Multivariate probit regression using (direct) maximum likelihood estimators
arthur charpentier
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Consider a random pair
of binary responses, i.e.
with
taking values 1 or 2. Assume that probability
can be function of some covariates .
The Gaussian vector latent structure
A standard model is based a latent Gaussian structure, i.e. there
exists some random vector
such that
if
is lower than a given threshold, and 1 otherwise.
As in standard probit models, assume that
where we can assume that
is a Gaussian
random vector. This assumption can be used to derive the
likelihood of a sample .
(the code is a bit long since I had trouble working properly with
matrices – or more precisely to vectorize my functions – so I used
loops… I am sure it is possible to write a better code).
It is possible to generate samples (based on that specific model) to
check that we can actually derive proper maximum likelihood estimators,
(the two columns on the right are latent observations, that cannot
be used since theoretically they are unobservable). Note that it is a simple regression, one of the
component is here only to bring some noise. First of all, let us look
at marginal probit regression
> reg1=glm(Y1~X1+X2,data=base,family=binomial)
> reg2=glm(Y2~X1+X2,data=base,family=binomial)
> summary(reg1)
Call:
glm(formula = Y1 ~ X1 + X2, family = binomial, data = base)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.90570 -0.50126 -0.00266 0.49162 2.78256
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.291725 0.267149 -16.065 <2e-16
X1 8.656836 0.510153 16.969 <2e-16 ***
X2 0.007375 0.090530 0.081 0.935
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 1386.29 on 999 degrees of freedom
Residual deviance: 726.48 on 997 degrees of freedom
AIC: 732.48
Number of Fisher Scoring iterations: 5
> summary(reg2)
Call:
glm(formula = Y2 ~ X1 + X2, family = binomial, data = base)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.74682 -0.51814 -0.00001 0.57969 2.58565
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 3.91709 0.24399 16.054 <2e-16 ***
X1 -7.89703 0.46277 -17.065 <2e-16 ***
X2 0.18360 0.08758 2.096 0.036 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 1386.29 on 999 degrees of freedom
Residual deviance: 777.61 on 997 degrees of freedom
AIC: 783.61
Number of Fisher Scoring iterations: 5
That procedure works well an can be extended to ordinal responses
(not only binary ones, or to three dimensional problems,
logV=function(beta){
BETA=matrix(beta[4:(3+ncol(Y)*ncol(X))],ncol(Y),ncol(X))
z=cbind(X%*%(BETA[1,]),X%*%(BETA[2,]),X%*%(BETA[3,]))
r12=beta[1]
r23=beta[2]
r31=beta[3]
s1=s2=s3=1
sigma=matrix(c(s1^2,r12*s1*s2,r31*s1*s3,
r12*s1*s2,s2^2,r23*s2*s3,
r31*s1*s3,r23*s2*s3,s3^2),3,3)
sigma1=matrix(c(s2^2,r23*s2*s3,
r23*s2*s3,s3^2),2,2)
sigma2=matrix(c(s1^2,r31*s1*s3,
r31*s1*s3,s3^2),2,2)
sigma3=matrix(c(s1^2,r12*s1*s2,
r12*s1*s2,s2^2),2,2)
a111=pmnorm(z[1,],rep(0,ncol(Y)),varcov=sigma)
for(i in 2:nrow(z)){a111=c(a111,pmnorm(z[i,],rep(0,ncol(Y)),varcov=sigma))}
a011=pmnorm(z[1,2:3],varcov=sigma1)-pmnorm(z[1,],varcov=sigma)
for(i in 2:nrow(z)){a011=c(a011,pmnorm(z[i,2:3],varcov=sigma1)-pmnorm(z[i,],varcov=sigma))}
a101=pmnorm(z[1,c(1,3)],varcov=sigma2)-pmnorm(z[1,],varcov=sigma)
for(i in 2:nrow(z)){a101=c(a101,pmnorm(z[i,c(1,3)],varcov=sigma2)-pmnorm(z[i,],varcov=sigma))}
a110=pmnorm(z[1,1:2],varcov=sigma3)-pmnorm(z[1,],varcov=sigma)
for(i in 2:nrow(z)){a110=c(a110,pmnorm(z[i,1:2],varcov=sigma3)-pmnorm(z[i,],varcov=sigma))}
a100=pnorm(z[1,1],sd=s1)-pmnorm(z[1,c(1,2)],varcov=sigma3)-pmnorm(z[1,c(1,3)],varcov=sigma2)+pmnorm(z[1,],rep(0,ncol(Y)),varcov=sigma)
for(i in 2:nrow(z)){a100=c(a100,pnorm(z[i,1],sd=s1)-pmnorm(z[i,c(1,2)],varcov=sigma3)-pmnorm(z[i,c(1,3)],varcov=sigma2)+pmnorm(z[i,],rep(0,ncol(Y)),varcov=sigma))}
a010=pnorm(z[1,2],sd=s2)-pmnorm(z[1,c(1,2)],varcov=sigma3)-pmnorm(z[1,c(2,3)],varcov=sigma1)+pmnorm(z[1,],rep(0,ncol(Y)),varcov=sigma)
for(i in 2:nrow(z)){a010=c(a010,pnorm(z[i,2],sd=s2)-pmnorm(z[i,c(1,2)],varcov=sigma3)-pmnorm(z[i,c(2,3)],varcov=sigma1)+pmnorm(z[i,],rep(0,ncol(Y)),varcov=sigma))}
a001=pnorm(z[1,3],sd=s3)-pmnorm(z[1,c(2,3)],varcov=sigma1)-pmnorm(z[1,c(1,3)],varcov=sigma2)+pmnorm(z[1,],rep(0,ncol(Y)),varcov=sigma)
for(i in 2:nrow(z)){a001=c(a001,pnorm(z[i,3],sd=s3)-pmnorm(z[i,c(2,3)],varcov=sigma1)-pmnorm(z[i,c(1,3)],varcov=sigma2)+pmnorm(z[i,],rep(0,ncol(Y)),varcov=sigma))}
a000=1-a111-a011-a101-a110-a001-a010-a100
a111[a111<=0]=1e-50
a110[a110<=0]=1e-50
a101[a101<=0]=1e-50
a011[a011<=0]=1e-50
a100[a100<=0]=1e-50
a010[a010<=0]=1e-50
a001[a001<=0]=1e-50
a000[a000<=0]=1e-50
-sum(((Y[,1]==0)&(Y[,2]==0)&(Y[,3]==0))*log(a111) +
((Y[,1]==1)&(Y[,2]==0)&(Y[,3]==0))*log(a011) +
((Y[,1]==0)&(Y[,2]==1)&(Y[,3]==0))*log(a101) +
((Y[,1]==0)&(Y[,2]==0)&(Y[,3]==1))*log(a110) +
((Y[,1]==1)&(Y[,2]==1)&(Y[,3]==0))*log(a001) +
((Y[,1]==1)&(Y[,2]==0)&(Y[,3]==1))*log(a010) +
((Y[,1]==0)&(Y[,2]==1)&(Y[,3]==1))*log(a100) +
((Y[,1]==1)&(Y[,2]==1)&(Y[,3]==1))*log(a000) )
}
A strong assumption in that
bivariate model is that residuals have a Gaussian structure. It is
possible to change that assumption
marginally: for instance if we use a logistic cumulative
distribution function, then we will have a bivariate logit regression
in terms of dependence structure: it is possible to consider
another copula than the gaussian one,
e.g. Gumbel’s copula (also called the bivariate
logistic copula), or Clayton’s
Here, the following code can be used to extend the model to non
Gaussian structures,
with here maths versus writing, with girls in red and boys in blue, where variables here are
female :
0: male
1: female
race :
1: hispanic
2: asian
3: african-amer
4: white
ses :
1: low
2: middle
3: high
schtyp : type of school
1: public
2: private
prog : type of program
1: general
2: academic
3: vocation
read : reading score
write : writing score
math : math score
science : science score
socst : social studies score
We can try to understand correlation between math and writing skills.
Covariates can be the sex of the child, and his reading skills. The question will
then be: are good students in maths and writing simply students that
can read well ?
Here the code is simply
> W=hsb2$write>=50
> M=hsb2$math>=50
> base=data.frame(Y1=W,Y2=M,
+ X1=hsb2$female,X2=hsb2$read)
>
> library(Zelig)
> REG= zelig(list(mu1=Y1~X1+X2,
+ mu2=Y2~X1+X2,
+ rho=~1),
+ model="bprobit",data=base)
> summary(REG)
Call:
zelig(formula = list(mu1 = Y1 ~ X1 + X2, mu2 = Y2 ~ X1 + X2,
rho = ~1), model = "bprobit", data = base)
Pearson Residuals:
Min 1Q Median 3Q Max
probit(mu1) -4.7518 -0.502594 0.15038 0.53038 1.8592
probit(mu2) -3.4243 -0.653537 0.23673 0.67011 2.6072
rhobit(rho) -4.9821 0.010481 0.13500 0.40776 2.9171
Coefficients:
Value Std. Error t value
(Intercept):1 -5.484711 0.787101 -6.96825
(Intercept):2 -4.061384 0.633781 -6.40818
(Intercept):3 1.332187 0.322175 4.13497
X1:1 1.125924 0.233550 4.82092
X1:2 0.167258 0.202498 0.82598
X2:1 0.103997 0.014662 7.09286
X2:2 0.082739 0.012026 6.88017
Number of linear predictors: 3
Names of linear predictors: probit(mu1), probit(mu2), rhobit(rho)
Dispersion Parameter for binom2.rho family: 1
Residual Deviance: 364.51 on 593 degrees of freedom
Log-likelihood: -182.255 on 593 degrees of freedom
Number of Iterations: 3
> (exp(summary(REG)@coef3[3])-1)/(exp(summary(REG)@coef3[3])+1)
[1] 0.5824045
with a remaining correlation among residuals of 0.58. So with only the
sex of the student, and his or her reading skill, we cannot explain the
correlation between maths and writing skills. With our previous code,
we have here
i.e. we obtain (almost) exactly the same estimators. But here I have
used as starting values for the optimization procedure the estimators
given by R. If we change them, hopefully we have a robust maximum
likelihood estimator,
> (OPT=optim(fn=logV,par=beta0/2,method="BFGS")$par)
(Intercept):1 (Intercept):2 X1:1 X1:2
0.58233360 -5.49428984 -4.06839571 1.12696594 0.16760347
X2:1 X2:2
0.10417767 0.08287409
There were 12 warnings (use warnings() to see them)
So once again, it is possible to optimize numerically a likelihood function, and it works.
