Performing Anova on Bootstrapped Estimates from Quantile Regression - r

So I'm using the quantreg package in R to conduct quantile regression analyses to test how the effects of my predictors vary across the distribution of my outcome.
FML <- as.formula(outcome ~ VAR + c1 + c2 + c3)
quantiles <- c(0.25, 0.5, 0.75)
q.Result <- list()
for (i in quantiles){
i.no <- which(quantiles==i)
q.Result[[i.no]] <- rq(FML, tau=i, data, method="fn", na.action=na.omit)
}
Then i call anova.rq which runs a Wald test on all the models and outputs a pvalue for each covariate telling me whether the effects of each covariate vary significantly across the distribution of my outcome.
anova.Result <- anova(q.Result[[1]], q.Result[[2]], q.Result[[3]], joint=FALSE)
Thats works just fine. However, for my particular data (and in general?), bootstrapping my estimates and their error is preferable. Which i conduct with a slight modification of the code above.
q.Result <- rqs(FML, tau=quantiles, data, method="fn", na.action=na.omit)
q.Summary <- summary(Q.mod, se="boot", R=10000, bsmethod="mcmb",
covariance=TRUE)
Here's where i get stuck. The quantreg currently cannot peform the anova (Wald) test on boostrapped estimates. The information files on the quantreg packages specifically states that "extensions of the methods to be used in anova.rq should be made" regarding the boostrapping method.
Looking at the details of the anova.rq method. I can see that it requires 2 components not present in the quantile model when bootstrapping.
1) Hinv (Inverse Hessian Matrix). The package information files specifically states "note that for se = "boot" there is no way to split the estimated covariance matrix into its sandwich constituent parts."
2) J which, according to the information files, is "Unscaled Outer product of gradient matrix returned if cov=TRUE and se != "iid". The Huber sandwich is cov = tau (1-tau) Hinv %*% J %*% Hinv. as for the Hinv component, there is no J component when se == "boot". (Note that to make the Huber sandwich you need to add the tau (1-tau) mayonnaise yourself.)"
Can i calculate or estimate Hinv and J from the bootstrapped estimates? If not what is the best way to proceed?
Any help on this much appreciated. This my first timing posting a question here, though I've greatly benefited from the answers to other peoples questions in the past.

For question 2: You can use R = for resampling. For example:
anova(object, ..., test = "Wald", joint = TRUE, score =
"tau", se = "nid", R = 10000, trim = NULL)
Where R is the number of resampling replications for the anowar form of the test, used to estimate the reference distribution for the test statistic.
Just a heads up, you'll probably get a better response to your questions if you only include 1 question per post.

Consulted with a colleague, and he confirmed that it was unlikely that Hinv and J could be 'reverse' computed from bootstrapped estimates. However we resolved that estimates from different taus could be compared using Wald test as follows.
From object rqs produced by
q.Summary <- summary(Q.mod, se="boot", R=10000, bsmethod="mcmb", covariance=TRUE)
you extract the bootstrapped Beta values for variable of interest in this case VAR, the first covariate in FML for each tau
boot.Bs <- sapply(q.Summary, function (x) x[["B"]][,2])
B0 <- coef(summary(lm(FML, data)))[2,1] # Extract liner estimate data linear estimate
Then compute wald statistic and get pvalue with number of quantiles for degrees of freedom
Wald <- sum(apply(boot.Bs, 2, function (x) ((mean(x)-B0)^2)/var(x)))
Pvalue <- pchisq(Wald, ncol(boot.Bs), lower=FALSE)
You also want to verify that bootstrapped Betas are normally distributed, and if you're running many taus it can be cumbersome to check all those QQ plots so just sum them by row
qqnorm(apply(boot.Bs, 1, sum))
qqline(apply(boot.Bs, 1, sum), col = 2)
This seems to be working, and if anyone can think of anything wrong with my solution, please share

Related

Quasi-Poisson mixed-effect model on overdispersed count data from multiple imputed datasets in R

I'm dealing with problems of three parts that I can solve separately, but now I need to solve them together:
extremely skewed, over-dispersed dependent count variable (the number of incidents while doing something),
necessity to include random effects,
lots of missing values -> multiple imputation -> 10 imputed datasets.
To solve the first two parts, I chose a quasi-Poisson mixed-effect model. Since stats::glm isn't able to include random effects properly (or I haven't figured it out) and lme4::glmer doesn't support the quasi-families, I worked with glmer(family = "poisson") and then adjusted the std. errors, z statistics and p-values as recommended here and discussed here. So I basically turn Poisson mixed-effect regression into quasi-Poisson mixed-effect regression "by hand".
This is all good with one dataset. But I have 10 of them.
I roughly understand the procedure of analyzing multiple imputed datasets – 1. imputation, 2. model fitting, 3. pooling results (I'm using mice library). I can do these steps for a Poisson regression but not for a quasi-Poisson mixed-effect regression. Is it even possible to A) pool across models based on a quasi-distribution, B) get residuals from a pooled object (class "mipo")? I'm not sure. Also I'm not sure how to understand the pooled results for mixed models (I miss random effects in the pooled output; although I've found this page which I'm currently trying to go through).
Can I get some help, please? Any suggestions on how to complete the analysis (addressing all three issues above) would be highly appreciated.
Example of data is here (repre_d_v1 and repre_all_data are stored in there) and below is a crucial part of my code.
library(dplyr); library(tidyr); library(tidyverse); library(lme4); library(broom.mixed); library(mice)
# please download "qP_data.RData" from the last link above and load them
## ===========================================================================================
# quasi-Poisson mixed model from single data set (this is OK)
# first run Poisson regression on df "repre_d_v1", then turn it into quasi-Poisson
modelSingle = glmer(Y ~ Gender + Age + Xi + Age:Xi + (1|Country) + (1|Participant_ID),
family = "poisson",
data = repre_d_v1)
# I know there are some warnings but it's because I share only a modified subset of data with you (:
printCoefmat(coef(summary(modelSingle))) # unadjusted coefficient table
# define quasi-likelihood adjustment function
quasi_table = function(model, ctab = coef(summary(model))) {
phi = sum(residuals(model, type = "pearson")^2) / df.residual(model)
qctab = within(as.data.frame(ctab),
{`Std. Error` = `Std. Error`*sqrt(phi)
`z value` = Estimate/`Std. Error`
`Pr(>|z|)` = 2*pnorm(abs(`z value`), lower.tail = FALSE)
})
return(qctab)
}
printCoefmat(quasi_table(modelSingle)) # done, makes sense
## ===========================================================================================
# now let's work with more than one data set
# object "repre_all_data" of class "mids" contains 10 imputed data sets
# fit model using with() function, then pool()
modelMultiple = with(data = repre_all_data,
expr = glmer(Y ~ Gender + Age + Xi + Age:Xi + (1|Country) + (1|Participant_ID),
family = "poisson"))
summary(pool(modelMultiple)) # class "mipo" ("mipo.summary")
# this has quite similar structure as coef(summary(someGLM))
# but I don't see where are the random effects?
# and more importantly, I wanted a quasi-Poisson model, not just Poisson model...
# ...but here it is not possible to use quasi_table function (defined earlier)...
# ...and that's because I can't compute "phi"
This seems reasonable, with the caveat that I'm only thinking about the computation, not whether this makes statistical sense. What I'm doing here is computing the dispersion for each of the individual fits and then applying it to the summary table, using a variant of the machinery that you posted above.
## compute dispersion values
phivec <- vapply(modelMultiple$analyses,
function(model) sum(residuals(model, type = "pearson")^2) / df.residual(model),
FUN.VALUE = numeric(1))
phi_mean <- mean(phivec)
ss <- summary(pool(modelMultiple)) # class "mipo" ("mipo.summary")
## adjust
qctab <- within(as.data.frame(ss),
{ std.error <- std.error*sqrt(phi_mean)
statistic <- estimate/std.error
p.value <- 2*pnorm(abs(statistic), lower.tail = FALSE)
})
The results look weird (dispersion < 1, all model results identical), but I'm assuming that's because you gave us a weird subset as a reproducible example ...

Standard Error of the Regression for NLS Model

I am currently working on a non-linear analysis of various datasets using nls model. On the other hand, I want to calculate the standard error of the regression of the nls model.
The formula of the standard error of regression:
n <- nrow(na.omit((data))
SE = (sqrt(sum(pv-av)^2)/(n-2))
where pv is the predicted value and av is the actual value.
I have a problem on calculating the standard error. Should I calculate the predicted value and actual value first? Are the values based on the dataset? Any help is highly appreciated. Thank You.
R provides this via sigma:
fm <- nls(demand ~ a + b * Time, BOD, start = list(a = 1, b = 1))
sigma(fm)
## [1] 3.085016
This would also work where deviance gives residual sum of squares.
sqrt(deviance(fm) / (nobs(fm) - length(coef(fm))))
## [1] 3.085016

Quick way to calculate a confidence interval after changing dispersion parameter

I'm teaching a modeling class in R. The students are all SAS users, and I have to create course materials that exactly match (when possible) SAS output. I'm working on the Poisson regression section and trying to match PROC GENMOD, with a "dscale" option that modifies the dispersion index so that the deviance/df==1.
Easy enough to do, but I need confidence intervals. I'd like to show the students how to do it without hand calculating them. Something akin to confint_default() or confint()
Data
skin_cancer <- data.frame(CASES=c(1,16,30,71,102,130,133,40,4,38,
119,221,259,310,226,65),
CITY=c(rep(0,8),rep(1,8)),
N=c(172875, 123065,96216,92051,72159,54722,
32185,8328,181343,146207,121374,111353,
83004,55932,29007,7583),
agegp=c(1:8,1:8))
skin_cancer$ln_n = log(skin_cancer$N)
The model
fit <- glm(CASES ~ CITY, family="poisson", offset=ln_n, data=skin_cancer)
Changing the dispersion index
summary(fit, dispersion= deviance(fit) / df.residual(fit)))
That gets me the "correct" standard errors (correct according to SAS). But obviously I can't run confint() on a summary() object.
Any ideas? Bonus points if you can tell me how to change the dispersion index within the model so I don't have to do it within the summary() call.
Thanks.
This is an interesting question, and slightly deeper than it seems.
The simplest potential answer is to use family="quasipoisson" instead of poisson:
fitQ <- update(fit, family="quasipoisson")
confint(fitQ)
However, this won't let you adjust the dispersion to be whatever you want; it specifically changes the dispersion to the estimate R calculates in summary.glm, which is based on the Pearson chi-squared (sum of squared Pearson residuals) rather than the deviance, i.e.
sum((object$weights * object$residuals^2)[object$weights > 0])/df.r
You should be aware that stats:::confint.glm() (which actually uses MASS:::confint.glm) computes profile confidence intervals rather than Wald confidence intervals (i.e., this is not just a matter of adjusting the standard deviations).
If you're satisfied with Wald confidence intervals (which are generally less accurate) you could hack stats::confint.default() as follows (note that the dispersion title is a little bit misleading, as this function basically assumes that the original dispersion of the model is fixed to 1: this won't work as expected if you use a model that estimates dispersion).
confint_wald_glm <- function(object, parm, level=0.95, dispersion=NULL) {
cf <- coef(object)
pnames <- names(cf)
if (missing(parm))
parm <- pnames
else if (is.numeric(parm))
parm <- pnames[parm]
a <- (1 - level)/2
a <- c(a, 1 - a)
pct <- stats:::format.perc(a, 3)
fac <- qnorm(a)
ci <- array(NA, dim = c(length(parm), 2L), dimnames = list(parm,
pct))
ses <- sqrt(diag(vcov(object)))[parm]
if (!is.null(dispersion)) ses <- sqrt(dispersion)*ses
ci[] <- cf[parm] + ses %o% fac
ci
}
confint_wald_glm(fit)
confint_wald_glm(fit,dispersion=2)

Wald Testing Bootstrapped Estimates in R

I've performed multiple regression (specifically quantile regression with multiple predictors using quantreg in R). I have estimated the standard error and confidence intervals based on bootstrapping the estimates. Now i want to test whether the estimates at different quantiles differ significantly from one another (Wald test would be preferable). How can i do this?
FML <- as.formula(outcome ~ VAR + c1 + c2 + c3)
quantiles <- c(0.25, 0.5, 0.75)
q.Result <- rqs(FML, tau=quantiles, data, method="fn", na.action=na.omit)
q.Summary <- summary(Q.mod, se="boot", R=10000, bsmethod="mcmb",
covariance=TRUE)
From q.Summary i've extracted the bootstrapped (ie 10000) estimates (ie vector of 10000 bootstrapped B values).
Note: In reality I'm not especially interested comparing the estimates from all my covariates (in FML), I'm primarily interested comparing the estimates for VAR. What is the best way to proceed?
Consulted with a colleague, and we resolved that estimates from different taus could be compared using Wald test as follows.
From object rqs produced by
q.Summary <- summary(Q.mod, se="boot", R=10000, bsmethod="mcmb", covariance=TRUE)
you extract the bootstrapped Beta values for variable of interest in this case VAR, the first covariate in FML for each tau
boot.Bs <- sapply(q.Summary, function (x) x[["B"]][,2])
B0 <- coef(summary(lm(FML, data)))[2,1] # Extract liner estimate data linear estimate
Then compute wald statistic and get pvalue with number of quantiles for degrees of freedom
Wald <- sum(apply(boot.Bs, 2, function (x) ((mean(x)-B0)^2)/var(x)))
Pvalue <- pchisq(Wald, ncol(boot.Bs), lower=FALSE)
You also want to verify that bootstrapped Betas are normally distributed, and if you're running many taus it can be cumbersome to check all those QQ plots so just sum them by row
qqnorm(apply(boot.Bs, 1, sum))
qqline(apply(boot.Bs, 1, sum), col = 2)
This seems to be working, and if anyone can think of anything wrong with my solution, please share

Finding critical values for the Pearson correlation coefficient

I'd like to use R to find the critical values for the Pearson correlation coefficient.
This has proved difficult to find in search engines since the standard variable for the Pearson correlation coefficient is itself r. In turn, I'm finding a lot of r critical value tables (rather than how to find this by using the statistical package R).
I'm looking for a function that will provide output like the following:
I'm comfortable finding the correlation with:
cor(x,y)
However, I'd also like to find the critical values.
Is there a function I can use to enter n (or degrees of freedom) as well as alpha in order to find the critical value?
The significance of a correlation coefficient, r, is determined by converting r to a t-statistic and then finding the significance of that t-value at the degrees of freedom that correspond to the sample size, n. So, you can use R to find the critical t-value and then convert that value back to a correlation coefficient to find the critical correlation coefficient.
critical.r <- function( n, alpha = .05 ) {
df <- n - 2
critical.t <- qt(alpha/2, df, lower.tail = F)
critical.r <- sqrt( (critical.t^2) / ( (critical.t^2) + df ) )
return(critical.r)
}
# Example usage: Critical correlation coefficient at sample size of n = 100
critical.r( 100 )
The general structure of hypothesis testing is kind of a mish-mash of two systems: Fisherian and Neyman-Pearson. Statisticians understand the differences but rarely does this get clearly presented in undergraduate stats classes. R was designed by and intended for statisticians as a toolbox, so they constructed a function named cor.test that will deliver a p-value (part of the Fisherian tradition) as well as a confidence interval for "r" (derived on the basis of the Neyman-Pearson formalism.) Fisher and Neyman had bitter disputes in their lifetime. The "critical value" terminology is part of the N-P testing strategy. It is equivalent to building a confidence interval and finding the particular statistic that reaches exactly a threshold value of 0.05 significance.
The code for constructing the inferential statistics in cor.test is available with:
methods(cor.test)
getAnywhere(cor.test.default)
# scroll down
method <- "Pearson's product-moment correlation"
#-----partial code----
r <- cor(x, y)
df <- n - 2L
ESTIMATE <- c(cor = r)
PARAMETER <- c(df = df)
STATISTIC <- c(t = sqrt(df) * r/sqrt(1 - r^2))
p <- pt(STATISTIC, df)
# ---- omitted some set up and error checking ----
# this is the confidence interval section------
z <- atanh(r)
sigma <- 1/sqrt(n - 3)
cint <- switch(alternative, less = c(-Inf, z + sigma *
qnorm(conf.level)), greater = c(z - sigma * qnorm(conf.level),
Inf), two.sided = z + c(-1, 1) * sigma * qnorm((1 +
conf.level)/2))
cint <- tanh(cint)
So now you know how R does it. Notice that there is no "critical value" mentioned. I suspect that your hope was to find some table where a tabulation of "r" and "df" was laid out displaying the minimum "r" that would reach a significance of 0.05 for a given 'df'. Such a table could be built but that's not how this particular toolbox is constructed. You should now have the tools to build it yourself.
I would do the same. But if you are using a Spearman correlation you need to convert t into r using a different formula.
just change the last line before the return in the function with this one:
critical.r <- sqrt(((critical.t^2) / (df)) + 1)

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