Fitting an exponential model with log(y) = a + b t or y = exp(a + b t)

 

Data

$$x=2010:2020$$

(11 points)

$$y=(10,10,15,20,30,60,100,120,200,300,400)$$

To simplify interpretation, the year is often centered:

$$t=x-2010=0,1,\dots ,10$$

---

1️⃣ Linear regression on log(y)

Model

$$\log (y)=\alpha +\beta t+\epsilon$$

Key assumption

• the error is additive on the log scale

• therefore multiplicative on the original scale

Fit (order of magnitude)

One typically obtains something like:

$$\log (y)\approx 2.2+0.36t$$

Back to the original scale

$$\widehat{y}=\exp (2.2+0.36t)$$

👉 regular exponential growth
👉 relative errors are roughly constant
👉 small values have as much weight as large ones

---

2️⃣ Direct nonlinear regression on y

Model

$$y=a{e}^{bt}+\epsilon$$

Key assumption

• the error is additive on y

• variance is assumed constant on the original scale

Typical fit

$$\widehat{y}\approx 9.5e^{0.39t}$$

Consequences

• large values (300, 400) strongly dominate the fit

• early years are often poorly fitted

• residuals are strongly heteroscedastic

---

3️⃣ Why the results differ (crucial point)

Even if the mean function is the same:

$$E[y\mid t]\propto e^{bt}$$

the error assumptions differ:

Model	Error type
$\log (y)=\alpha +\beta t+\epsilon$	multiplicative
$y=a{e}^{bt}+\epsilon$	additive

➡️ these are not the same least-squares problem

---

4️⃣ Intuitive illustration with data

Look at successive ratios of $y$:

Year	y	Ratio
2010	10	–
2011	10	1.0
2012	15	1.5
2013	20	1.33
2014	30	1.5
2015	60	2.0
2016	100	1.67
2017	120	1.2
2018	200	1.67
2019	300	1.5
2020	400	1.33

➡️ ratios are relatively stable, not absolute differences
➡️ a multiplicative assumption is reasonable
➡️ using log(y) is therefore statistically justified

---

5️⃣ Beware of retransformation bias

If you compute:

$$\widehat{y}=\exp (\widehat{\log (y)})$$

then:

$$E[y]\mathrm{\ne }\exp (E[\log (y)])$$

Classical correction (smearing estimator):

$$\widehat{y}=\exp (\widehat{\mu })\times \exp ({\sigma }^2\mathrm{/}2)$$

This becomes important when doing prediction.

---

6️⃣ Key takeaway

👉 Putting log(y) in a linear regression is not equivalent to fitting an exponential model directly to y because the error structure changes completely

Here is clean, reproducible R code using these data, showing the two approaches side by side and how to handle retransformation correctly.

## Data
x <- 2010:2020
y <- c(10, 10, 15, 20, 30, 60, 100, 120, 200, 300, 400)

## Centered time variable
t <- x - 2010

## -------------------------------
## 1) Linear regression on log(y)
## -------------------------------
m_log <- lm(log(y) ~ t)
summary(m_log)

## Fitted values on original scale (naive back-transformation)
yhat_log_naive <- exp(predict(m_log))

## Smearing correction (Duan, 1983)
sigma2 <- mean(residuals(m_log)^2)
smearing <- exp(sigma2 / 2)

yhat_log_corrected <- yhat_log_naive * smearing

## -------------------------------
## 2) Nonlinear regression on y
## -------------------------------
m_nls <- nls(
  y ~ a * exp(b * t),
  start = list(a = 10, b = 0.3)
)
summary(m_nls)

yhat_nls <- fitted(m_nls)

## -------------------------------
## 3) Compare fits
## -------------------------------
plot(
  x, y,
  log = "y",
  pch = 16,
  xlab = "Year",
  ylab = "y (log scale)",
  main = "Log-linear vs nonlinear exponential fit"
)

lines(x, yhat_log_corrected, col = "blue", lwd = 2)
lines(x, yhat_nls, col = "red", lwd = 2)

legend(
  "topleft",
  legend = c("Observed", "log(y) ~ t (smearing)", "y ~ a exp(bt)"),
  pch = c(16, NA, NA),
  lwd = c(NA, 2, 2),
  col = c("black", "blue", "red")
)

## -------------------------------
## 4) Residual diagnostics
## -------------------------------
par(mfrow = c(1, 2))

plot(
  fitted(m_log),
  residuals(m_log),
  pch = 16,
  main = "Residuals: log-linear model",
  xlab = "Fitted log(y)",
  ylab = "Residuals"
)
abline(h = 0, lty = 2)

plot(
  yhat_nls,
  y - yhat_nls,
  pch = 16,
  main = "Residuals: nonlinear model",
  xlab = "Fitted y",
  ylab = "Residuals"
)
abline(h = 0, lty = 2)

How to interpret the output

• lm(log(y) ~ t)

  • assumes multiplicative error

  • stabilizes variance

  • gives equal weight to early and late years

• nls(y ~ a * exp(b * t))

  • assumes additive error

  • large values dominate the fit

  • often shows heteroscedastic residuals

• The smearing correction is essential if you want unbiased predictions on the original scale from a log-linear model.

  Duan, N. (1983). Smearing estimate: A nonparametric retransformation method. Journal of the American Statistical Association, 78(383), 605-610. https://doi.org/10.2307/2288126