Functional forms: log-lin, lin-log and log-log models

What if the relationship is not a straight line?

A straight line fits some relationships badly. If a company’s revenue compounds at a steady percentage, a scatter against time bends upward, and the residuals of a linear fit trace a U. The remedy is to change the functional form : replace Y, X or both with their logarithms. The natural logarithm transformation turns multiplicative growth into additive steps, which a straight line handles. Logs are the usual choice; other transformations such as squares, reciprocals and differences (changes instead of levels) can also straighten a relationship.

The regression machinery is unchanged. Everything from the earlier lessons (least squares, the coefficient of determination R², t-tests, the standard error of the estimate, $s_e$) works on the transformed variables. What changes is what the slope means and how you predict.

Which side of the equation is in logs?logs straighten a curved scatter

• Neither: lin-lin Use it the ordinary straight-line model
• Y only: log-lin Memorize ln Y = b₀ + b₁X
  • Use for constant percentage growth over time
  • Slope relative change in Y per unit X
• X only: lin-log Memorize Y = b₀ + b₁ ln X
  • Use for diminishing effects
  • Slope change in Y per relative change in X
• Both: log-log Memorize ln Y = b₀ + b₁ ln X
  • Use for constant-elasticity relations
  • Slope elasticity, % change in Y per 1% in X

Which form for which relationship?

Form	A change in X of	changes Y by	Use when	Our example
Log-lin	one unit	a constant percentage, about b̂₁ × 100% (exactly e^(b̂₁) − 1)	steady growth or decay	revenue compounding each year; a balance shrinking by a constant share each period
Lin-log	1%	a constant amount, about b̂₁ ÷ 100	each extra unit of X adds less (saturation)	margin and sales volume
Log-log	1%	b̂₁ percent	a constant elasticity	advertising and sales; price and quantity demanded

What is the log-lin model?

In the log-lin model (also called log-linear) only Y is in logs.

$$\ln Y_i = b_0 + b_1 X_i + \varepsilon_i$$

The slope is the relative change in Y for a one-unit change in X. For small values, $b_1 \times 100$ is the percentage growth; the exact growth is $e^{b_1} - 1$. This is the model for compound growth over time, with X as the time period.

To predict Y you must reverse the log:

$$\hat{Y} = e^{\hat{b}_0 + \hat{b}_1 X}$$

For a change of more than one unit, compound the slope: $e^{\hat{b}_1 \Delta X} - 1$.

$$\%\Delta Y = e^{\hat{b}_1 \Delta X} - 1$$

What is the lin-log model?

In the lin-log model (also called linear-log) only X is in logs. Equal percentage rises in X give equal absolute rises in Y, so the effect of each extra unit of X fades as X grows.

$$Y_i = b_0 + b_1 \ln X_i + \varepsilon_i$$

The slope is the change in Y for a relative change in X: a 1% rise in X changes Y by about $b_1 \times 0.01$.

Percent or percentage points? When Y is itself a rate in percent, such as a margin, a linear model’s slope is a change in percentage points, and so is a lin-log model’s: a 1% rise in units changes the margin by 1.5 ÷ 100 = 0.015 percentage points. Only a model with ln Y gives a percent change in Y. Keep percent versus percentage point straight: a margin rising from 8% to 9% is a rise of 1 percentage point, or 12.5% in percent terms.

What is the log-log model?

In the log-log model both variables are in logs.

$$\ln Y_i = b_0 + b_1 \ln X_i + \varepsilon_i$$

The slope is an elasticity : the percentage change in Y for a 1% change in X. It is the standard way to estimate a demand curve with constant elasticity.

How do you choose the form?

Start from economic or financial theory about how Y should respond to X, then check goodness of fit and the residual patterns. Look at the scatter and the residual plot of a linear fit. A curved pattern in the residuals is the signal. Try the transformation that matches the economics: compounding growth suggests log-lin, fading effects suggest lin-log, constant percentage responses suggest log-log. The right form leaves residuals that look like random noise.

You can compare candidate models using R², F and $s_e$, but only when the dependent variable is the same. The R² of a model for ln Y and the R² of a model for Y describe the variation of different things, so they cannot be compared directly.

Exam tip Name the form by where the logs are: log on Y only is log-lin, on X only is lin-log, on both is log-log. Then say what the slope is a change in.

Go deeper Why compound growth is linear in logs

If revenue grows by the same factor each year, $Y_t = Y_0 (1+g)^t$, then $\ln Y_t = \ln Y_0 + t\ln(1+g)$. That is a straight line in $t$ with slope $\ln(1+g)$, the continuously compounded growth rate you met in the rates and returns lessons. Regressing the log of revenue on time estimates that rate directly, and $e^{b_1} - 1$ converts it to the ordinary annual growth.

Go deeper Elasticities in economics

The price elasticity of demand is the percentage change in quantity for a 1% change in price. A log-log demand regression estimates it as a constant across all prices, which is why the form appears in economics. A slope of −1.2 means a 1% price rise cuts quantity by about 1.2%, and demand is elastic.