Logistic Regression: Model comparison

Announcements

• HW 04 due April 10 - released later today

• Team Feedback (email from TEAMMATES) due Tuesday, April 8 at 11:59pm (check email)

• Next project milestone: Analysis and draft in April 11 lab

• Statistics experience due April 22

Questions from this week’s content?

Topics

• Comparing models using AIC and BIC

• Test of significance for a subset of predictors

Computational setup

library(tidyverse)
library(tidymodels)
library(pROC)      
library(knitr)
library(kableExtra)

# set default theme in ggplot2
ggplot2::theme_set(ggplot2::theme_bw())

Risk of coronary heart disease

This data set 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.

• high_risk:

  • 1: High risk of having heart disease in next 10 years
  • 0: Not high risk of having heart disease in next 10 years

• age: Age at exam time (in years)

• totChol: Total cholesterol (in mg/dL)

• currentSmoker: 0 = nonsmoker, 1 = smoker

• education: 1 = Some High School, 2 = High School or GED, 3 = Some College or Vocational School, 4 = College

Modeling risk of coronary heart disease

Using age, totChol, and currentSmoker

term	estimate	std.error	statistic	p.value	conf.low	conf.high
(Intercept)	-6.673	0.378	-17.647	0.000	-7.423	-5.940
age	0.082	0.006	14.344	0.000	0.071	0.094
totChol	0.002	0.001	1.940	0.052	0.000	0.004
currentSmoker1	0.443	0.094	4.733	0.000	0.260	0.627

Review: ROC Curve + Model fit

# A tibble: 1 × 3
  .metric .estimator .estimate
  <chr>   <chr>          <dbl>
1 roc_auc binary         0.697

Review: Classification

We will use a threshold of 0.2 to classify observations

Review: Classification

1. Compute the misclassification rate.

2. Compute sensitivity and explain what it means in the context of the data.

3. Compute specificity and explain what it means in the context of the data.

Model comparison

Which model do we choose?

Model 1

term	estimate
(Intercept)	-6.673
age	0.082
totChol	0.002
currentSmoker1	0.443

Model 2

term	estimate
(Intercept)	-6.456
age	0.080
totChol	0.002
currentSmoker1	0.445
education2	-0.270
education3	-0.232
education4	-0.035

Log-Likelihood

Recall the log-likelihood function

$$\begin{array}{rl}\log L& (\mathit{\beta }|x_1,\dots ,x_n,y_1,\dots ,y_n)\\ & =\sum _{i=1}^n[y_i\log ({\pi }_i)+(1-y_i)\log (1-{\pi }_i)]\end{array}$$

where ${\pi }_i=\frac{\exp \{x_i^{\mathsf{T}}\mathit{\beta }\}}{1+\exp \{x_i^{\mathsf{T}}\mathit{\beta }\}}$

AIC & BIC

Estimators of prediction error and relative quality of models:

. . .

Akaike’s Information Criterion (AIC)¹: $$AIC=-2\log L+2(p+1)$$

. . .

Schwarz’s Bayesian Information Criterion (BIC)²: $$BIC=-2\log L+\log (n)\times (p+1)$$

AIC & BIC

$$\begin{array}{rl}& AIC=-2\log L+2(p+1)\\ & BIC=-2\log L+\log (n)\times (p+1)\end{array}$$

. . .

First Term: Decreases as p increases

AIC & BIC

$$\begin{array}{rl}& AIC=-2\log L+2(p+1)\\ & BIC=-2\log L+\log (n)\times (p+1)\end{array}$$

Second term: Increases as p increases

Using AIC & BIC

$$\begin{array}{rl}& AIC=-2\log L+2(p+1)\\ & BIC=-2\log L+\log (n)\times (p+1)\end{array}$$

• Choose model with the smaller value of AIC or BIC

• If $n\ge 8$, the penalty for BIC is larger than that of AIC, so BIC tends to favor more parsimonious models (i.e. models with fewer terms)

AIC from the glance() function

Let’s look at the AIC for the model that includes age, totChol, and currentSmoker

glance(high_risk_fit)$AIC

[1] 3232.812

. . .

Calculating AIC

- 2 * glance(high_risk_fit)$logLik + 2 * (3 + 1)

[1] 3232.812

Comparing the models using AIC

Let’s compare the full and reduced models using AIC.

glance(high_risk_fit_reduced)$AIC

[1] 3232.812

glance(high_risk_fit_full)$AIC

[1] 3231.6

Based on AIC, which model would you choose?

Comparing the models using BIC

Let’s compare the full and reduced models using BIC

glance(high_risk_fit_reduced)$BIC

[1] 3258.074

glance(high_risk_fit_full)$BIC

[1] 3275.807

Based on BIC, which model would you choose?

Drop-in-deviance test

Drop-in-deviance test

We will use a drop-in-deviance test (aka Likelihood Ratio Test) to test

• the overall statistical significance of a logistic regression model

• the statistical significance of a subset of coefficients in the model

Deviance

The deviance is a measure of the degree to which the predicted values are different from the observed values (compares the current model to a “saturated” model)

In logistic regression,

$$D=-2\log L$$

$D\sim {\chi }_{n-p-1}^2$ ( $D$ follows a Chi-square distribution with $n-p-1$ degrees of freedom)³

Note: $n-p-1$ a the degrees of freedom associated with the error in the model (like residuals)

${\chi }^2$ distribution

Test for overall significance

We can test the overall significance for a logistic regression model, i.e., whether there is at least one predictor with a non-zero coefficient

$$\begin{array}{rl}& H_0:{\beta }_1=\cdots ={\beta }_p=0\\ & H_a:{\beta }_j\ne 0\text{ for at least one }j\end{array}$$

. . .

The drop-in-deviance test for overall significance compares the fit of a model with no predictors to the current model.

Drop-in-deviance test statistic

Let $L_0$ and $L_a$ be the likelihood functions of the model under $H_0$ and $H_a$, respectively. The test statistic is

$$\begin{array}{rl}G=D_0-D_a& =(-2\log L_0)-(-2\log L_a)\\ & =-2(\log L_0-\log L_a)\\ & =-2\sum _{i=1}^n[y_i\log (\frac{{\hat{\pi }}^0}{{\hat{\pi }}_i^a})+(1-y_i)\log (\frac{1-{\hat{\pi }}^0}{1-{\hat{\pi }}_i^a})]\end{array}$$

where ${\hat{\pi }}^0$ is the predicted probability under $H_0$ and ${\hat{\pi }}_i^a=\frac{\exp \{x_i^{\mathsf{T}}\mathit{\beta }\}}{1+\exp \{x_i^{\mathsf{T}}\mathit{\beta }\}}$ is the predicted probability under $H_a$ ⁴

Drop-in-deviance test statistic

$$G=-2\sum _{i=1}^n[y_i\log (\frac{{\hat{\pi }}^0}{{\hat{\pi }}_i^a})+(1-y_i)\log (\frac{1-{\hat{\pi }}^0}{1-{\hat{\pi }}_i^a})]$$

. . .

• When $n$ is large, $G\sim {\chi }_p^2$, ( $G$ follows a Chi-square distribution with $p$ degrees of freedom)

• The p-value is calculated as $P({\chi }^2>G)$

• Large values of $G$ (small p-values) indicate at least one ${\beta }_j$ is non-zero

Heart disease model: drop-in-deviance test

$$\begin{array}{rl}& H_0:{\beta }_{age}={\beta }_{totChol}={\beta }_{currentSmoker}=0\\ & H_a:{\beta }_j\ne 0\text{ for at least one }j\end{array}$$

. . .

Fit the null model (we’ve already fit the alternative model)

null_model <- glm(high_risk ~ 1, data = heart_disease, family = "binomial")

term	estimate	std.error	statistic	p.value
(Intercept)	-1.72294	0.0436342	-39.486	0

Heart disease model: drop-in-deviance test

Calculate the log-likelihood for the null and alternative models

(L_0 <- glance(null_model)$logLik)

[1] -1737.735

(L_a <- glance(high_risk_fit)$logLik)

[1] -1612.406

. . .

Calculate the likelihood ratio test statistic

(G <- -2 * (L_0 - L_a))

[1] 250.6572

. . .

Heart disease model: likelihood ratio test

Calculate the p-value

(p_value <- pchisq(G, df = 3, lower.tail = FALSE))

[1] 4.717158e-54

. . .

Conclusion

The p-value is small, so we reject $H_0$. The data provide evidence that at least one predictor in the model has a non-zero coefficient.

Why use overall test?

Why do we use a test for overall significance instead of just looking at the test for individual coefficients?⁵

. . .

Suppose we have a model such that $p=100$ and $H_0:{\beta }_1=\cdots ={\beta }_{100}=0$ is true

. . .

• About 5% of the p-values for individual coefficients will be below 0.05 by chance.

• So we expect to see 5 small p-values if even no linear association actually exists.

• Therefore, it is very likely we will see at least one small p-value by chance.

• The overall test of significance does not have this problem. There is only a 5% chance we will get a p-value below 0.05, if a relationship truly does not exist.

Test a subset of coefficients

Testing a subset of coefficients

• Suppose there are two models:

  • Reduced Model: includes predictors $x_1,\dots ,x_q$

  • Full Model: includes predictors $x_1,\dots ,x_q,x_{q+1},\dots ,x_p$

• We can use a drop-in-deviance test to determine if any of the new predictors are useful

. . .

$$\begin{array}{rl}& H_0:{\beta }_{q+1}=\cdots ={\beta }_p=0\\ & H_a:{\beta }_j\ne 0\text{ for at least one }j\end{array}$$

Drop-in-deviance test

$$\begin{array}{rl}& H_0:{\beta }_{q+1}=\cdots ={\beta }_p=0\\ & H_a:{\beta }_j\ne 0\text{ for at least one }j\end{array}$$

. . .

The test statistic is

$$\begin{array}{rl}G=D_{reduced}-D_{full}& =(-2\log L_{reduced})-(-2\log L_{full})\\ & =-2(\log L_{reduced}-\log L_{full})\end{array}$$

. . .

The p-value is calculated using a ${\chi }_{\mathrm{\Delta }df}^2$ distribution, where $\mathrm{\Delta }df$ is the number of parameters being tested (the difference in number of parameters between the full and reduced model).

Example: Include education?

Should we include education in the model?

• Reduced model: age, totChol, currentSmoker

• Full model: age, totChol, currentSmoker , education

. . .

$$\begin{array}{rl}& H_0:{\beta }_{ed2}={\beta }_{ed3}={\beta }_{ed4}=0\\ & H_a:{\beta }_j\ne 0\text{ for at least one }j\end{array}$$

Example: Include education?

reduced_model <- glm(high_risk ~ age + totChol + currentSmoker, 
              data = heart_disease, family = "binomial")

full_model <- glm(high_risk ~ age + totChol + currentSmoker + education, 
              data = heart_disease, family = "binomial")

. . .

Calculate deviances

(deviance_reduced <- -2 * glance(reduced_model)$logLik)

[1] 3224.812

(deviance_full <- -2 * glance(full_model)$logLik)

[1] 3217.6

. . .

Calculate test statistic

(G <- deviance_reduced - deviance_full)

[1] 7.212113

Example: Include education?

Calculate p-value

pchisq(G, df = 3, lower.tail = FALSE)

[1] 0.06543567

. . .

What is your conclusion? Would you include education in the model that already has age, totChol, currentSmoker?

Drop-in-deviance test in R

Conduct the drop-in-deviance test using the anova() function in R with option test = "Chisq"

anova(reduced_model, full_model, test = "Chisq") |> 
  tidy() |> 
  kable(digits = 3)

term	df.residual	residual.deviance	df	deviance	p.value
high_risk ~ age + totChol + currentSmoker	4082	3224.812	NA	NA	NA
high_risk ~ age + totChol + currentSmoker + education	4079	3217.600	3	7.212	0.065

Add interactions with currentSmoker?

term	df.residual	residual.deviance	df	deviance	p.value
high_risk ~ age + totChol + currentSmoker	4082	3224.812	NA	NA	NA
high_risk ~ age + totChol + currentSmoker + currentSmoker * age + currentSmoker * totChol	4080	3222.377	2	2.435	0.296

Questions from this week’s content?

Recap

• Introduced model comparison for logistic regression using

  • AIC and BIC

  • Drop-in-deviance test

References

Wilks, SS. 1935. “The Likelihood Test of Independence in Contingency Tables.” The Annals of Mathematical Statistics 6 (4): 190–96.

Footnotes

1. Akaike, Hirotugu. “A new look at the statistical model identification.” IEEE transactions on automatic control 19.6 (1974): 716-723.

2. Schwarz, Gideon. “Estimating the dimension of a model.” The annals of statistics (1978): 461-464.

3. See Wilks (1935) for explanation of why -2 is included.

4. See Wilks (1935) for explanation of why -2 is included.

5. Example from Introduction to Statistical Learning