Question

I need to generate a loop for 1,000 Monte Carlo iterations in R. I have two randomly generated standard normal variables for this with sample sizes of 100.

Answer 1

If you don't insist on using the MonteCarlo package (which is new to me and looks interesting) you could wrap your code into a function.

First, creating a random matrix with only the desired 90%-row. Then, after specifying the amount, you can do the iterations and pack them into a list with lapply(). At the end just create an other matrix with your (rounded) statistics.

Something like this:

set.seed(54897)  # setting seed for sake of reproducibility
probability.sequence <- seq(0.1, 0.90, 0.1)

# iteration function
mx <- function(X, n) {
  random.sample <- sample(1:1e2, 1e3, replace = TRUE) 
  a = matrix(random.sample, nrow = 10, ncol = 10, byrow = TRUE)  
  b <- apply(a[ ,1:10], 2, SMA, n = 3)
  c <- pmax(a, b)
  d <- apply(c, 2, quantile, probs = probability.sequence,  na.rm = TRUE)
  return(d[9, ])  # returns only the desired 90 % row
}

# amount of iterations
amount <- 1e3 

# iteration
mx.lst <- lapply(1:amount, mx)

# matrix for statistics
r = matrix(NA, nrow = 1, ncol = 4, 
           dimnames = list(NULL, c("mean", "variance", "Upper CL", "Lower CL")))
r[, 1] <- (mean(unlist(lapply(mx.lst, mean))))  # mean
r[, 2]  <- sqrt(var(unlist(lapply(mx.lst, mean))))  # variance
r[, 3]  <- r[, 1] - 1.96 * sqrt(r[, 2])  # lower CI 95 %
r[, 4]  <- r[, 1] + 1.96 * sqrt(r[, 2])  # upper CI 95 %

# output
(monte.carlo.aggregate.results <- round(r, 0))
#      mean variance Upper CL Lower CL
# [1,]   82        3       78       86

str(monte.carlo.aggregate.results)
# num 1, 1:4] 82 3 78 86
# - attr(*, "dimnames")=List of 2
# ..$ : NULL
# ..$ : chr [1:4] "mean" "variance" "Upper CL" "Lower CL"

# speed test
amount <- 1e4
system.time(mx.lst <- lapply(1:amount, mx))
#     user      system     elapsed 
#    48.21        0.00       48.31 

Note: please check yourself if the used formula for the confidence interval fits to your needs.