Logistic regression

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# Logistic regression
## Prediction
### Prof. Maria Tackett

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## [Click for PDF of slides](21-logistic-prediction.pdf)

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## Topics

- Calculating predicted probabilities from the logistic regression model

- Using the predicted probabilities to make a "yes/no" decision for a given observation

- Assessing model performance using 
  - Confusion matrix
  - ROC curve

---

## Risk of coronary heart disease

.midi[This dataset is from an ongoing cardiovascular study on residents of the town of Framingham, Massachusetts. We want to examine the relationship between various health characteristics and the risk of having heart disease in the next 10 years.]

.midi[.vocab[`high_risk`]: 1 = High risk, 0 = Not high risk]

.midi[.vocab[`age`]: Age at exam time (in years)]

.midi[.vocab[`totChol`]: Total cholesterol (in mg/dL)]

.midi[.vocab[`currentSmoker`]: 0 = nonsmoker; 1 = smoker]

---

## Modeling risk of coronary heart disease

|term           | estimate| std.error| statistic| p.value| conf.low| conf.high|
|:--------------|--------:|---------:|---------:|-------:|--------:|---------:|
|(Intercept)    |   -6.638|     0.372|   -17.860|   0.000|   -7.374|    -5.917|
|age            |    0.082|     0.006|    14.430|   0.000|    0.071|     0.093|
|totChol        |    0.002|     0.001|     2.001|   0.045|    0.000|     0.004|
|currentSmoker1 |    0.457|     0.092|     4.951|   0.000|    0.277|     0.639|

---

## Using the model for prediction

We are often interested in predicting whether a given observation will have a "yes" response

To do so

--
- Use the logistic regression model to calculate the predicted log-odds that an observation has a "yes" response.

- Then, use the log-odds to calculate the predicted probability of a "yes" response.

- Then, use the predicted probabilities to classify the observation as having a "yes" or "no" response.

---

## Calculating the predicted probability

`$$\small{\log\Big(\frac{\hat{\pi}_i}{1-\hat{\pi}_i}\Big) = \hat{\beta}_0 + \hat{\beta}_1 x_i}$$`

`$$\small{\Rightarrow \exp\bigg\{\log\Big(\frac{\hat{\pi}_i}{1-\hat{\pi}_i}\Big)\bigg\} = \exp\{\hat{\beta}_0 + \hat{\beta}_1 x_i\}}$$`
--

`$$\small{\Rightarrow \frac{\hat{\pi}_i}{1-\hat{\pi}_i} = \exp\{\hat{\beta}_0 + \hat{\beta}_1 x_i\}}$$`
--

`$$\small{\Rightarrow \hat{\pi}_i = \frac{\exp\{\hat{\beta}_0 + \hat{\beta}_1 x_i\}}{1+\exp\{\hat{\beta}_0 + \hat{\beta}_1 x_i\}}}$$`

---

## `$\hat{\pi}$` vs. `$\widehat{\text{log-odds}}$`

$$\hat{\pi}_i = \frac{\exp(\hat{\beta}_0 + \hat{\beta}_1 x_i)}{1 + \exp(\hat{\beta}_0 + \hat{\beta}_1 x_i)} = \frac{\exp(\widehat{\text{log-odds})}}{1 + \exp(\widehat{\text{log-odds}})} $$

---

## Predicted response for a patient

Suppose a patient comes in who is 60 years old, does not currently smoke, and has a total cholesterol of 263 mg/dL.

Predicted log-odds that this person is high risk for coronary heart disease in the next 10 years:

`$$\widehat{\text{log-odds}} = -6.638 + 0.082 \times 60 + 0.002 \times 263 + 0.457 \times 0 = -1.192$$`
--

The probability this patient is high risk for coronary heart disease in the next 10 years:

`$$\widehat{\text{probability}} = \frac{\exp\{-1.192\}}{1 + \exp\{-1.192\}} = 0.233$$`
---

## Predictions in R

```r
x0 <- data_frame(age = 60, totChol = 263,
 currentSmoker = factor(0))
```

--
.pull-left[

**Predicted log-odds**

```r
predict(risk_m, x0) 
```

```
##         1 
## -1.214193
```
]

.pull-right[
**Predicted probability**

```r
predict(risk_m, x0, 
      type = "response") 
```

```
##       1 
## 0.22896
```
]

---

## Is this patient high risk?

The probability the patient is at risk for coronary heart disease is 0.229.

.question[
Based on this probability, would you consider this patient as being high risk for getting coronary heart disease in the next 10 years? Why or why not?
]

---

## Confusion Matrix

- We can use the predicted probability to predict the outcome for a given observation
  - In other words, we can classify the observations into two groups: "yes" and "no"

- **How**: Establish a threshold such that `$y=1$` if predicted probability is greater than the threshold `$(y = 0 \text{ otherwise})$`

- To assess the accuracy of our predictions, we can make a table of the observed (actual) response versus the predicted response.
  + This table is the .vocab[confusion matrix]
  
---

## Confusion Matrix

Suppose we use 0.3 as the threshold to classify observations.

If `$\hat{\pi}_i > 0.3$`, then risk_predict = "Yes". Otherwise, risk_predict = "No".

|high_risk |risk_predict |    n|
|:---------|:------------|----:|
|0         |No           | 3339|
|0         |Yes          |  216|
|1         |No           |  530|
|1         |Yes          |  105|

---

## Confusion matrix

|high_risk |risk_predict |    n|
|:---------|:------------|----:|
|0         |No           | 3339|
|0         |Yes          |  216|
|1         |No           |  530|
|1         |Yes          |  105|

.question[ 
What proportion of observations were misclassified? This is called the .vocab[misclassification rate].

]

---

## Confusion matrix: 2 X 2 table

In practice, you often see the confusion matrix presented as a 2 `$\times$` 2 table as shown below:

|high_risk |   No| Yes|
|:---------|----:|---:|
|0         | 3339| 216|
|1         |  530| 105|

.question[
What is the disadvantage of relying on a single confusion matrix to assess the accuracy of the model?
]
---

### Receiver Operating Characteristic (ROC) curve

---

## Sensitivity & Specificity

- Sensitivity: Proportion of observations with `$y=1$` that have predicted probability above a specified threshold
 + Called **true positive rate** (y-axis)

- Specificity: Proportion of observations with `$y=0$` that have predicted probability below a specified threshold
 + (1 - specificity) called **false positive rate** (x-axis)

- What we want:

⬆️ High sensitivity

⬇️ Low values of 1-specificity

---

## ROC curve in R

```r
library(yardstick)

# Need to put 1 as the first level
risk_m_aug <- risk_m_aug %>%
 mutate(high_risk = fct_relevel(high_risk, c("1", "0")))

# calculate sensitivity and specificity at each threshold
roc_curve_data <- risk_m_aug %>%
 roc_curve(high_risk, .fitted)

# plot roc curve
autoplot(roc_curve_data)
```

---

## ROC curve

---

## Area under curve (AUC)

We can use the area under the curve (AUC) as one way to assess how well the logistic model fits the data
- `$AUC = 0.5$` very bad fit (no better than a coin flip)
- `$AUC$` close to 1: good fit

```r
risk_m_aug %>%
  roc_auc(high_risk, .fitted) %>%
  pull(.estimate)
```

```
## [1] 0.6955
```

---

## Which threshold would you choose?

.question[
A doctor plans to use the results from your model to help select patients for a new heart disease prevention program. She asks you which threshold would be best to select patients for this program. Based on the ROC curve from the previous slide, which threshold would you recommend to the doctor? Why?
]

---

## Recap

- Calculating predicted probabilities from the logistic regression model

- Using the predicted probabilities to make a "yes/no" decision for a given observation

- Assessing model performance using 
  - Confusion matrix
  - ROC curve