I am trying to compute robust/cluster standard errors after using mlogit() to fit a Multinomial Logit (MNL) in a Discrete Choice problem. Unfortunately, I suspect I am having problems with it because I am using data in long format (this is a must in my case), and getting the error #Error in ef/X : non-conformable arrays after sandwich::vcovHC( , "HC0").
The Data
For illustration, please gently consider the following data. It represents data from 5 individuals (id_ind ) that choose among 3 alternatives (altern). Each of the five individuals chose three times; hence we have 15 choice situations (id_choice). Each alternative is represented by two generic attributes (x1 and x2), and the choices are registered in y (1 if selected, 0 otherwise).
df <- read.table(header = TRUE, text = "
id_ind id_choice altern x1 x2 y
1 1 1 1 1.586788801 0.11887832 1
2 1 1 2 -0.937965347 1.15742493 0
3 1 1 3 -0.511504401 -1.90667519 0
4 1 2 1 1.079365680 -0.37267925 0
5 1 2 2 -0.009203032 1.65150370 1
6 1 2 3 0.870474033 -0.82558651 0
7 1 3 1 -0.638604013 -0.09459502 0
8 1 3 2 -0.071679538 1.56879334 0
9 1 3 3 0.398263302 1.45735788 1
10 2 4 1 0.291413453 -0.09107974 0
11 2 4 2 1.632831160 0.92925495 0
12 2 4 3 -1.193272276 0.77092623 1
13 2 5 1 1.967624379 -0.16373709 1
14 2 5 2 -0.479859282 -0.67042130 0
15 2 5 3 1.109780885 0.60348187 0
16 2 6 1 -0.025834772 -0.44004183 0
17 2 6 2 -1.255129594 1.10928280 0
18 2 6 3 1.309493274 1.84247199 1
19 3 7 1 1.593558740 -0.08952151 0
20 3 7 2 1.778701074 1.44483791 1
21 3 7 3 0.643191170 -0.24761157 0
22 3 8 1 1.738820924 -0.96793288 0
23 3 8 2 -1.151429915 -0.08581901 0
24 3 8 3 0.606695064 1.06524268 1
25 3 9 1 0.673866953 -0.26136206 0
26 3 9 2 1.176959443 0.85005871 1
27 3 9 3 -1.568225496 -0.40002252 0
28 4 10 1 0.516456176 -1.02081089 1
29 4 10 2 -1.752854918 -1.71728381 0
30 4 10 3 -1.176101700 -1.60213536 0
31 4 11 1 -1.497779616 -1.66301234 0
32 4 11 2 -0.931117325 1.50128532 1
33 4 11 3 -0.455543630 -0.64370825 0
34 4 12 1 0.894843784 -0.69859139 0
35 4 12 2 -0.354902281 1.02834859 0
36 4 12 3 1.283785176 -1.18923098 1
37 5 13 1 -1.293772990 -0.73491317 0
38 5 13 2 0.748091387 0.07453705 1
39 5 13 3 -0.463585127 0.64802031 0
40 5 14 1 -1.946438667 1.35776140 0
41 5 14 2 -0.470448172 -0.61326604 1
42 5 14 3 1.478763383 -0.66490028 0
43 5 15 1 0.588240775 0.84448489 1
44 5 15 2 1.131731049 -1.51323232 0
45 5 15 3 0.212145247 -1.01804594 0
")
The problem
Consequently, we can fit an MNL using mlogit() and extract their robust variance-covariance as follows:
library(mlogit)
library(sandwich)
mo <- mlogit(formula = y ~ x1 + x2|0 ,
method ="nr",
data = df,
idx = c("id_choice", "altern"))
sandwich::vcovHC(mo, "HC0")
#Error in ef/X : non-conformable arrays
As we can see there is an error produced by sandwich::vcovHC, which says that ef/X is non-conformable. Where X <- model.matrix(x) and ef <- estfun(x, ...). After looking through the source code on the mirror on GitHub I spot the problem which comes from the fact that, given that the data is in long format, ef has dimensions 15 x 2 and X has 45 x 2.
My workaround
Given that the show must continue, I am computing the robust and cluster standard errors manually using some functions that I borrow from sandwich and I adjusted to accommodate the Stata's output.
> Robust Standard Errors
These lines are inspired on the sandwich::meat() function.
psi<- estfun(mo)
k <- NCOL(psi)
n <- NROW(psi)
rval <- (n/(n-1))* crossprod(as.matrix(psi))
vcov(mo) %*% rval %*% vcov(mo)
# x1 x2
# x1 0.23050261 0.09840356
# x2 0.09840356 0.12765662
Stata Equivalent
qui clogit y x1 x2 ,group(id_choice) r
mat li e(V)
symmetric e(V)[2,2]
y: y:
x1 x2
y:x1 .23050262
y:x2 .09840356 .12765662
> Clustered Standard Errors
Here, given that each individual answers 3 questions is highly likely that there is some degree of correlation among individuals; hence cluster corrections should be preferred in such situations. Below I compute the cluster correction in this case and I show the equivalence with the Stata output of clogit , cluster().
id_ind_collapsed <- df$id_ind[!duplicated(mo$model$idx$id_choice,)]
psi_2 <- rowsum(psi, group = id_ind_collapsed )
k_cluster <- NCOL(psi_2)
n_cluster <- NROW(psi_2)
rval_cluster <- (n_cluster/(n_cluster-1))* crossprod(as.matrix(psi_2))
vcov(mo) %*% rval_cluster %*% vcov(mo)
# x1 x2
# x1 0.1766707 0.1007703
# x2 0.1007703 0.1180004
Stata equivalent
qui clogit y x1 x2 ,group(id_choice) cluster(id_ind)
symmetric e(V)[2,2]
y: y:
x1 x2
y:x1 .17667075
y:x2 .1007703 .11800038
The Question:
I would like to accommodate my computations within the sandwich ecosystem, meaning not computing the matrices manually but actually using the sandwich functions. Is it possible to make it work with models in long format like the one described here? For example, providing the meat and bread objects directly to perform the computations? Thanks in advance.
PS: I noted that there is a dedicated bread function in sandwich for mlogit, but I could not spot something like meat for mlogit, but anyways I am probably missing something here...
Why vcovHC does not work for mlogit
The class of HC covariance estimators can just be applied in models with a single linear predictor where the score function aka estimating function is the product of so-called "working residuals" and a regressor matrix. This is explained in some detail in the Zeileis (2006) paper (see Equation 7), provided as vignette("sandwich-OOP", package = "sandwich") in the package. The ?vcovHC also pointed to this but did not explain it very well. I have improved this in the documentation at http://sandwich.R-Forge.R-project.org/reference/vcovHC.html now:
The function meatHC is the real work horse for estimating the meat of HC sandwich estimators - the default vcovHC method is a wrapper calling sandwich and bread. See Zeileis (2006) for more implementation details. The theoretical background, exemplified for the linear regression model, is described below and in Zeileis (2004). Analogous formulas are employed for other types of models, provided that they depend on a single linear predictor and the estimating functions can be represented as a product of “working residual” and regressor vector (Zeileis 2006, Equation 7).
This means that vcovHC() is not applicable to multinomial logit models as they generally use separate linear predictors for the separate response categories. Similarly, two-part or hurdle models etc. are not supported.
Basic "robust" sandwich covariance
Generally, for computing the basic Eicker-Huber-White sandwich covariance matrix estimator, the best strategy is to use the sandwich() function and not the vcovHC() function. The former works for any model with estfun() and bread() methods.
For linear models sandwich(..., adjust = FALSE) (default) and sandwich(..., adjust = TRUE) correspond to HC0 and HC1, respectively. In a model with n observations and k regression coefficients the former standardizes with 1/n and the latter with 1/(n-k).
Stata, however, divides by 1/(n-1) in logit models, see:
Different Robust Standard Errors of Logit Regression in Stata and R. To the best of my knowledge there is no clear theoretical reason for using specifically one or the other adjustment. And already in moderately large samples, this makes no difference anyway.
Remark: The adjustment with 1/(n-1) is not directly available in sandwich() as an option. However, coincidentally, it is the default in vcovCL() without specifying a cluster variable (i.e., treating each observation as a separate cluster). So this is a convenient "trick" if you want to get exactly the same results as Stata.
Clustered covariance
This can be computed "as usual" via vcovCL(..., cluster = ...). For mlogit models you just have to consider that the cluster variable just needs to be provided once (as opposed to stacked several times in long format).
Replicating Stata results
With the data and model from your post:
vcovCL(mo)
## x1 x2
## x1 0.23050261 0.09840356
## x2 0.09840356 0.12765662
vcovCL(mo, cluster = df$id_choice[1:15])
## x1 x2
## x1 0.1766707 0.1007703
## x2 0.1007703 0.1180004
Related
I have this table abd I want to analyse it statistically.
table(sci$category, sci$true_group)
mono sim_rus_nen suc_balanced suc_nen_rus suc_rus_nen
generalization 9 3 9 4 3
description 35 16 15 13 17
scheme 2 1 1 1 2
syncretism 5 3 7 16 2
tautology 2 2 2 3 3
substitution 1 0 0 0 0
indefinite 7 5 5 6 9
no_answer 30 17 18 13 19
So I decided to apply Fisher's exact test. But I have this error (although it's OK with chiq.square)
fisher.test(table(sci$category, sci$true_group))
Error in fisher.test(my_tab) : Bug in fexact3, it[i=6]=0: negative
key -1099365618 (kyy=91)
How can I fix this?
For larger contingency table/ counts, it gets resource intensive, to count all the worst cases to arrive at the p-value (seems to be that error).
So its convenient to simulate the p-values for tables larger than (2x2):
df <- table(sci$category, sci$true_group)
fisher.test(df, simulate.p.value = TRUE, B = 1e6)
Fisher's Exact Test for Count Data with simulated p-value (based on 1e+06 replicates)
data: df
p-value = 0.1054
alternative hypothesis: two.sided
PS: Choosing between Fishers exact test vs Chisq-test is a whole another discussion. I would refer you to this cross validated post for clarity: Alternatives to chisq-test
I have a multilevel repeated measures dataset of around 300 patients each with up to 10 repeated measures predicting troponin rise. There are other variables in the dataset, but I haven't included them here.
I am trying to use nlme to create a random slope, random intercept model where effects vary between patients, and effect of time is different in different patients. When I try to introduce a first-order covariance structure to allow for the correlation of measurements due to time I get the following error message.
Error in `coef<-.corARMA`(`*tmp*`, value = value[parMap[, i]]) : Coefficient matrix not invertible
I have included my code and a sample of the dataset, and I would be very grateful for any words of wisdom.
#baseline model includes only the intercept. Random slopes - intercept varies across patients
randomintercept <- lme(troponin ~ 1,
data = df, random = ~1|record_id, method = "ML",
na.action = na.exclude,
control = list(opt="optim"))
#random intercept and time as fixed effect
timeri <- update(randomintercept,.~. + day)
#random slopes and intercept: effect of time is different in different people
timers <- update(timeri, random = ~ day|record_id)
#model covariance structure. corAR1() first order autoregressive covariance structure, timepoints equally spaced
armodel <- update(timers, correlation = corAR1(0, form = ~day|record_id))
Error in `coef<-.corARMA`(`*tmp*`, value = value[parMap[, i]]) : Coefficient matrix not invertible
Data:
record_id day troponin
1 1 32
2 0 NA
2 1 NA
2 2 NA
2 3 8
2 4 6
2 5 7
2 6 7
2 7 7
2 8 NA
2 9 9
3 0 14
3 1 1167
3 2 1935
4 0 19
4 1 16
4 2 29
5 0 NA
5 1 17
5 2 47
5 3 684
6 0 46
6 1 45440
6 2 47085
7 0 48
7 1 87
7 2 44
7 3 20
7 4 15
7 5 11
7 6 10
7 7 11
7 8 197
8 0 28
8 1 31
9 0 NA
9 1 204
10 0 NA
10 1 19
You can fit this if you change your optimizer to "nlminb" (or at least it works with the reduced data set you posted).
armodel <- update(timers,
correlation = corAR1(0, form = ~day|record_id),
control=list(opt="nlminb"))
However, if you look at the fitted model, you'll see you have problems - the estimated AR1 parameter is -1 and the random intercept and slope terms are correlated with r=0.998.
I think the problem is with the nature of the data. Most of the data seem to be in the range 10-50, but there are excursions by one or two orders of magnitude (e.g. individual 6, up to about 45000). It might be hard to fit a model to data this spiky. I would strongly suggest log-transforming your data; the standard diagnostic plot (plot(randomintercept)) looks like this:
whereas fitting on the log scale
rlog <- update(randomintercept,log10(troponin) ~ .)
plot(rlog)
is somewhat more reasonable, although there is still some evidence of heteroscedasticity.
The AR+random-slopes model fits OK:
ar.rlog <- update(rlog,
random = ~day|record_id,
correlation = corAR1(0, form = ~day|record_id))
## Linear mixed-effects model fit by maximum likelihood
## ...
## Random effects:
## Formula: ~day | record_id
## Structure: General positive-definite, Log-Cholesky parametrization
## StdDev Corr
## (Intercept) 0.1772409 (Intr)
## day 0.6045765 0.992
## Residual 0.4771523
##
## Correlation Structure: ARMA(1,0)
## Formula: ~day | record_id
## Parameter estimate(s):
## Phi1
## 0.09181557
## ...
A quick glance at intervals(ar.rlog) shows that the confidence intervals on the autoregressive parameter are (-0.52,0.65), so it may not be worth keeping ...
With the random slopes in the model the heteroscedasticity no longer seems problematic ...
plot(rlog,sqrt(abs(resid(.)))~fitted(.),type=c("p","smooth"))
I am trying to estimate the probability of winning or losing an account, and I'd like to do this using Bayesian Methods. I'm not really that familiar with these methods, but I think I understand the general idea.
I know some information about losses and wins. Wins are usually characterized by some combination of activities; losses are usually characters by a different combination of activities. I'd like to be able to get some posterior probability of whether or not a new observation will be won or lost based on the current number of activities that are associated with that account.
Here is an example of my data: (This is just a sample for simplicity)
Email Call Callback Outcome
14 9 2 1
3 2 4 0
16 14 2 0
15 1 3 1
5 2 2 0
1 1 0 0
10 3 5 0
2 0 1 0
17 8 4 1
3 15 2 0
17 1 3 0
10 7 5 0
10 2 3 0
8 0 0 1
14 10 3 0
1 9 3 1
5 10 3 1
13 5 1 0
9 4 4 0
So from here I know that 30% of the observations have an outcome of 1 (win) and 70% have an outcome of 0 (loss). Let's say that I want to use the other columns to get a probability of win/loss for a new observation which may have a small number of events (emails, calls, and callbacks) associated with it.
Now let's say that I want to use the counts/proportions of the different events as priors for a new observation. This is where I start getting tripped up. My thinking is to create a dirichlet distribution for wins and losses, so two separate distributions, one for wins and one for losses. Using the counts/proportions of events for each outcome as the priors. I guess I'm not sure how to do this in R. I think my course of action would be estimate a dirichlet distribution (since I have 3 variables) for each outcome using maximum likelihood. I've been trying to use the dirichlet.simul and dirichlet.mle functions from the sirt package in R. I'm not sure if I need to simulate one first?
Another issue is once I have this distribution, it's unclear to me how to get a posterior distribution of a new observation. I've read several papers and can't seem to find a straightforward process on how to do this. (Or maybe there's some holes in my understanding). Any pushes in the right direction would be greatly appreciated.
This is the code I've tried so far:
### FOR WON ACCOUNTS
set.seed(789)
N <- 6
probs <- c(0.535714286, 0.330357143, 0.133928571 )
alpha <- probs
alpha <- matrix( alpha , nrow=N , ncol=length(alpha) , byrow=TRUE )
x <- dirichlet.simul( alpha )
dirichlet.mle(x)
$alpha
[1] 0.3385607 0.2617939 0.1972898
$alpha0
[1] 0.7976444
$xsi
[1] 0.4244507 0.3282088 0.2473405
### FOR LOST ACCOUNTS
set.seed(789)
N2 <- 14
probs2 <- c(0.528037383,0.308411215,0.163551402 )
alpha2 <- probs2
alpha2 <- matrix( alpha2 , nrow=N , ncol=length(alpha2) , byrow=TRUE )
x2 <- dirichlet.simul( alpha2 )
dirichlet.mle(x2)
$alpha
[1] 0.3388486 0.2488771 0.2358043
$alpha0
[1] 0.8235301
$xsi
[1] 0.4114587 0.3022077 0.2863336
Not sure if this is a correct approach or how to get posteriors from here. I realize all the outputs look similar across won/lost accounts. I just used some simulated data to represent what I'm working with.
I am looking for multivariate detrending under common trend of a time series data in R.
Time series data sample:
> head(d)
T x1 x2 x3 x4
1 1 2 4 3 1
2 2 3 5 4 4
3 3 6 6 6 6
4 4 8 9 10 7
5 5 10 13 20 9
I would like to detrend the above multivariate time series dataset d under common trend. I hope I am clear in explaining the problem that I am facing.
Thanks!
You can use multivariate regression to solve for constants. Because the betas are the same, (i.e. beta in Y=x*beta is an n by 2 matrix with identical rows), you need to account for that constraint. However, you can just string all the Ys together for this.
dvec=as.numeric(d)
n=dim(d)[1]
ncol=dim(d)[2]
x=rep(1:n,ncol)
model<-lm(dvec~x)
Then you can do
d=matrix(model$residuals,nrow=n)
I'm a PhD student of genetics and I am trying do association analysis of some genetic data using linear regression. In the table below I'm regressing each 'trait' against each 'SNP' There is also a interaction term include as 'var'
I've only used R for 2 weeks and I don't have any programming background so please explain any help provided as I want to understand.
This is a sample of my data:
Sample ID var trait 1 trait 2 trait 3 SNP1 SNP2 SNP3
77856517 2 188 3 2 1 0 0
375689755 8 17 -1 -1 1 -1 -1
392513415 8 28 14 4 1 1 1
393612038 8 85 14 6 1 1 0
401623551 8 152 11 -1 1 0 0
348466144 7 -74 11 6 1 0 0
77852806 4 81 16 6 1 1 0
440614343 8 -93 8 0 0 1 0
77853193 5 3 6 5 1 1 1
and this is the code I've been using for a single regression:
result1 <-lm(trait1~SNP1+var+SNP1*var, na.action=na.exclude)
I want to run a loop where every trait is tested against each SNP.
I've been trying to modify codes I've found online but I always run into some error that I don't understand how to solve.
Thank you for any and all help.
Personally I don't find the problem so easy. Specially for an R novice.
Here a solution based on creating dynamically the regression formula.
The idea is to use paste function to create different formula terms, y~ x + var + x * var then coercing the result string tp a formula using as.formula. Here y and x are the formula dynamic terms: y in c(trait1,trai2,..) and x in c(SNP1,SNP2,...). Of course here I use lapply to loop.
lapply(1:3,function(i){
y <- paste0('trait',i)
x <- paste0('SNP',i)
factor1 <- x
factor2 <- 'var'
factor3 <- paste(x,'var',sep='*')
listfactor <- c(factor1,factor2,factor3)
form <- as.formula(paste(y, "~",paste(listfactor,collapse="+")))
lm(formula = form, data = dat)
})
I hope someone come with easier solution, ore more R-ish one:)
EDIT
Thanks to #DWin comment , we can simplify the formula to just y~x*var since it means y is modeled by x,var and x*var
So the code above will be simplified to :
lapply(1:3,function(i){
y <- paste0('trait',i)
x <- paste0('SNP',i)
LHS <- paste(x,'var',sep='*')
form <- as.formula(paste(y, "~",LHS)
lm(formula = form, data = dat)
})