# 7.2 - Model Diagnostics

7.2 - Model Diagnostics

While the goodness-of-fit statistics can tell us how well a particular model fits the data, they don't tell us much about why a model may fit poorly. Do the model assumptions hold? To assess the lack of fit we need to look at regression diagnostics.

Let us begin with a bit of a review of Linear Regression diagnostics. The standard linear regression model is given by:

$$y_i \sim N(\mu_i,\sigma^2)$$

$$\mu_i =x_i^T \beta$$

The two crucial features of this model are

1. the assumed mean structure, $$\mu_i =x_i^T \beta$$ , and
2. the assumed constant variance $$\sigma^2$$ (homoscedasticity).

The most common diagnostic tool is the residuals, the difference between the estimated and observed values of the dependent variable. There are other useful regression diagnostics, e.g. measures of leverage and influence, but for now, our focus will be on the estimated residuals.

The most common way to check these assumptions is to fit the model and then plot the residuals versus the fitted values $$\hat{y}_i=x_i^T \hat{\beta}$$ .

• If the model assumptions are correct, the residuals should fall within an area representing a horizontal band, like this,
• If the residuals have been standardized in some fashion (i.e., scaled by an estimate of $$\sigma$$), then we would expect most of them to have values within $$\pm2$$ or $$\pm3$$; residuals outside of this range are potential outliers.
• If the plot reveals some kind of curvature—for example, like this,

it suggests a failure of the mean model; the true relationship between $$\mu_i$$, and the covariates might not be linear.

• If the variability in the residuals is not constant as we move from left to right—for example, if the plot looks like this,

or like this,

then the variance $$V(Y_i)$$ is not constant but changes as the mean $$\mu_i$$ changes.

## Analogous plots for logistic regression.

The logistic regression model says that the mean of $$Y_i$$ is

$$\mu_i=n_i \pi_i$$

where

$$\log\left(\dfrac{\pi_i}{1-\pi_i}\right)=x_i^T \beta$$

and that the variance of $$Y_i$$ is

$$V(Y_i)=n_i \pi_i(1-\pi_i)$$.

After fitting the model, we can calculate the Pearson residuals:

$$r_i=\dfrac{y_i-\hat{\mu}_i}{\sqrt{\hat{V}(Y_i)}}=\dfrac{y_i-n_i\hat{\pi}_i}{\sqrt{n_i\hat{\pi}_i(1-\hat{\pi}_i)}}$$

If the $$n_i$$s are relatively large, these act like standardized residuals in linear regression. To see what's happening, we can plot them against the linear predictors:

$$\hat{\eta}_i=\log\left(\dfrac{\pi_i}{1-\pi_i}\right)=x_i^T \hat{\beta}_i$$

which are the estimated log-odds of success, for cases $$i = 1, \ldots , N$$.

• If the fitted logistic regression model was true, we would expect to see a horizontal band with most of the residuals falling within $$\pm 3$$:
• If the $$n_i$$s are small, then the plot might not have such a nice pattern, even if the model is true. In the extreme case of ungrouped data (with all $$n_i$$s equal to 1), this plot becomes uninformative. From now on, we will suppose that the $$n_i$$s are not too small so that the plot is at least somewhat meaningful.
• If outliers are present—that is, if a few residuals or even one residual is substantially larger than $$\pm 3$$, then $$X^2$$ and $$G^2$$ may be much larger than the degrees of freedom. In that situation, the lack of fit can be attributed to outliers, and the large residuals will be easy to find in the plot.

#### Curvature

• If the plot of Pearson residuals versus the linear predictors reveals curvature—for example, like this,

then it suggests that the mean model

$$\log\left(\dfrac{\pi_i}{1-\pi_i}\right)=x_i^T \beta$$ (2)

has been misspecified in some fashion. That is, it could be that one or more important covariates do not influence the log-odds of success in a linear fashion. For example, it could be that a covariate $$x$$ ought to be replaced by a transformation such as $$\sqrt{x}$$, $$\log x$$, $$x^2$$, etc.

To see whether individual covariates ought to enter in the logit model in a non-linear fashion, we could plot the empirical logits

$$\log\left(\dfrac{y_i+1/2}{n_i-y_i+1/2}\right)$$ (3)

versus each covariate in the model.

• If one of the plots looks substantially non-linear, we might want to transform that particular covariate.
• If many of them are nonlinear, however, it may suggest that the link function has been mis-specified—i.e., that the left-hand side of (2) should not involve a logit transformation but some other function.

Changing the link function will change the interpretation of the coefficients entirely; the $$\beta$$s will no longer be log-odds ratios. But, depending on what the link function is, they might still have a nice interpretation. For example, in a model with a log link, $$\log\pi_i=x_i^T \beta$$, an exponentiated coefficient $$\exp(\beta_j)$$ becomes a relative risk.

#### Test for correctness of the link function

Hinkley (1985) suggests a nice, easy test to see whether the link function is plausible:

$$\hat{\eta}_i=x_i^T \hat{\beta}$$

• Fit the model and save the estimated linear predictors
• Add $$\eta^2_i$$ to the model as a new predictor and see if it's significant.

A significant result indicates that the link function is misspecified. A nice feature of this test is that it applies even to ungrouped data ($$n_i$$s equal to one), for which residual plots are uninformative.

#### Non-constant Variance

Suppose that the residual plot shows non-constant variance as we move from left to right:

Another way to detect non-constancy of variance is to plot the absolute values of the residuals versus the linear predictors and look for a non-zero slope:

Non-constant variance in the Pearson residuals means that the assumed form of the variance function,

$$V(y_i)=n_i \pi_i (1-\pi_i)$$

is wrong and cannot be corrected by simply introducing a scale factor for overdispersion. Overdispersion and changes to the variance function will be discussed later.

## Example

The SAS online help documentation provides the following quantal assay dataset. In this table, $$x_i$$ refers to the log-dose, $$n_i$$ is the number of subjects exposed, and $$y_i$$ is the number who responded.

 $$x_i$$ $$y_i$$ $$n_i$$ 2.68 10 31 2.76 17 30 2.82 12 31 2.90 7 27 3.02 23 26 3.04 22 30 3.13 29 31 3.20 29 30 3.21 23 30

If we fit a simple logistic regression model, we will find that the coefficient for $$x_i$$ is highly significant, but the model doesn't fit. The plot of Pearson residuals versus the fitted values resembles a horizontal band, with no obvious curvature or trends in the variance. This seems to be a classic example of overdispersion. Since there's only a single covariate, a good place to start is to plot the empirical logits as defined in equation (3) above versus $$x$$.

This is basically the same thing as a scatterplot of $$Y$$ versus $$x$$ in the context of ordinary linear regression. This plot becomes more meaningful as the $$n_i$$s grow. With ungrouped data (all $$n_i$$ = 1), the empirical logits will only take two possible values—$$\log(1/3)$$ and $$\log(3/1$$\)—and the plot will not be very useful.

Here is part of the SAS program file assay.sas:

options nocenter nodate nonumber linesize=72;
data assay;
input logconc y n;
emplogit=log((y+.5)/(n-y+.5));
cards;
2.68  10  31
2.76  17  30
2.82  12  31
2.90   7  27
3.02  23  26
3.04  22  30
3.13  29  31
3.20  29  30
3.21  23  30
;
run;
axis1 label=('Log-concentration');
axis2 label=('Empirical Logit');
proc gplot data=assay;
title 'Empirical logits vs. x';
plot emplogit * logconc / haxis=axis1 vaxis=axis2;
run; 

The relationship between the logits and $$x$$ seems linear. Let's fit the logistic regression and see what happens.

*no adjustment for overdispersion (scale=none);
proc logistic data=assay;
model y/n= logconc / scale=none;
output out=out1 xbeta=xb reschi=reschi;
run;

axis1 label=('Linear predictor');
axis2 label=('Pearson Residual');
proc gplot data=out1;
title 'Residual plot';
plot reschi * xb / haxis=axis1 vaxis=axis2;
run;  

In plot reschi * xb, reschi are the Pearson residuals, and xb the linear predictor.

The output reveals that the coefficient of $$x$$ is highly significant, but the model does not fit.

Deviance and Pearson Goodness-of-Fit Statistics
Criterion Value DF Value/DF Pr > ChiSq
Deviance 29.3462 7 4.1923 0.0001
Pearson 28.5630 7 4.0804 0.0002

Number of events/trials observations: 9

Analysis of Maximum Likelihood Estimates
Parameter DF Estimate Standard
Error
Wald
Chi-Square
Pr > ChiSq
Intercept 1 -15.8331 2.4371 42.2072 <.0001
logconc 1 5.5776 0.8319 44.9521 <.0001

Here is the R code assay.R that corresponds to the SAS program assay.sas:


#### Logistic regression

r=c(10,17,12,7,23,22,29,29,23)
n=c(31,30,31,27,26,30,31,30,30)
logconc=c(2.68,2.76,2.82,2.90,3.02,3.04,3.13,3.20,3.21)
counts=cbind(r,n-r)
result=glm(counts~logconc,family=binomial("logit"))
summary(result,correlation=TRUE,symbolic.cor = TRUE)
result$coefficients #### plot residuals vs. linear predictor plot(residuals(result, type="pearson"),result$linear.predictors)

#### plot logconc vs. empirical logits

emplogit=log((r+0.5)/(n-r+0.5))
plot(logconc,emplogit)
plot(result)


With plot(result), R will produce four diagnostic plots, including a residual plot, a QQ plot, a scale-location plot, and a residual vs leverage plot as well.

The output reveals that the coefficient of $$x$$ is highly significant, but the model does not fit.


Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -15.8339     2.4371  -6.497 8.20e-11 ***
logconc       5.5778     0.8319   6.705 2.02e-11 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 83.631  on 8  degrees of freedom
Residual deviance: 29.346  on 7  degrees of freedom
AIC: 62.886


Here is the residual plot from R output:

The residuals plots in both SAS and R above do not show any obvious curvature or trends in the variance. And there are no other predictors that are good candidates for inclusion in the model. (It can be shown that a quadratic term for log-concentration will not improve the fit.) So it is quite reasonable to attribute the problem to overdispersion.

## Influence

Regression diagnostics can also tell us how influential each observation is to the fit of the logistic regression model. We can evaluate the numerical values of these statistics and/or consider their graphical representation (like residual plots in linear regression).

Some measures of influence:

• Pearson and Deviance Residuals identify observations not well explained by the model.
• Hat Matrix Diagonal detects extremely large points in the design space. These are often labeled as "leverage" or "$$h_i$$" and are related to standardized residuals. A general rule says that if $$h_i > 2p/n$$ or $$> 3p/n$$, the points is influential. Here "p" is the number of parameters in the model and "n" is the number of observations. For the donner data example with the main effects model, $$2(3)/45=0.13$$. There are two points possibly influential.
• The regression diagnostics work is very similar to linear regression.

The LOGISTIC Procedure

• DFBETA assesses the effect of an individual observation on the estimated parameter of the fitted model. It’s calculated for each observation for each parameter estimate. It's a difference in the parameter estimated due to deleting the corresponding observation. For example, for the main effects model there are 3 such plots (one for intercept, one for age, and one for sex). Here is the one for AGE, showing that observation #35 is potentially influential.

The LOGISTIC Procedure

• C and CBAR provide scalar measures of the influence of the individual observations on the parameter estimates, similar to the Cook distance in linear regression. The same rules in linear regression hold here too. For example, observation #35 again seems influential (see the rest of donner.lst).
• DIFDEV and DIFCHISQ detect observations that heavily add to the lack of fit, i.e. there is a large disagreement between the fitted and observed values. DIFDEV measures the change in the deviance ($$G^2$$) as an individual observation is deleted. DIFCHISQ measures the change in $$X^2$$.

In SAS, under PROC LOGISTIC, INFLUENCE option and IPLOTS option will output these diagnostics. In R see, influence() function.

As pointed out before, the standardized residual values are considered to be indicative of lack of fit if their absolute value exceeds 3 (some sources are more conservative and take 2 as the threshold value).

The Index plot, in which the residuals are plotted against the indices of the observations, can help us determine outliers, and/or systematic patterns in variation and thus assess the lack of fit. In our example, all Pearson and Deviance residuals fall within $$-2$$ and $$+2$$; thus, there does not seem to be any outliers or particular heavy influential points; see donner.html for the IPLOTS.

For our example, here is the IPLOT of the deviance residuals vs. the case number.

The LOGISTIC Procedure

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility