Simulating logistic regression from saved estimates in R - r

I have a bit of an issue. I am trying to develop some code that will allow me to do the following: 1) run a logistic regression analysis, 2) extract the estimates from the logistic regression analysis, and 3) use those estimates to create another logistic regression formula that I can use in a subsequent simulation of the original model. As I am, relatively new to R, I understand I can extract these coefficients 1-by-1 through indexing, but it is difficult to "scale" this to models with different numbers of coefficients. I am wondering if there is a better way to extract the coefficients and setup the formula. Then, I would have to develop the actual variables, but the development of these variables would have to be flexible enough for any number of variables and distributions. This appears to be easily done in Mplus (example 12.7 in the Mplus manual), but I haven't figured this out in R. Here is the code for as far as I have gotten:
#generating the data
set.seed(1)
gender <- sample(c(0,1), size = 100, replace = TRUE)
age <- round(runif(100, 18, 80))
xb <- -9 + 3.5*gender + 0.2*age
p <- 1/(1 + exp(-xb))
y <- rbinom(n = 100, size = 1, prob = p)
#grabbing the coefficients from the logistic regression model
matrix_coef <- summary(glm(y ~ gender + age, family = "binomial"))$coefficients
the_estimates <- matrix_coef[,1]
the_estimates
the_estimates[1]
the_estimates[2]
the_estimates[3]
I just cannot seem to figure out how to have R create the formula with the variables (x's) and the coefficients from the original model in a flexible manner to accommodate any number of variables and different distributions. This is not class assignment, but a necessary piece for the research that I am producing. Any help will be greatly appreciated, and please, treat this as a teaching moment. I really want to learn this.

I'm not 100% sure what your question is here.
If you want to simulate new data from the same model with the same predictor variables, you can use the simulate() method:
dd <- data.frame(y, gender, age)
## best practice when modeling in R: take the variables from a data frame
model <- glm(y ~ gender + age, data = dd, family = "binomial")
simulate(model)
You can create multiple replicates by specifying the nsim= argument (or you can simulate anew every time through a for() loop)
If you want to simulate new data from a different set of predictor variables, you have to do a little bit more work (some model types in R have a newdata= argument, but not GLMs alas):
## simulate new model matrix (including intercept)
simdat <- cbind(1,
gender = rbinom(100, prob = 0.5, size = 1),
age = sample(18:80, size = 100, replace = TRUE))
## extract inverse-link function
invlink <- family(model)$linkinv
## sample new values
resp <- rbinom(n = 100, size = 1, prob = invlink(simdat %*% coef(model)))
If you want to do this later from coefficients that have been stored, substitute the retrieved coefficient vector for coef(model) in the code above.
If you want to flexibly construct formulas, reformulate() is your friend — but I don't see how it fits in here.
If you want to (say) re-fit the model 1000 times to new responses simulated from the original model fit (same coefficients, same predictors: i.e. a parametric bootstrap), you can do something like this.
nsim <- 1000
res <- matrix(NA, ncol = length(coef(model)), nrow = nsim)
for (i in 1:nsim) {
## simulate returns a list (in this case, of length 1);
## extract the response vector
newresp <- simulate(model)[[1]]
newfit <- update(model, newresp ~ .)
res[i,] <- coef(newfit)
}
You don't have to store coefficients - you can extract/compute whatever model summaries you like (change the number of columns of res appropriately).

Let’s say your data matrix including age and gender, or whatever predictors, is X. Then you can use X on the right-hand side of your glm formula, get xb_hat <- X %*% the_estimates (or whatever other data matrix replacing X as long as it has same columns) and plug xb_hat into whatever link function you want.

Related

Lagged Residual as Independent Variable in R

I am building a factor model to estimate future equity returns. I'd like to include an autoregressive residual term in this model. I'd like to have yesterday's error (the difference between yesterday's predicted return and actual return) to be included in the regression as an independent variable. What type of autoregressive model is this called? I've searched through various time series econometrics texts and have not found this particular model described. My current solution in R is to rerun the regression at every discrete time step (t), and manually include yesterday's residual, but I am curious if there is a more efficient method or package that does this.
Below is some sample code without the residual term included:
Data:
# fake data
set.seed(333)
df <- data.frame(seq(as.Date("2017/1/1"), as.Date("2017/2/19"), "days"),
matrix(runif(50*506), nrow = 50, ncol = 506))
names(df) <- c("Date", paste0("var", 1:503), c("mktrf", "smb", "hml"))
Then I store my necessary variables for regression:
1.All the dep var
x = df[,505:507]
2.All the indep var
y <- df[,2:504]
4.Fit all the models
list_models_AR= lapply(y, function(y)
with(x, lm(y ~ mktrf + smb + hml , na.action = na.exclude)))
It’s a ARIMA(0, 0, 1), with regressors model

Obtaining Standardized coefficients from "rstanarm" package in R?

I was wondering if it might be possible (and perhaps recommended) to obtain standardized coefficients from stan_glm() in the rstanarm package? (did not find anything specific in the documentation)
Can I just standardize all variables as in normal regression? (see below)
Example:
library("rstanarm")
fit <- stan_glm(wt ~ vs*gear, data = mtcars)
Standardization:
design <- wt ~ vs*gear
vars <- all.vars(design)
stand.vars <- lapply(mtcars[, vars], scale)
fit <- stan_glm(stand.vars, data = mtcars)
I would not say that it is affirmatively recommended, but I would recommend that you not subtract the sample mean and divide by the sample standard deviation of the outcome because the estimation uncertainty in those two statistics will not be propagated to the posterior distribution.
Standardizing the predictors is more debatable. You can do it, but it makes doing posterior prediction with new data harder because you have to remember to subtract the old means from the new data and divide by the old standard deviations.
The most computationally efficient approach is to leave the variables as they are but specify the non-default argument QR = TRUE, especially if you are not going to modify the default (normal) priors on the coefficients anyway.
You can then standardize the posterior coefficients after-the-fact if standardized coefficients are of interest. To do so, you can do
X <- model.matrix(fit)
sd_X <- apply(X, MARGIN = 2, FUN = sd)[-1]
sd_Y <- apply(posterior_predict(fit), MARGIN = 1, FUN = sd)
beta <- as.matrix(fit)[ , 2:ncol(X), drop = FALSE]
b <- sweep(sweep(beta, MARGIN = 2, STATS = sd_X, FUN = `*`),
MARGIN = 1, STATS = sd_Y, FUN = `/`)
summary(b)
However, standardizing regression coefficients just gives the illusion of comparability across variables and says nothing about how germane a one standard deviation difference is, particularly for dummy variables. If your question is really whether manipulating this predictor or that predictor is going to make a bigger difference on the outcome variable, then simply simulate those manipulations like
PPD_0 <- posterior_predict(fit)
nd <- model.frame(fit)
nd[ , 2] <- nd[ , 2] + 1 # for example
PPD_1 <- posterior_predict(fit, newdata = nd)
summary(c(PPD_1 - PPD_0))
and repeat that process for other manipulations of interest.

Linear Mixed-Effects Models for a big spatial auto-correlated dataset

So, I am working with a big dataset (55965 points). I am trying to run a LME accounting for correlation. But R will return me this
Error: 'sumLenSq := sum(table(groups)^2)' = 3.13208e+09 is too large.
Too large or no groups in your correlation structure?
I can not subset it since I need all the points. My questions are:
Is there some setting I can change in the function?
If not, is there any other package with similar function that would run such a big dataset?
Here is a reproducible example:
require(nlme)
my.data<- matrix(data = 0, nrow = 55965, ncol = 3)
my.data<- as.data.frame(my.data)
dummy <- rep(1, 55965)
my.data$dummy<- dummy
my.data$V1<- seq(780, 56744)
my.data$V2<- seq(1:55965)
my.data$X<- seq(49.708, 56013.708)
my.data$Y<-seq(-12.74094, -55977.7409)
null.model <- lme(fixed = V1~ V2, data = my.data, random = ~ 1 | dummy, method = "ML")
spatial_model <- update(null.model, correlation = corGaus(1, form = ~ X + Y), method = "ML")
Since you have assigned a grouping factor with only one level, there are no groups in the data, which is what the error message reports. If you just want to account for spatial autocorrelation, with no other random effects, use gls from the same package.
Edit: A further note on 2 different approaches to modelling spatial autocorrelation: The corrGauss (and other corrSpatial type functions) implement spatial correlation models for regression residuals, which is different from, say, a spatial random effect added to the model based on county/district/grid identity.

R: obtain coefficients&CI from bootstrapping mixed-effect model results

The working data looks like:
set.seed(1234)
df <- data.frame(y = rnorm(1:30),
fac1 = as.factor(sample(c("A","B","C","D","E"),30, replace = T)),
fac2 = as.factor(sample(c("NY","NC","CA"),30,replace = T)),
x = rnorm(1:30))
The lme model is fitted as:
library(lme4)
mixed <- lmer(y ~ x + (1|fac1) + (1|fac2), data = df)
I used bootMer to run the parametric bootstrapping and I can successfully obtain the coefficients (intercept) and SEs for fixed&random effects:
mixed_boot_sum <- function(data){s <- sigma(data)
c(beta = getME(data, "fixef"), theta = getME(data, "theta"), sigma = s)}
mixed_boot <- bootMer(mixed, FUN = mixed_boot_sum, nsim = 100, type = "parametric", use.u = FALSE)
My first question is how to obtain the coefficients(slope) of each individual levels of the two random effects from the bootstrapping results mixed_boot ?
I have no problem extracting the coefficients(slope) from mixed model by using augment function from broom package, see below:
library(broom)
mixed.coef <- augment(mixed, df)
However, it seems like broom can't deal with boot class object. I can't use above functions directly on mixed_boot.
I also tried to modify the mixed_boot_sum by adding mmList( I thought this would be what I am looking for), but R complains as:
Error in bootMer(mixed, FUN = mixed_boot_sum, nsim = 100, type = "parametric", :
bootMer currently only handles functions that return numeric vectors
Furthermore, is it possible to obtain CI of both fixed&random effects by specifying FUN as well?
Now, I am very confused about the correct specifications for the FUN in order to achieve my needs. Any help regarding to my question would be greatly appreciated!
My first question is how to obtain the coefficients(slope) of each individual levels of the two random effects from the bootstrapping results mixed_boot ?
I'm not sure what you mean by "coefficients(slope) of each individual level". broom::augment(mixed, df) gives the predictions (residuals, etc.) for every observation. If you want the predicted coefficients at each level I would try
mixed_boot_coefs <- function(fit){
unlist(coef(fit))
}
which for the original model gives
mixed_boot_coefs(mixed)
## fac1.(Intercept)1 fac1.(Intercept)2 fac1.(Intercept)3 fac1.(Intercept)4
## -0.4973925 -0.1210432 -0.3260958 0.2645979
## fac1.(Intercept)5 fac1.x1 fac1.x2 fac1.x3
## -0.6288728 0.2187408 0.2187408 0.2187408
## fac1.x4 fac1.x5 fac2.(Intercept)1 fac2.(Intercept)2
## 0.2187408 0.2187408 -0.2617613 -0.2617613
## ...
If you want the resulting object to be more clearly named you can use:
flatten <- function(cc) setNames(unlist(cc),
outer(rownames(cc),colnames(cc),
function(x,y) paste0(y,x)))
mixed_boot_coefs <- function(fit){
unlist(lapply(coef(fit),flatten))
}
When run through bootMer/confint/boot::boot.ci these functions will give confidence intervals for each of these values (note that all of the slopes facW.xZ are identical across groups because the model assumes random variation in the intercept only). In other words, whatever information you know how to extract from a fitted model (conditional modes/BLUPs [ranef], predicted intercepts and slopes for each level of the grouping variable [coef], parameter estimates [fixef, getME], random-effects variances [VarCorr], predictions under specific conditions [predict] ...) can be used in bootMer's FUN argument, as long as you can flatten its structure into a simple numeric vector.

Generating predictive simulations from a multilevel model with random intercepts

I am struggling to understand how, in R, to generate predictive simulations for new data using a multilevel linear regression model with a single set of random intercepts. Following the example on pp. 146-147 of this text, I can execute this task for a simple linear model with no random effects. What I can't wrap my head around is how to extend the set-up to accommodate random intercepts for a factor added to that model.
I'll use iris and some fake data to show where I'm getting stuck. I'll start with a simple linear model:
mod0 <- lm(Sepal.Length ~ Sepal.Width, data = iris)
Now let's use that model to generate 1,000 predictive simulations for 250 new cases. I'll start by making up those cases:
set.seed(20912)
fakeiris <- data.frame(Sepal.Length = rnorm(250, mean(iris$Sepal.Length), sd(iris$Sepal.Length)),
Sepal.Width = rnorm(250, mean(iris$Sepal.Length), sd(iris$Sepal.Length)),
Species = sample(as.character(unique(iris$Species)), 250, replace = TRUE),
stringsAsFactors=FALSE)
Following the example in the aforementioned text, here's what I do to get 1,000 predictive simulations for each of those 250 new cases:
library(arm)
n.sims = 1000 # set number of simulations
n.tilde = nrow(fakeiris) # set number of cases to simulate
X.tilde <- cbind(rep(1, n.tilde), fakeiris[,"Sepal.Width"]) # create matrix of predictors describing those cases; need column of 1s to multiply by intercept
sim.fakeiris <- sim(mod0, n.sims) # draw the simulated coefficients
y.tilde <- array(NA, c(n.sims, n.tilde)) # build an array to hold results
for (s in 1:n.sims) { y.tilde[s,] <- rnorm(n.tilde, X.tilde %*% sim.fakeiris#coef[s,], sim.fakeiris#sigma[s]) } # use matrix multiplication to fill that array
That works fine, and now we can do things like colMeans(y.tilde) to inspect the central tendencies of those simulations, and cor(colMeans(y.tilde), fakeiris$Sepal.Length) to compare them to the (fake) observed values of Sepal.Length.
Now let's try an extension of that simple model in which we assume that the intercept varies across groups of observations --- here, species. I'll use lmer() from the lme4 package to estimate a simple multilevel/hierarchical model that matches that description:
library(lme4)
mod1 <- lmer(Sepal.Length ~ Sepal.Width + (1 | Species), data = iris)
Okay, that works, but now what? I run:
sim.fakeiris.lmer <- sim(mod1, n.sims)
When I use str() to inspect the result, I see that it is an object of class sim.merMod with three components:
#fixedef, a 1,000 x 2 matrix with simulated coefficients for the fixed effects (the intercept and Sepal.Width)
#ranef, a 1,000 x 3 matrix with simulated coefficients for the random effects (the three species)
#sigma, a vector of length 1,000 containing the sigmas associated with each of those simulations
I can't wrap my head around how to extend the matrix construction and multiplication used for the simple linear model to this situation, which adds another dimension. I looked in the text, but I could only find an example (pp. 272-275) for a single case in a single group (here, species). The real-world task I'm aiming to perform involves running simulations like these for 256 new cases (pro football games) evenly distributed across 32 groups (home teams). I'd greatly appreciate any assistance you can offer.
Addendum. Stupidly, I hadn't looked at the details on simulate.merMod() in lme4 before posting this. I have now. It seems like it should do the trick, but when I run simulate(mod0, nsim = 1000, newdata = fakeiris), the result has only 150 rows. The values look sensible, but there are 250 rows (cases) in fakeiris. Where is that 150 coming from?
One possibility is to use the predictInterval function from the merTools package. The package is about to be submitted to CRAN, but the current developmental release is available for download from GitHub,
install.packages("devtools")
devtools::install_github("jknowles/merTools")
To get the median and a 95% credible interval of 100 simulations:
mod1 <- lmer(Sepal.Length ~ Sepal.Width + (1 | Species), data = iris)
out <- predictInterval(mod1, newdata=fakeiris, level=0.95,
n.sims=100, stat="median")
By default, predictInterval includes the residual variation, but you can
turn that feature off with:
out2 <- predictInterval(mod1, newdata=fakeiris, level=0.95,
n.sims=100, stat="median",
include.resid.var=FALSE)
Hope this helps!
This might help: it doesn't use sim(), but instead uses mvrnorm() to draw the new coefficients from the sampling distribution of the fixed-effect parameters, uses a bit of internal machinery (setBeta0) to reassign the internal values of the fixed-effect coefficients. The internal values of the random effect coefficients are automatically resampled by simulate.merMod using the default argument re.form=NA. However, the residual variance is not resampled -- it is held fixed across the simulations, which isn't 100% realistic.
In your use case, you would specify newdata=fakeiris.
library(lme4)
mod1 <- lmer(Sepal.Length ~ Sepal.Width + (1 | Species), data = iris)
simfun <- function(object,n=1,newdata=NULL,...) {
v <- vcov(object)
b <- fixef(object)
betapars <- MASS::mvrnorm(n,mu=b,Sigma=v)
npred <- if (is.null(newdata)) {
length(predict(object))
} else nrow(newdata)
res <- matrix(NA,npred,n)
for (i in 1:n) {
mod1#pp$setBeta0(betapars[i,])
res[,i] <- simulate(mod1,newdata=newdata,...)[[1]]
}
return(res)
}
ss <- simfun(mod1,100)

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