# Introduction

## Column

### Log Transformations

Log transformations can be useful when a variable is very right-skewed, or multiplicative effects are desired over additive. However, interpretation can be challenging.

We are always discussing the natural log (ln), i.e. log base e.

## Column

### Multiplicative vs Percent Change

Note that multiplicative changes can be expressed as percent changes and vice-versa.

If we multiply $$X$$ by 1.1, the resultant $$1.1X$$ is 10% larger than $$X$$. E.g. 16.5 is 10% larger than 15.

If we multiply $$X$$ by .7, the resultant $$.7X$$ is 30% lower than $$X$$. E.g. 7 is 30% smaller than 10.

# Logged Outcome

## Column

### Discussion

#### tldr

A 1 unit change in a predictor is associated with a $$\textrm{exp}(\hat{\beta})$$ mulitplicative change in $$Y$$, or a $$100(\textrm{exp}(\hat{\beta}) - 1)\%$$ change in $$Y$$.

Examples:

• If $$\hat{\beta}$$ is .2, a 1 unit increase in $$X$$ is associated with a $$\textrm{exp}(.2)\ \approx 1.22$$ multiplicative change in $$Y$$, or a 22% increase.
• If $$\hat{\beta}$$ is -.4, a 1 unit increase in $$X$$ is associated with a a $$\textrm{exp}(-.4)\ \approx .67$$ multiplicative change in $$Y$$, or a 33% decrease.

#### Theory

Assume our regression equation is

$E(Y|X = x) = \beta_0 + \beta_1x.$

If we regress on the log of $$Y$$ instead,

$E(\log(Y)|X = x) = \beta_0 + \beta_1x.$

By Taylor expansion,

$\log(E(X)) \approx E(\log(X)).$

Therefore we can write \begin{align*} E(Y|X = x + 1) & = \textrm{exp}\left(\beta_0 + \beta_1(x + 1)\right) \\ & = \textrm{exp}\left(\beta_0 + \beta_1x + \beta_1\right) \\ & = \textrm{exp}\left(\beta_0 + \beta_1x\right)\textrm{exp}(\beta_1) \\ & = E(Y|X = x)\textrm{exp}(\beta_1) \end{align*}

## Column

### Example

data(mtcars)
(m <- lm(log(disp) ~ drat, data = mtcars))

Call:
lm(formula = log(disp) ~ drat, data = mtcars)

Coefficients:
(Intercept)         drat
8.2782      -0.8323  

Therefore a 1-unit increase in drat is associated with an $$\textrm{exp}(-0.832) = 0.435$$ multiplicative change in disp, corresponding to a 56.5% decrease.

To test this, we predict the ratio in predicted outcome with some values of drat, and that value increased by 1. Note: We exponentiate the predicted values to get them on the outcome scale.

exp(predict(m, newdata = data.frame(drat = 5)))/exp(predict(m, newdata = data.frame(drat = 4)))
        1
0.4350567 
exp(predict(m, newdata = data.frame(drat = 30)))/exp(predict(m, newdata = data.frame(drat = 29)))
        1
0.4350567 

# Logged Predictor

## Column

### Discussion

#### tldr

A $$k\%$$ change in a predictor is associated with $$\hat{\beta}\log\left(1 + \frac{k}{100}\right)$$ change in the outcome.

Examples:

• If $$\hat{\beta}$$ is 2, a $$10\%$$ increase in $$X$$ is associated with a $$2\log\left(1 + \frac{10}{100}\right) = 2\log(1.1) \approx 0.19$$ increase in $$Y$$.
• If $$\hat{\beta}$$ is -1.5, a $$20\%$$ decrease in $$X$$ is associated with a $$-1.5\log\left(1 + \frac{-20}{100}\right) = -1.5\log(.8) \approx 0.15$$ increase in $$Y$$.

#### Theory

Assume our regression equation is

$E(Y|X = x) = \beta_0 + \beta_1x.$

If we include $$\log(X)$$ instead, we have

$E(Y|X = x) = \beta_0 + \beta_1\log(x).$

Consider when $$X = cX$$ where $$c$$ is some constant (e.g. 2 for a doubling of $$X$$ or 1.3 for a 30% increase in $$X$$).

$E(Y|X = cx) = \beta_0 + \beta_1\log(cx).$

Therefore if we look at the difference in expectation,

$E(Y|X = cx) - E(Y|X = x) = \beta_1(\log(cx) - \log(x)) = \beta_1\log(c).$

#### Approximation

If your percent change is small (e.g. a few percent) then you can approximate the change. This is because

$log(1 + x) \approx x,$

when $$x$$ is close to 0. So to approximate what effect a 1% change in $$X$$ would have, simply multiple $$\hat{\beta}$$ by that value; $$0.1\hat{\beta}$$. This works reliably well up to $$\pm3\%$$, moderately up to $$\pm5\%$$ and gets much worse beyond that.

## Column

### Example

data(mtcars)
(m <- lm(drat ~ log(disp), data = mtcars))

Call:
lm(formula = drat ~ log(disp), data = mtcars)

Coefficients:
(Intercept)    log(disp)
7.2301      -0.6875  

Therefore a 25% increase in disp is associated with a $$-0.688\log(1.25) = -0.153$$ change in drat.

To test this, we predict the difference in predicted outcome with some values of disp, and that value increaed by 25%.
predict(m, newdata = data.frame(disp = 5)) - predict(m, newdata = data.frame(disp = 5*1.25))
        1
0.1534182 
predict(m, newdata = data.frame(disp = 30)) - predict(m, newdata = data.frame(disp = 30*1.25))
        1
0.1534182 

# Both Logged

## Column

### Discussion

#### tldr

A $$k\%$$ change in a predictor is associated with a $$\left(1 + \frac{k}{100}\right)^{\hat{\beta}}$$ multiplicative change in the outcome.

Examples:

• If $$\hat{\beta}$$ is 2, a $$10\%$$ increase in $$X$$ is associated with a $$\left(1 + \frac{10}{100}\right)^2 = 1.1^2 \approx 1.21 = 21\%$$ increaes in $$Y$$.
• If $$\hat{\beta}$$ is -1.5, a $$20\%$$ decrease in $$X$$ is associated with a $$\left(1 + \frac{-20}{100}\right)^{-1.5} = .8^{-1.5} \approx 1.40 = 40\%$$ increase in $$Y$$.

To-do.

## Column

### Example

data(mtcars)
(m <- lm(log(drat) ~ log(disp), data = mtcars))

Call:
lm(formula = log(drat) ~ log(disp), data = mtcars)

Coefficients:
(Intercept)    log(disp)
2.2763      -0.1905  

Therefore a 25% increase in disp is associated with a $$1.25^{-0.191} = 0.958$$ multiplicative change in drat, corresponding to a 4.2% decrease.

To test this, we predict the difference in predicted outcome with some values of disp, and that value increaed by 25%.
predict(m, newdata = data.frame(disp = 5)) - predict(m, newdata = data.frame(disp = 5*1.25))
         1
0.04251857 
predict(m, newdata = data.frame(disp = 30)) - predict(m, newdata = data.frame(disp = 30*1.25))
         1
0.04251857 `