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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.
Please again accept my apologies for my little knowledge in R. I'm, trying to get better! I'm a biologist and my statistical knowledge is sadly low
I have the following data set:
Perc_Reacting,Pulses,IndMutant,Proportion
93,1,1,0.93
81,2,1,0.81
73,3,1,0.73
64,4,1,0.64
73,5,1,0.73
68,6,1,0.68
64,7,1,0.64
65,8,1,0.65
50,9,1,0.5
68,10,1,0.68
57,11,1,0.57
50,12,1,0.5
62,13,1,0.62
44,14,1,0.44
54,15,1,0.54
56,16,1,0.56
50,17,1,0.5
42,18,1,0.42
42,19,1,0.42
29,20,1,0.29
96,1,0,0.96
100,2,0,1
92,3,0,0.92
96,4,0,0.96
92,5,0,0.92
92,6,0,0.92
84,7,0,0.84
96,8,0,0.96
91,9,0,0.91
82,10,0,0.82
86,11,0,0.86
82,12,0,0.82
91,13,0,0.91
85,14,0,0.85
83,15,0,0.83
70,16,0,0.7
74,17,0,0.74
64,18,0,0.64
68,19,0,0.68
78,20,0,0.78
The first and last rows are the same, one expressed in % an the other in a 1-0 proportion
I need to run a Beta regression model, but when I try to create the model an error jumps:
model.beta<-betareg(C_elegans$Proportion~C_elegans$Pulses)
Error in betareg(C_elegans$Proportion ~ C_elegans$Pulses) :
invalid dependent variable, all observations must be in (0, 1)
Could you help me to create a beta regression model for this data and how to make relevant plots to show it fits good?
Also I need to propose a linear regression model for this data, can anyone let me know how you think it could be done better?
Here are the results of fitting the last three columns to a flat surface plane equation "Proportion = a + (b * Pulses) + (c * IndMutant)" with parameters a = 1.0468289473684214E+00, b = -1.8650375939849695E-02, and c = -2.5850000000000006E-01 yielding R-squared = 0.876 and RMSE = 0.064.
(here "absolute error" means "not relative error")
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 am trying to evaluate themodel fit of several regressions in R, and I have run into a problem I have had multiple times now: the log-likelihood of my Poisson regression is infinite.
I'm using a non-integer dependent variable (Note: I know what I'm doing in this regard), and I'm wondering if maybe that's the problem. However, I don't get an infinite log-likelihood when running the regression with glm.nb.
Code to reproduce the issue is below.
Edit: the problem appears to go away when I coerce the DV to integer. Any idea how to get log likelihood from Poissons with non-integer DVs?
# Input Data
so_data <- data.frame(dv = c(21.0552722691125, 24.3061351414885, 7.84658638053276,
25.0294679770848, 15.8064731063311, 10.8171744654056, 31.3008088413026,
2.26643928259238, 18.4261153345417, 5.62915828161753, 17.0691184593063,
1.11959635820499, 30.0154935602592, 23.0000809735738, 28.4389825676123,
27.7678405415711, 23.7108405071757, 23.5070651053276, 14.2534787168392,
15.2058525068363, 19.7449094187771, 2.52384709295823, 29.7081691356397,
32.4723790240354, 19.2147002673637, 61.7911384519901, 10.5687170234821,
23.9047421013736, 18.4889651451222, 13.0360878554798, 15.1752866581849,
11.5205948111817, 31.3539840929108, 31.7255952728076, 25.3034625215724,
5.00013988265465, 30.2037887018226, 1.86123112349445, 3.06932041603219,
22.6739418581257, 6.33738321053804, 24.2933951601142, 14.8634827414491,
31.8302947881089, 34.8361908525564, 1.29606416941288, 13.206844629927,
28.843579313401, 25.8024295609021, 14.4414831628722, 18.2109680632694,
14.7092063453463, 10.0738043919183, 28.4124482962025, 27.1004208775326,
1.31350378236957, 14.3009307888745, 1.32555197766214, 2.70896028922312,
3.88043749517381, 3.79492216916016, 19.4507965653633, 32.1689088941444,
2.61278585713499, 41.6955885902228, 2.13466761675063, 30.4207256294235,
24.8231524369244, 20.7605955978196, 17.2182798298094, 2.11563574288652,
12.290778250655, 0.957467139696772, 16.1775287334746))
# Run Model
p_mod <- glm(dv ~ 1, data = so_data, family = poisson(link = 'log'))
# Be Confused
logLik(p_mod)
Elaborating on #ekstroem's comment: the Poisson distribution is only supported over the non-negative integers (0, 1, ...). So, technically speaking, the probability of any non-integer value is zero -- although R does allow for a little bit of fuzz, to allow for round-off/floating-point representation issues:
> dpois(1,lambda=1)
[1] 0.3678794
> dpois(1.1,lambda=1)
[1] 0
Warning message:
In dpois(1.1, lambda = 1) : non-integer x = 1.100000
> dpois(1+1e-7,lambda=1) ## fuzz
[1] 0.3678794
It is theoretically possible to compute something like a Poisson log-likelihood for non-integer values:
my_dpois <- function(x,lambda,log=FALSE) {
LL <- -lambda+x*log(lambda)-lfactorial(x)
if (log) LL else exp(LL)
}
but I would be very careful - some quick tests with integrate suggest it integrates to 1 (after I fixed the bug in it), but I haven't checked more carefully that this is really a well-posed probability distribution. (On the other hand, some reasonable-seeming posts on CrossValidated suggest that it's not insane ...)
You say "I know what I'm doing in this regard"; can you give some more of the context? Some alternative possibilities (although this is steering into CrossValidated territory) -- the best answer depends on where your data really come from (i.e., why you have "count-like" data that are non-integer but you think should be treated as Poisson).
a quasi-Poisson model (family=quasipoisson). (R will still not give you log-likelihood or AIC values in this case, because technically they don't exist -- you're supposed to do inference on the basis of the Wald statistics of the parameters; see e.g. here for more info.)
a Gamma model (probably with a log link)
if the data started out as count data that you've scaled by some measure of effort or exposure), use an appropriate offset model ...
a generalized least-squares model (nlme::gls) with an appropriate heteroscedasticity specification
Poisson log-likelihood involves calculating log(factorial(x)) (https://www.statlect.com/fundamentals-of-statistics/Poisson-distribution-maximum-likelihood). For values larger than 30 it has to be done using Stirling's approximation formula in order to avoid exceeding the limit of computer arithmetic. Sample code in Python:
# define a likelihood function. https://www.statlect.com/fundamentals-of- statistics/Poisson-distribution-maximum-likelihood
def loglikelihood_f(lmba, x):
#Using Stirling formula to avoid calculation of factorial.
#logfactorial(n) = n*ln(n) - n
n = x.size
logfactorial = x*np.log(x+0.001) - x #np.log(factorial(x))
logfactorial[logfactorial == -inf] = 0
result =\
- np.sum(logfactorial) \
- n * lmba \
+ np.log(lmba) * np.sum(x)
return result
I am applying guantile regression for my data-set (using R). It is easy to produce the nice scatterplot-image with different quantile regression lines
(taus <- c(0.05,0.25,0.75,0.95)).
Problem occurs when I want to produce p-values (in order to see statistical significance of each regression line) for each one of these quantiles. For median quantile (tau=0.5) this is not problematic, but when it comes to for example tau=0.25, I get following error message:
>QRmodel<-rq(y~x,tau=0.25,model=T)
>summary(QRmodel,se="nid")
Error in summary.rq(QRmodel, se = "nid") : tau - h < 0: error in summary.rq
What could be the reason for this?
Also: Is it recommendable to mention p-values and coefficients regarding the results of quantile regression model or could it be enough to show just the plot-picture and discuss the results based on that picture?
Best regards, frustrated person
A good way to learn what's going on in these sorts of debugging situations is to find the relevant portion of code that is throwing the error. If you type 'summary.rq' at the console, you'll see the code for the function summary.rq. Scanning through it you'll find the section where it calculates se's using the "nid" method, starting with this code:
else if (se == "nid") {
h <- bandwidth.rq(tau, n, hs = hs)
if (tau + h > 1)
stop("tau + h > 1: error in summary.rq")
if (tau - h < 0)
stop("tau - h < 0: error in summary.rq")
bhi <- rq.fit.fnb(x, y, tau = tau + h)$coef
blo <- rq.fit.fnb(x, y, tau = tau - h)$coef
So what's happening here is that in order to calculate the se's, the function first need to calculate a bandwidth, h, and the quantreg model is refit for tau +/- h. For tau's near 0 or 1, there's a possibility that adding or subtracting the bandwidth 'h' will lead to a tau below 0 or greater than 1, which isn't good, so the function stops.
You have a couple of options:
1.) Try a different se method (bootstrapping?)
2.) Modify the summary.rq code yourself to force it to use either max(tau,0) or min(tau,1) in the instances where the bandwidth pushes tau out of bounds. (There could be serious theoretical reasons why this is a bad idea; not advised unless you know what you're doing.)
3.) You could try to read up on the theory behind the calculation of these se's so you'd have a better idea of when they might work well or not. This might shed some light on why you're running into errors with values of tau near 0 or 1.
Try summary(QRmodel,se="boot")
Have a look at the help for summary.rq as well!