Variable transformations

# Variable transformations
### Prof. Maria Tackett

---

## [Click here for PDF of slides](11-transformations.pdf)

---

## Topics

- Log transformation on the response

- Log transformation on the predictor

---

## Respiratory Rate vs. Age

- A high respiratory rate can potentially indicate a respiratory infection in children. In order to determine what indicates a "high" rate, we first want to understand the relationship between a child's age and their respiratory rate.

- The data contain the respiratory rate for 618 children ages 15 days to 3 years.

- **Variables**: 
    - <font class="vocab">`Age`</font>: age in months
    - <font class="vocab">`Rate`</font>: respiratory rate (breaths per minute)

---

## Rate vs. Age

---

## Rate vs. Age

<table>
 <thead>
  <tr>
   <th style="text-align:left;"> term </th>
   <th style="text-align:right;"> estimate </th>
   <th style="text-align:right;"> std.error </th>
   <th style="text-align:right;"> statistic </th>
   <th style="text-align:right;"> p.value </th>
   <th style="text-align:right;"> conf.low </th>
   <th style="text-align:right;"> conf.high </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:left;"> (Intercept) </td>
   <td style="text-align:right;"> 47.052 </td>
   <td style="text-align:right;"> 0.504 </td>
   <td style="text-align:right;"> 93.317 </td>
   <td style="text-align:right;"> 0 </td>
   <td style="text-align:right;"> 46.062 </td>
   <td style="text-align:right;"> 48.042 </td>
  </tr>
  <tr>
   <td style="text-align:left;"> Age </td>
   <td style="text-align:right;"> -0.696 </td>
   <td style="text-align:right;"> 0.029 </td>
   <td style="text-align:right;"> -23.684 </td>
   <td style="text-align:right;"> 0 </td>
   <td style="text-align:right;"> -0.753 </td>
   <td style="text-align:right;"> -0.638 </td>
  </tr>
</tbody>
</table>

---

## Log transformation on the response

---

## Need to transform `$Y$`

- Typically, a "fan-shaped" residual plot indicates the need for a transformation of the response variable `$y$`
  + .vocab[log(Y)] is the most straightforward to interpret

- When building a model: 
  + Choose a transformation and build the model on the transformed data
  + Reassess the residual plots
  + If the residuals plots did not sufficiently improve, try a new transformation!

---

## Log transformation on `$Y$`

- If we apply a log transformation to the response variable, we want to estimate the parameters for the model...
.alert[
`$$\widehat{\log(Y)} = \hat{\beta}_0 + \hat{\beta}_1 X$$`
]

- We want to interpret the model in terms of `$y$` not `$\log(Y)$`, so we write all interpretations in terms of

.alert[
`$$y = \exp\{\hat{\beta}_0 + \hat{\beta}_1 X\} = \exp\{\hat{\beta}_0\}\exp\{\hat{\beta}_1X\}$$`
]

---

## Mean and logs

Suppose we have a set of values

```r
x <- c(3, 5, 6, 8, 10, 14, 19)
```

```r
log_x <- log(x)
mean(log_x)
```

```
## [1] 2.066476
```
]
]

```r
xbar <- mean(x)
log(xbar)
```

```
## [1] 2.228477
```
]
]
---

## Median and logs

```r
x <- c(3, 5, 6, 8, 10, 14, 19)
```

```r
log_x <- log(x)
median(log_x)
```

```
## [1] 2.079442
```
]
]
--

```r
median_x <- median(x)
log(median_x)
```

```
## [1] 2.079442
```
]
]

---

## Mean, Median, and log

`$$\overline{\log(x)} \neq \log(\bar{x})$$`

```r
mean(log_x) == log(xbar)
```

```
## [1] FALSE
```

`$$\text{Median}(\log(x)) = \log(\text{Median}(x))$$`

```r
median(log_x) == log(median_x)
```

```
## [1] TRUE
```

---

## Mean and median of `$\log(Y)$`

- Recall that `$y= \beta_0 + \beta_1 x_i$` is the **mean** value of `$y$` at the given value `$x_i$`. This doesn't hold when we log-transform `$y$`

- The mean of the logged values is **not** equal to the log of the mean value. Therefore at a given value of `$x$`

.alert[
`$$\begin{aligned}\exp\{\text{Mean}(\log(y))\} \neq \text{Mean}(y) \\[5pt]
\Rightarrow \exp\{\beta_0 + \beta_1 x\} \neq \text{Mean}(y) \end{aligned}$$`
]

---

## Mean and median of `$\log(y)$`

- However, the median of the logged values **is** equal to the log of the median value. Therefore,

- If the distribution of `$\log(y)$` is symmetric about the regression line, for a given value `$x_i$`,
.alert[
`$$\text{Median}(\log(y)) = \text{Mean}(\log(y))$$`
]
---

## Interpretation with log-transformed `$y$`

- Given the previous facts, if `$\widehat{\log(Y)} = \hat{\beta}_0 + \hat{\beta}_1 x$`, then 
.alert[
`$$\text{Median}(\hat{Y}) = \exp\{\hat{\beta}_0\}\exp\{\hat{\beta}_1 x\}$$`
]

- <font class="vocab">Intercept:</font> When `$X=0$`, the median of `$Y$` is expected to be `$\exp\{\hat{\beta}_0\}$`
<br>

- <font class="vocab">Slope: </font>For every one unit increase in `$X$`, the median of `$Y$` is expected to multiply by a factor of `$\exp\{\hat{\beta}_1\}$`

---

## log(Rate) vs. Age

---

## log(Rate) vs. Age

---

## log(Rate) vs. Age

|term        | estimate| std.error| statistic| p.value| conf.low| conf.high|
|:-----------|--------:|---------:|---------:|-------:|--------:|---------:|
|(Intercept) |    3.845|     0.013|   304.500|       0|     3.82|     3.870|
|Age         |   -0.019|     0.001|   -25.839|       0|    -0.02|    -0.018|

<br>

.vocab[Intercept]: The median respiratory rate for a new born child is expected to be 46.759 (exp{3.845}) breaths per minute.

.vocab[Slope]: For each additional month in a child's age, the respiratory rate is expected to multiply by a factor of 0.981 (exp{-0.019}).

---

## Confidence interval for `$\beta_j$`

- The confidence interval for the coefficient of `$X$` describing its relationship with `$\log(Y)$` is

`$$\hat{\beta}_j \pm t^* SE(\hat{\beta_j})$$`

- The confidence interval for the coefficient of `$x$` describing its relationship with `$Y$` is

---

## Coefficient of `Age`

<table>
 <thead>
  <tr>
   <th style="text-align:left;"> term </th>
   <th style="text-align:right;"> estimate </th>
   <th style="text-align:right;"> std.error </th>
   <th style="text-align:right;"> statistic </th>
   <th style="text-align:right;"> p.value </th>
   <th style="text-align:right;"> conf.low </th>
   <th style="text-align:right;"> conf.high </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:left;"> (Intercept) </td>
   <td style="text-align:right;"> 3.845 </td>
   <td style="text-align:right;"> 0.013 </td>
   <td style="text-align:right;"> 304.500 </td>
   <td style="text-align:right;"> 0 </td>
   <td style="text-align:right;"> 3.82 </td>
   <td style="text-align:right;"> 3.870 </td>
  </tr>
  <tr>
   <td style="text-align:left;"> Age </td>
   <td style="text-align:right;"> -0.019 </td>
   <td style="text-align:right;"> 0.001 </td>
   <td style="text-align:right;"> -25.839 </td>
   <td style="text-align:right;"> 0 </td>
   <td style="text-align:right;"> -0.02 </td>
   <td style="text-align:right;"> -0.018 </td>
  </tr>
</tbody>
</table>

.vocab[We are 95% confident that for each additional month in age, the respiratory rate will multiply by a factor of 0.98 to 0.982 (exp{-0.02} to exp{-0.018}).]

---

## Log transformation on the predictor

---

## Log Transformation on `$X$`

Try a transformation on `$X$` if the scatterplot shows some curvature but the variance is constant for all values of `$X$`

---

## Model with Transformation on `$X$`

- <font class="vocab">Intercept: </font> When `$\log(X)=0$`, `$(X=1)$`, `$Y$` is expected to be `$\hat{\beta}_0$` (i.e. the mean of `$y$` is `$\hat{\beta}_0$`)

- <font class="vocab">Slope: </font> When `$X$` is multiplied by a factor of `$\mathbf{C}$`, the mean of `$Y$` is expected to change by `$\boldsymbol{\hat{\beta}_1}\mathbf{\log(C)}$` units
    - *Example*: when `$X$` is multiplied by a factor of 2, `$y$` is expected to change by `$\boldsymbol{\hat{\beta}_1}\mathbf{\log(2)}$` units

---

## Rate vs. log(Age)

---

## Rate vs. log(Age)

|term        | estimate| std.error| statistic| p.value| conf.low| conf.high|
|:-----------|--------:|---------:|---------:|-------:|--------:|---------:|
|(Intercept) |   50.135|     0.632|    79.330|       0|   48.893|    51.376|
|log_age     |   -5.982|     0.263|   -22.781|       0|   -6.498|    -5.467|

<br>

.vocab[Intercept]: The expected (mean) respiratory rate for children who are 1 month old (log(1) = 0) is 50.135	breaths per minute.

.vocab[Slope]: If a child's age doubles, we expect their respiratory rate to decrease by 4.146 (-5.982*log(2)) breaths per minute.

---

See [Log Transformations in Linear Regression](https://github.com/sta210-sp20/supplemental-notes/blob/master/log-transformations.pdf) for more details about interpreting regression models with log-transformed variables.

---

## Recap

- Log transformation on the response

- Log transformation on the predictor