How to test for simultaneous equality of choosen coefficients in logit or probit model ? What is the standard approach and what is the state of art approach ?


3 Answers 3


Wald test

One standard approach is the Wald test. This is what the Stata command test does after a logit or probit regression. Let's see how this works in R by looking at an example:

mydata <- read.csv("http://www.ats.ucla.edu/stat/data/binary.csv") # Load dataset from the web
mydata$rank <- factor(mydata$rank)
mylogit <- glm(admit ~ gre + gpa + rank, data = mydata, family = "binomial") # calculate the logistic regression


             Estimate Std. Error z value Pr(>|z|)    
(Intercept) -3.989979   1.139951  -3.500 0.000465 ***
gre          0.002264   0.001094   2.070 0.038465 *  
gpa          0.804038   0.331819   2.423 0.015388 *  
rank2       -0.675443   0.316490  -2.134 0.032829 *  
rank3       -1.340204   0.345306  -3.881 0.000104 ***
rank4       -1.551464   0.417832  -3.713 0.000205 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Say, you want to test the hypothesis $\beta_{gre}=\beta_{gpa}$ vs. $\beta_{gre}\neq \beta_{gpa}$. This is equivalent of testing $\beta_{gre} - \beta_{gpa} = 0$. The Wald test statistic is:

$$ W=\frac{(\hat{\beta}-\beta_{0})}{\widehat{\operatorname{se}}(\hat{\beta})}\sim \mathcal{N}(0,1) $$


$$ W^2 = \frac{(\hat{\theta}-\theta_{0})^2}{\operatorname{Var}(\hat{\theta})}\sim \chi_{1}^2 $$

Our $\widehat{\theta}$ here is $\beta_{gre} - \beta_{gpa}$ and $\theta_{0}=0$. So all we need is the standard error of $\beta_{gre} - \beta_{gpa}$. We can calculate the standard error with the Delta method:

$$ \hat{se}(\beta_{gre} - \beta_{gpa})\approx \sqrt{\operatorname{Var}(\beta_{gre}) + \operatorname{Var}(\beta_{gpa}) - 2\cdot \operatorname{Cov}(\beta_{gre},\beta_{gpa})} $$

So we also need the covariance of $\beta_{gre}$ and $\beta_{gpa}$. The variance-covariance matrix can be extracted with the vcov command after running the logistic regression:

var.mat <- vcov(mylogit)[c("gre", "gpa"),c("gre", "gpa")]

colnames(var.mat) <- rownames(var.mat) <- c("gre", "gpa")

              gre           gpa
gre  1.196831e-06 -0.0001241775
gpa -1.241775e-04  0.1101040465

Finally, we can calculate the standard error:

se <- sqrt(1.196831e-06 + 0.1101040465 -2*-0.0001241775)
[1] 0.3321951

So your Wald $z$-value is

wald.z <- (gre-gpa)/se
[1] -2.413564

To get a $p$-value, just use the standard normal distribution:

[1] 0.01579735

In this case we have evidence that the coefficients are different from each other. This approach can be extended to more than two coefficients.

Using multcomp

This rather tedious calculations can be conveniently done in R using the multcomp package. Here is the same example as above but done with multcomp:


glht.mod <- glht(mylogit, linfct = c("gre - gpa = 0"))


Linear Hypotheses:
               Estimate Std. Error z value Pr(>|z|)  
gre - gpa == 0  -0.8018     0.3322  -2.414   0.0158 *


A confidence interval for the difference of the coefficients can also be calculated:

Quantile = 1.96
95% family-wise confidence level

Linear Hypotheses:
               Estimate lwr     upr    
gre - gpa == 0 -0.8018  -1.4529 -0.1507

For additional examples of multcomp, see here or here.

Likelihood ratio test (LRT)

The coefficients of a logistic regression are found by maximum likelihood. But because the likelihood function involves a lot of products, the log-likelihood is maximized which turns the products into sums. The model that fits better has a higher log-likelihood. The model involving more variables has at least the same likelihood as the null model. Denote the log-likelihood of the alternative model (model containing more variables) with $LL_{a}$ and the log-likelihood of the null model with $LL_{0}$, the likelihood ratio test statistic is:

$$ D=2\cdot (LL_{a} - LL_{0})\sim \chi_{df1-df2}^{2} $$

The likelihood ratio test statistic follows a $\chi^{2}$-distribution with the degrees of freedom being the difference in number of variables. In our case, this is 2.

To perform the likelihood ratio test, we also need to fit the model with the constraint $\beta_{gre}=\beta_{gpa}$ to be able to compare the two likelihoods. The full model has the form $$\log\left(\frac{p_{i}}{1-p_{i}}\right)=\beta_{0}+\beta_{1}\cdot \mathrm{gre} + \beta_{2}\cdot \mathrm{gpa}+\beta_{3}\cdot \mathrm{rank_{2}} + \beta_{4}\cdot \mathrm{rank_{3}}+\beta_{5}\cdot \mathrm{rank_{4}}$$. Our constraint model has the form: $$\log\left(\frac{p_{i}}{1-p_{i}}\right)=\beta_{0}+\beta_{1}\cdot (\mathrm{gre} + \mathrm{gpa})+\beta_{2}\cdot \mathrm{rank_{2}} + \beta_{3}\cdot \mathrm{rank_{3}}+\beta_{4}\cdot \mathrm{rank_{4}}$$.

mylogit2 <- glm(admit ~ I(gre + gpa) + rank, data = mydata, family = "binomial")

In our case, we can use logLik to extract the log-likelihood of the two models after a logistic regression:

L1 <- logLik(mylogit)
'log Lik.' -229.2587 (df=6)

L2 <- logLik(mylogit2)
'log Lik.' -232.2416 (df=5)

The model containing the constraint on gre and gpa has a slightly higher log-likelihood (-232.24) compared to the full model (-229.26). Our likelihood ratio test statistic is:

D <- 2*(L1 - L2)
[1] 16.44923

We can now use the CDF of the $\chi^{2}_{2}$ to calculate the $p$-value:

1-pchisq(D, df=1)
[1] 0.01458625

The $p$-value is very small indicating that the coefficients are different.

R has the likelihood ratio test built in; we can use the anova function to calculate the likelhood ratio test:

anova(mylogit2, mylogit, test="LRT")

Analysis of Deviance Table

Model 1: admit ~ I(gre + gpa) + rank
Model 2: admit ~ gre + gpa + rank
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
1       395     464.48                       
2       394     458.52  1   5.9658  0.01459 *
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Again, we have strong evidence that the coefficients of gre and gpa are significantly different from each other.

Score test (aka Rao's Score test aka Lagrange multiplier test)

The Score function $U(\theta)$ is the derivative of the log-likelihood function ($\text{log} L(\theta|x)$) where $\theta$ are the parameters and $x$ the data (the univariate case is shown here for illustration purposes):

$$ U(\theta) = \frac{\partial \text{log} L(\theta|x)}{\partial \theta} $$

This is basically the slope of the log-likelihood function. Further, let $I(\theta)$ be the Fisher information matrix which is the negative expectation of the second derivative of the log-likelihood function with respect to $\theta$. The score test statistics is:

$$ S(\theta_{0})=\frac{U(\theta_{0}^{2})}{I(\theta_{0})}\sim\chi^{2}_{1} $$

The score test can also be calculated using anova (the score test statistics is called "Rao"):

anova(mylogit2, mylogit,  test="Rao")

Analysis of Deviance Table

Model 1: admit ~ I(gre + gpa) + rank
Model 2: admit ~ gre + gpa + rank
  Resid. Df Resid. Dev Df Deviance    Rao Pr(>Chi)  
1       395     464.48                              
2       394     458.52  1   5.9658 5.9144  0.01502 *
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The conclusion is the same as before.


An interesting relationship between the different test statistics when the model is linear is (Johnston and DiNardo (1997): Econometric Methods): Wald $\geq$ LR $\geq$ Score.

  • 1
    $\begingroup$ I wonder why the reduced model simply excludes gre and gpa? Isn't that testing $\beta_1=\beta_2=0$, not $\beta_1=\beta_2$? To me, to correctly test $\beta_1=\beta_2$, we need to keep gre and gpa and meanwhile impose $\beta_{\text{gre}}=\beta_{\text{gpa}}$. $\endgroup$ Commented May 11, 2015 at 9:07
  • 1
    $\begingroup$ @SibbsGambling Good catch! I updated my answer accordingly. $\endgroup$ Commented May 11, 2015 at 16:06
  • $\begingroup$ Is this limited to continuous predictors only, or could I - for instance - also see whether two levels of a categorical variable are significantly different? Let's say, is the difference between rank3 and rank4 significant? $\endgroup$
    – Daniel
    Commented Jun 4, 2015 at 17:16
  • 1
    $\begingroup$ @Daniel Yes, this approach can also be used for levels of a categorical variable. The multcomp packages makes it particularly easy. For example, try this: glht.mod <- glht(mylogit, linfct = c("rank3 - rank4= 0")). But a much easier way would be to make rank3 the reference level (using mydata$rank <- relevel(mydata$rank, ref="3")) and then just use the normal regression output. Each level of the factor is compared to the reference level. The p-value for rank4 would be the desired comparison. $\endgroup$ Commented Jun 4, 2015 at 20:31
  • 1
    $\begingroup$ @Daniel The p-values from the model output (changed reference level) and glht are the same for me (about $0.591$). Regarding your second question: linfct = c("rank3 - rank4= 0") tests only one linear hypothesis whereas mcp(rank="Tukey") tests all 6 pairwise comparisons of rank. So the p-values have to be adjusted for multiple comparisons. This means that the p-values using Tukey's test are generally higher than the single comparison. $\endgroup$ Commented Jun 5, 2015 at 6:31

You did not specify your variables, if they are binary or something else. I think you talk about binary variables. There also exist multinomial versions of the probit and logit model.

In general, you can use the complete trinity of test approaches, i.e.




Each test uses different test-statistics. The standard approach would be to take one of the three tests. All three can be used to do joint tests.

The LR test uses the differnce of the log-likelihood of a restricted and the unrestricted model. So the restricted model is the model, in which the specified coefficients are set to zero. The unrestricted is the "normal" model. The Wald test has the advantage, that only the unrestriced model is estimated. It basically asks, if the restriction is nearly satisfied if it is evaluated at the unrestriced MLE. In case of the Lagrange-Multiplier test only the restricted model has to be estimated. The restricted ML estimator is used to calculate the score of the unrestricted model. This score will be usually not zero, so this discrepancy is the basis of the LR test. The LM-Test can in your context also be used to test for heteroscedasticity.


The standard approaches are the Wald test, the likelihood ratio test and the score test. Asymptotically they should be the same. In my experience the likelihood ratio tests tends to perform slightly better in simulations on finite samples, but the cases where this matters would be in very extreme (small sample) scenarios where I would take all of these tests as a rough approximation only. However, depending on your model (number of covariates, presence of interaction effects) and your data (multicolinearity, the marginal distribution of your dependent variable), the "wonderful kingdom of Asymptotia" can be well approximated by a surprisingly small number of observations.

Below is an example of such a simulation in Stata using the Wald, likelihood ratio and score test in a sample of only 150 observations. Even in such a small sample the three test produce fairly similar p-values and the sampling distribution of the p-values when the null hypothesis is true does seem to follow a uniform distribution as it should (or at least the deviations from the uniform distribution are no larger than one would expect due to the randomness inherrit in a Monte Carlo experiment).

clear all
set more off

// data preparation
sysuse nlsw88, clear

gen byte edcat = cond(grade <  12, 1,     ///
                 cond(grade == 12, 2, 3)) ///
                 if grade < .
label define edcat 1 "less than high school" ///
                   2 "high school"           ///
                   3 "more than high school"
label value edcat edcat
label variable edcat "education in categories"

// create cascading dummies, i.e.
// edcat2 compares high school with less than high school
// edcat3 compares more than high school with high school
gen byte edcat2 = (edcat >= 2) if edcat < .
gen byte edcat3 = (edcat >= 3) if edcat < .

keep union edcat2 edcat3 race south
bsample 150 if !missing(union, edcat2, edcat3, race, south)

// constraining edcat2 = edcat3 is equivalent to adding 
// a linear effect (in the log odds) of edcat
constraint define 1 edcat2 = edcat3

// estimate the constrained model
logit union edcat2 edcat3 i.race i.south, constraint(1)

// predict the probabilities
predict pr
gen byte ysim = .
gen w = .

program define sim, rclass
    // create a dependent variable such that the null hypothesis is true
    replace ysim = runiform() < pr

    // estimate the constrained model
    logit ysim edcat2 edcat3 i.race i.south, constraint(1)
    est store constr

    // score test
    tempname b0
    matrix `b0' = e(b)
    logit ysim edcat2 edcat3 i.race i.south, from(`b0') iter(0)
    matrix chi = e(gradient)*e(V)*e(gradient)'
    return scalar p_score = chi2tail(1,chi[1,1])

    // estimate unconstrained model
    logit ysim edcat2 edcat3 i.race i.south 
    est store full

    // Wald test
    test edcat2 = edcat3
    return scalar p_Wald = r(p)

    // likelihood ratio test
    lrtest full constr
    return scalar p_lr = r(p)

simulate p_score=r(p_score) p_Wald=r(p_Wald) p_lr=r(p_lr), reps(2000) : sim
simpplot p*, overall reps(20000) scheme(s2color) ylab(,angle(horizontal))

enter image description here

  • 2
    $\begingroup$ score test is a different name for what @jen-bohold called a Lagrange multiplier (LM) test. $\endgroup$ Commented May 15, 2013 at 9:44
  • $\begingroup$ Nice answer (+1). I especially like the effort of the simulation. I didn't know how to calculate the score test in Stata. Thanks. $\endgroup$ Commented May 20, 2013 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.