Question

Is there a single function, similar to “runif”, “rnorm” and the like which will produce simulated predictions for a linear model? I can code it on my own, but the code is ugly and I assume that this is something someone has done before.

slope = 1.5
intercept = 0
x = as.numeric(1:10)
e = rnorm(10, mean=0, sd = 1)
y = slope * x + intercept + e
fit = lm(y ~ x, data = df)
newX = data.frame(x = as.numeric(11:15))

What I’m interested in is a function that looks like the line below:

sims = rlm(1000, fit, newX)

That function would return 1000 simulations of y values, based on the new x variables.

Answer 1

Showing that Gavin Simpson’s suggestion of modifying stats:::simulate.lm is a viable one.

## Modify stats:::simulate.lm by inserting some tracing code immediately
## following the line that reads "ftd <- fitted(object)" 
trace(what = stats:::simulate.lm,
      tracer = quote(ftd <- list(...)[["XX"]]),
      at = list(6))

## Prepare the data and 'fit' object 
df <- data.frame(x =x<-1:10, y = 1.5*x + rnorm(length(x)))
fit <- lm(y ~ x, data = df)

## Define new covariate values and compute their predicted/fitted values
newX <- 8:1
newFitted <- predict(fit, newdata = data.frame(x = newX))

## Pass in fitted via the argument 'XX'
simulate(fit, nsim = 4, XX = newFitted)
#        sim_1     sim_2       sim_3     sim_4
# 1 11.0910257 11.018211 10.95988582 13.398902
# 2 12.3802903 10.589807 10.54324607 11.728212
# 3  8.0546746  9.925670  8.14115433  9.039556
# 4  6.4511230  8.136040  7.59675948  7.892622
# 5  6.2333459  3.131931  5.63671024  7.645412
# 6  3.7449859  4.686575  3.45079655  5.324567
# 7  2.9204519  3.417646  2.05988078  4.453807
# 8 -0.5781599 -1.799643 -0.06848592  0.926204

That works, but this is a cleaner (and likely better) approach:

## A function for simulating at new x-values
simulateX <- function(object, nsim = 1, seed = NULL, X, ...) {
    object$fitted.values <- predict(object, X)
    simulate(object = object, nsim = nsim, seed = seed, ...)
}
    
## Prepare example data and a fit object
df <- data.frame(x =x<-1:10, y = 1.5*x + rnorm(length(x)))
fit <- lm(y ~ x, data = df)

## Supply new x-values in a data.frame of the form expected by
## the newdata= argument of predict.lm()
newX <- data.frame(x = 8:1)
   
## Try it out
simulateX(fit,  nsim = 4, X = newX)
#       sim_1     sim_2     sim_3     sim_4
# 1 11.485024 11.901787 10.483908 10.818793
# 2 10.990132 11.053870  9.181760 10.599413
# 3  7.899568  9.495389 10.097445  8.544523
# 4  8.259909  7.195572  6.882878  7.580064
# 5  5.542428  6.574177  4.986223  6.289376
# 6  5.622131  6.341748  4.929637  4.545572
# 7  3.277023  2.868446  4.119017  2.609147
# 8  1.296182  1.607852  1.999305  2.598428