I am trying to run a ivprobit regression on individual crime and demographic variables. The dependent variable is annualassault and I have 24 independent variables regarding to the demographic. The r.h.s endogenous variable is sogs_total, and the instruments are drivetime and distance. I got the error: the leading minor of order 4 is not positive definite.
I reduced my independent variables to only 5 because I think I might be over fitting the model (too many variables), but I still get the same error.
library(ivprobit)
ivprobit.assault = ivprobit(annualassault ~ age +sex+ education_1+ education_2+
education_3+ education_4+ education_5+ education_6+ education_7
+maritalstatus_0+maritalstatus_1+maritalstatus_2+maritalstatus_3+maritalstatus_4+
a_nondm+d_nonam+m_nonad+ ad_nonm+am_nond
+dm_nona+ adm |sogs_total|drivetime + distance, newdata)
ivprobit.assault2 = ivprobit(annualassault ~ age +sex+ education_1+maritalstatus_1+
am_nond|sogs_total|drivetime + distance, newdata)
Error in chol.default(mat) :
the leading minor of order 4 is not positive definite
Related
I am getting errors of mismatched parameters for one of the predictor variables in my JAGS code and this is not producing convergence for 2 of my variables.
I am analyzing a data set of 1000 patients (observations) who had epilepsy and those who did not (1/0). Other variables are categorical - gender (male, female, LGBTQ), drug type use (diclofenac (1), bicarbonate (2) and thiazide (3)). Then, there is a continuous variable (BMI, ranging from 10:40). Below is my code. What could be wrong with it:
"model{
for (i in 1:n)
{
epilepsy[i]~dbern(prob[i])
logit(prob[i])<- intercept+gendereffect[gender[i]] + druguseeffect[Druguse[i]]+ BMIeffect*BMI[i]+
interaction*gender[i]*Druguse[i]}
intercept~dnorm(0,precision)
gendereffect[1]<-0
gendereffect[2]~dnorm(0,10^6)
gendereffect[3]~dnorm(0,10^6)
druguseeffect[1]<-0
druguseeffect[2] ~ dnorm(0,10^6)
druguseeffect[3]~dnorm(0,10^6)
intercept~dbeta(1,1)
BMIeffect~dnorm(0,10^6)
interaction~dnorm(0,10^6)
precision~dgamma(10,10)
#data# druguse, gender, BMI,n,epilepsy
#inits# intercept,BMIeffect,gendereffect,druguseeffect,interaction,precision
#monitor# gendereffect,BMIeffect,druguseeffect,interaction,precision
}"
Note: I have a list created. And on R, I converted druguse and gender to integers, both ranging 1:3. But the model is not running.
Question: Have a look at data set Two.csv. It contains a potentially dependent binary variable Y , and
two potentially independent variables {X1, X2} for each unit of measurement.
(a) Read data set Two.csv into R and have a look at the format of the dependent variable.
Discuss three models which might be appropriate in this data situation. Discuss which
aspects speak in favor of each model, and which aspects against.
(b) Suppose variable Y measures financial ratings A : y = 1, B : y = 2, and C : y = 3, that
is, the creditworthiness A: high, B: intermediate, C: low for unit of measurement firm
i. Model Y by means of an ordered Logit model as a function of {X1,X2} and estimate
your model by means of a built-in command.
(c) Explain the proportional odds-assumption and test whether the assumption is critical
in the context of the data set at hand.
##a) Read data set Two.csv into R and have a look at the format of the dependent variable.
O <- read.table("C:/Users/DELL/Downloads/ExamQEIII2021/Two.csv",header=TRUE,sep=";")
str(O)
dim(O)
View(O)
##b)
library(oglmx)
ologit<- oglmx(y~x1+x2,data=O, link="logit",constantMEAN = FALSE, constantSD = FALSE,
delta=0,threshparam =NULL)
results.ologis <- ologit.reg(y~x1+x2,data=O)
summary(results.ologis)
## x1 1.46251
## x2 -0.45391
margins.oglmx(results.ologis,ascontinuous = FALSE) #Build in command for AMElogit
##c) Explain the proportional odds-assumption and test whether the assumption is critical
#in the context of the data set at hand.
#ordinal Logit WITH proportional odds(PO)
library(VGAM)
a <- vglm(y~x1+x2,family=cumulative(parallel=TRUE),data=O)
summary(a)
#ordinal Logit WITHOUT proportional odds [a considers PO and c doesn't]
c <- vglm(y~x1+x2,family=cumulative(parallel=FALSE),data=O)
summary(c)
pchisq(deviance(a)-deviance(c),df.residual(a)-df.residual(c),lower.tail=FALSE)
## 0.4936413 ## No significant difference in the variance left unexplained. Cannot
#confirm that PO assumption is critical.
#small model
LLa <- logLik(a)
#large model
LLc <- logLik(c)
#2*LLc-2*
df.residual(c)
df.residual(a) #or similarly, via a Likelihood Ratio test.
# or, if you are unsure about the number of degrees of freedom
LL<- 2*(LLc -LLa)
1-pchisq(LL,df.residual(a)-df.residual(c))
## 0.4936413 [SAME AS ## No sign. line]
##Conclusion: Likelihood do not differ significantly with the assumption of non PO.
For my Bachelor's thesis I am trying to apply a linear median regression model on constant sum data from a survey (see formula from A.Blass (2008)). It is an attempt to recreate the probability elicitation approach proposed by A. Blass et al (2008) - Using Elicited Choice Probabilities to Estimate Random Utility Models: Preferences for Electricity Reliability
My dependent variable is the log-odds transformation of the constant sum allocations. Calculated using the following formula:
PE_raw <- PE_raw %>% group_by(sys_RespNum, Task) %>% mutate(LogProb = c(log(Response[1]/Response[1]),
log(Response[2]/Response[1]),
log(Response[3]/Response[1])))
My independent variables are delivery costs, minimum order quantity and delivery window, each categorical variables with levels 0, 1, 2 and 3. Here, level 0 represent the none-option.
Data snapshot
I tried running the following quantile regression (using R's quantreg package):
LAD.factor <- rq(LogProb ~ factor(`Delivery costs`) + factor(`Minimum order quantity`) + factor(`Delivery window`) + factor(NoneOpt), data=PE_raw, tau=0.5)
However, I ran into the following error indicating singularity:
Error in rq.fit.br(x, y, tau = tau, ...) : Singular design matrix
I ran a linear regression and applied R's alias function for further investigation. This informed me of three cases of perfect multicollinearity:
minimum order quantity 3 = delivery costs 1 + delivery costs 2 + delivery costs 3 - minimum order quantity 1 - minimum order quantity 2
delivery window 3 = delivery costs 1 + delivery costs 2 + delivery costs 3 - delivery window 1 - delivery window 2
NoneOpt = intercept - delivery costs 1 - delivery costs 2 - delivery costs 3
In hindsight these cases all make sense. When R dichotomizedthe categorical variables you get these results by construction as, delivery costs 1 + delivery costs 2 + delivery costs 3 = 1 and minimum order quantity 1 + minimum order quantity 2 + minimum order quantity 3 = 1. Rewriting gives the first formula.
It looks like a classic dummy trap. In an attempt to workaround this issue I tried to manually dichotomize the data and used the following formula:
LM.factor <- rq(LogProb ~ Delivery.costs_1 + Delivery.costs_2 + Minimum.order.quantity_1 + Minimum.order.quantity_2 + Delivery.window_1 + Delivery.window_2 + factor(NoneOpt), data=PE_dichomitzed, tau=0.5)
Instead of an error message I now got the following:
Warning message:
In rq.fit.br(x, y, tau = tau, ...) : Solution may be nonunique
When using the summary function:
> summary(LM.factor)
Error in base::backsolve(r, x, k = k, upper.tri = upper.tri, transpose = transpose, :
singular matrix in 'backsolve'. First zero in diagonal [2]
In addition: Warning message:
In summary.rq(LM.factor) : 153 non-positive fis
Is anyone familiar with this issue? I am looking for alternative solutions. Perhaps I am making mistakes using the rq() function, or the data might be misrepresented.
I am grateful for any input, thank you in advance.
Reproducible example
library(quantreg)
#### Raw dataset (PE_raw_SO) ####
# quantile regression (produces singularity error)
LAD.factor <- rq(
LogProb ~ factor(`Delivery costs`) +
factor(`Minimum order quantity`) + factor(`Delivery window`) +
factor(NoneOpt),
data = PE_raw_SO,
tau = 0.5
)
# linear regression to check for singularity
LM.factor <- lm(
LogProb ~ factor(`Delivery costs`) +
factor(`Minimum order quantity`) + factor(`Delivery window`) +
factor(NoneOpt),
data = PE_raw_SO
)
alias(LM.factor)
# impose assumptions on standard errors
summary(LM.factor, se = "iid")
summary(LM.factor, se = "boot")
#### Manually created dummy variables to get rid of
#### collinearity (PE_dichotomized_SO) ####
LAD.di.factor <- rq(
LogProb ~ Delivery.costs_1 + Delivery.costs_2 +
Minimum.order.quantity_1 + Minimum.order.quantity_2 +
Delivery.window_1 + Delivery.window_2 + factor(NoneOpt),
data = PE_dichotomized_SO,
tau = 0.5
)
summary(LAD.di.factor) #backsolve error
# impose assumptions (unusual results)
summary(LAD.di.factor, se = "iid")
summary(LAD.di.factor, se = "boot")
# linear regression to check for singularity
LM.di.factor <- lm(
LogProb ~ Delivery.costs_1 + Delivery.costs_2 +
Minimum.order.quantity_1 + Minimum.order.quantity_2 +
Delivery.window_1 + Delivery.window_2 + factor(NoneOpt),
data = PE_dichotomized_SO
)
alias(LM.di.factor)
summary(LM.di.factor) #regular results, all significant
Link to sample data + code: GitHub
The Solution may be nonunique behaviour is not unusual when doing quantile regressions with dummy explanatory variables.
See, e.g., the quantreg FAQ:
The estimation of regression quantiles is a linear programming
problem. And the optimal solution may not be unique.
A more intuitive explanation for what is happening is given by Roger Koenker (the author of quantreg) on r-help back in 2006:
When computing the median from a sample with an even number of
distinct values there is inherently some ambiguity about its value:
any value between the middle order statistics is "a" median.
Similarly, in regression settings the optimization problem solved by
the "br" version of the simplex algorithm, modified to do general
quantile regression identifies cases where there may be non
uniqueness of this type. When there are "continuous" covariates this
is quite rare, when covariates are discrete then it is relatively
common, atleast when tau is chosen from the rationals. For univariate
quantiles R provides several methods of resolving this sort of
ambiguity by interpolation, "br" doesn't try to do this, instead
returning the first vertex solution that it comes to.
Your second warning -- "153 non-positive fis" -- is a warning related to how the local densities are calculated by rq. Occasionally, it could be possible that local densities of the quantile regression function end up being negative (which is obviously impossible). If this happens, rq automatically sets them to zero. Again, quoting from the FAQ:
This is generally harmless, leading to a somewhat conservative
(larger) estimate of the standard errors, however if the reported
number of non-positive fis is large relative to the sample size then
it is an indication of misspecification of the model.
I am trying to learn hierarchical models in R and I have generated some sample data for myself. I am having trouble with the correct syntax for coding a multilevel regression problem.
I generated some data for salaries in a Business school. I made the salaries depend linearly on the number of years of employment and the total number of publications by the faculty member. The faculty are in various departments and I made the base salary(intercept) different for each department and also the yearly hike(slopes) different for each department. This way, I have the intercept (base salary) and slope(w.r.t experience in number of years) of the salary depend on the nested level (department) and slope w.r.t another explanatory variable (Publications) not depend on the nested level. What would be the correct syntax to model this in R?
here's my data
Data <-data.frame(Sl_No = c(1:40),
+ Dept = as.factor(sample(c("Mark","IT","Fin"),40,replace = TRUE)),
+ Years = round(runif(40,1,10)))
pubs <-round(Data$Years*runif(40,1,3))
Data$Pubs <- pubs
lookup_table<-data.frame(Dept = c("Mark","IT","Fin","Strat","Ops"),
+ base = c(100000,140000,150000,150000,120000),
+ slope = c(6000,5000,3000,2000,4000))
Data <- merge(Data,lookup_table,by = 'Dept')
salary <-Data$base+Data$slope*Data$Years+Data$Pubs*10000+rnorm(length(Data$Dept))*10000
Data$base<-NULL
Data$slope<-NULL
I have tried the following:
1)
multilevel_model<-lmer(Salary~1|Dept+Pubs+Years|Dept, data = Data)
Error in model.matrix.default(eval(substitute(~foo, list(foo = x[[2]]))), :
model frame and formula mismatch in model.matrix()
2)
multilevel_model<-lmer(`Salary`~ Dept + `Pubs`+`Years`|Dept , data = Data)
boundary (singular) fit: see ?isSingular
I want to see the estimates of the salary intercept and yearly hike by Dept and the estimate of the effect of publication as a standalone (pooled). Right now I am not getting the code to work at all.
I know the base salary and the yearly hike by dept and the effect of a publication (since I generated it).
Dept base Slope
Fin 150000 3000
Mark 100000 6000
Ops 120000 4000
IT 140000 5000
Strat 150000 2000
Every publication increases the salary by 10,000.
ANSWER:
Thanks to #Ben 's answer here I think the correct model is
multilevel_model<-lmer(Salary~(1|Dept)+ Pubs +(0+Years|Dept), data = Data)
This gives me the following fixed effects by running
summary(multilevel_model)
Fixed effects:
Estimate Std. Error t value
(Intercept) 131667.4 10461.0 12.59
Pubs 10235.0 550.8 18.58
Correlation of Fixed Effects:
Pubs -0.081
The Department level coefficients are as follows:
coef(multilevel_model)
$Dept
Years (Intercept) Pubs
Fin 3072.5133 148757.6 10235.02
IT 5156.6774 136710.7 10235.02
Mark 5435.8301 102858.3 10235.02
Ops 3433.1433 118287.1 10235.02
Strat 963.9366 151723.1 10235.02
These are pretty good estiamtes of the original values. Now I need to learn to assess "how good" they are. :)
(1)
multilevel_model<-lmer(`Total Salary`~ 1|Dept +
`Publications`+`Years of Exp`|Dept , data = sample_data)
I can't immediately diagnose why this gives a syntax error, but parentheses are generally recommended around random-effect terms because the | operator has high precedence in formulas. Thus the response/right-hand side (RHS) formula
~ (1|Dept) + (`Publications`+`Years of Exp`|Dept)
might work, except that it would be problematic because both terms contain the same intercept term: if you wanted to do this you'd probably need
~ (1|Dept) + (0+`Publications`+`Years of Exp`|Dept)
(2)
~ Dept + `Publications`+`Years of Exp`|Dept
It doesn't really make any sense to put the same variable (Dept) on both the left- and right-hand sides of the bar.
You should probably use
~ pubs + years_exp + (1 + years_exp|Dept)
Since in principle the effect of publication could vary across departments, the maximal model would be
~ pubs + years_exp + (1 + pubs + years_exp|Dept)
It rarely makes sense to include a random effect without its corresponding fixed effect.
Note that you may get singular fits even if you have the right model; see the ?isSingular man page.
if the 18 observations listed above represent your whole data set, it's very likely too small to fit the maximal model successfully. Rule of thumb is that you need 10-20 observations per parameter estimated, and the maximal model has (intercept + 2 fixed-effect params + (3*4)/2=6 random-effect parameters) = 9 parameters. (Since it's simulated, you can easily simulate a big data set ...)
I'd recommend renaming variables in your data frame so you don't have to fuss with backtick-protecting variable names with spaces in them ...
The GLMM FAQ has more on model specification
I wanted to use the function sem (with the package lavaan) on my data in R :
Model1<- 'Transfer~Amotivation+Gender+Age
Amotivation~Gender+Age
transfer are 4 questions with a 5 point likert scale
Amotivation: 4 questions with a 5 pint likert scale
Gender: 0 (=male) and 1 (=female)
Age: just the different ages
And i got next error:
in getDataFull (data= data, group = group, grow.label = group.label,:
lavaan WARNING: some observed variances are (at least) a factor 100 times larger than others; please rescale
Is anybody familiar with this error? Does it influence my results? Do I have to change anything? I really don't know what this error means.
Your scales are not equivalent. Your gender variables are constrained to be either 0 or 1. Amotivation is constrained to be between 1 and 5, but age is even less constrained. I created some sample data for gender, age, and amotivation. You can see that the variance for the age variable is over 4,000 times higher than the variance for gender, and about 500 times higher than sample amotivation data.
gender <- c(0,1,1,1,0,0,1,1,0,1,1,0,0,1,1,1)
age <- c(18,42,87,12,24,26,98,84,23,12,95,44,54,23,10,16)
set.seed(42)
amotivation <- rnorm(16, 3, 1.5)
var(gender) # 0.25 variance
var(age) # 1017.27 variance
var(amotivation) # 2.21 variance
I'm not sure how the unequal variances influence your results, or if you need to do anything at all. To make your age variable more closely match the amotivation scale, you could transform the data so that it's also on a 5 point scale.
newage <- age/max(age)*5
var(newage) # 2.65 variance
You could try running the analysis both ways (using your original data and the transformed data) and see if there are differences.