization

Since we utilized a linear regression model, the predicted values will by definition fall on a straight line. Since the raw data does not display a linear relationship, the model will probably not do a good job of predicting future values.

Log Linear Trend Models

Time series data, particularly financial time series, often display exponential growth (growth with continuous compounding). Positive exponential growth means that the random variable (i.e., the time series) tends to increase at some constant rate of growth. If we plot the data, the observations will form a convex curve.

Negative exponential growth means that the data tends to decrease at some constant rate of decay, and the plotted time series will be a concave curve.

When a series exhibits exponential growth, it can be modeled as:

y_{t} = e^{b_{0} + b_{1} t}

This model defines y, the dependent variable, as an exponential function of time, the independent variable. Rather than try to fit the nonlinear data with a linear (straight line) regression, we take the natural log of both sides of the equation and arrive at the log linear model. This is frequently used when time series data exhibit exponential growth.

ln (y_{t}) = ln (e^{b_{0} + b_{1} t}) = b_{0} + b_{1} t

Now that the equation has been transformed from an exponential to a linear function, we can use a linear regression technique to model the series. The use of the transformed data produces a linear trend line with a better fit for the data and increases the predictive ability of the model.

Worked example: Log linear trend model

An analyst estimates a log linear trend model using quarterly revenue data (in millions of $) from the first quarter of 2012 to the fourth quarter of 2023 for JP Northfield, Inc.:

ln (revenue_{t}) = b_{0} + b_{1} t + ε_{t}, t = 1, 2, \dots, 48

The results are shown in the following table.

Variable	Coefficient	Standard error	t statistic
Intercept	4.00	0.05	80.0
Trend	0.09	0.01	9.0

Calculate JP Northfield’s predicted revenues in the first quarter of 2024.

Solution

In the first quarter of 2024, t is equal to 49 because the sample has 48 observations.

ln (revenue_{49}) = 4.00 + 0.09 (49) = 8.41

revenue_{49} = e^{8.41} = $4, 492 million

The first answer you get in this calculation is the natural log of the revenue forecast. In order to turn the natural log into a revenue figure, you use the 2nd function of the LN key ( $e^{x}$ ) on your BA II Plus: enter 8.41 and press [2nd] $e^{x}$ = 4,492 million.

Factors that Determine Which Model is Best

To determine if a linear or log linear trend model should be used, the analyst should plot the data. A linear trend model may be appropriate if the data points appear to be equally distributed above and below the regression line. Inflation rate data can often be modeled with a linear trend model.

If, on the other hand, the data plots with a non linear (curved) shape, then the residuals from a linear trend model will be persistently positive or negative for a period of time. In this case, the log linear model may be more suitable. In other words, when the residuals from a linear trend model are serially correlated, a log linear trend model may be more appropriate.

By taking the log of the y variable, a regression line can better fit the data. Financial data (e.g., stock indices and stock prices) and company sales data are often best modeled with log linear models.

Figure 2.1 shows a time series that is best modeled with a log linear trend model rather than a linear trend model.

Figure 2.1: Linear vs. Log Linear Trend Models

The left panel is a plot of data that exhibits exponential growth along with a linear trend line. The panel on the right is a plot of the natural logs of the original data and a representative log linear trend line. The log linear model fits the transformed data better than the linear trend model and, therefore, yields more accurate forecasts.

The bottom line is that when a variable grows at a constant rate, a log linear model is most appropriate. When the variable increases over time by a constant amount, a linear trend model is most appropriate.

Limitations of Trend Models

Recall that one of the assumptions underlying linear regression is that the residuals are uncorrelated with each other. A violation of this assumption is referred to as serial correlation or autocorrelation.

In the preceding discussion, we suggested that a log linear trend model would be better than a linear trend model when the variable exhibits a constant growth rate. However, it may be the case that even a log linear model is not appropriate in the presence of serial correlation. In this case, we will want to turn to an autoregressive model.

Recall from the previous topic review that the Durbin Watson statistic (DW) is used to detect autocorrelation. For a trend model without serial correlation DW should be approximately equal to 2.0. A DW significantly different from 2.0 suggests that the residual terms are correlated.

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