I want to create a random sample of length n from a normal multivariate distribution

Question

I have a problem in understanding this question, especially this part:

"generate a random sample of length n from a normal multivariate"

This is what I have done using the R package mvtnorm:

my_function <- function(n=1,k){
mean = rep(0,k)
sigma = diag(length(mean))
rmvnorm(n, mean, sigma,method=c("eigen", "svd", "chol"), pre0.9_9994 = FALSE)}

my_function(3)

This way the output of my function is, in this example, a vector of three numbers which in my mind is A sample of length 3. Is this correct? Is there another way to do this?

I am asking this question because I have to do the same without using the library rmvnorm and with a normal bivariate.

I just managed to write the density function:

dbivnorm <- function(x,y,mux=0,muy=0,sigmax=1,sigmay=1,rho=0){
  (2*pi)^(-1) * ((1-rho^2)*sigmax^2*sigmay^2)^(-.5) *
    exp( -((x-mux)^2/(sigmax^2) -2*rho*((x-mux)/sigmax *(y-muy)/sigmay) +
             (y-muy)^2/(sigmay^2))/(2*(1-rho^2)) )
}

But now I am not sure how to proceed.

Answer 1

Usually, a good starting point to understand how to run an R function is to read the attached documentation:

which also contains an example as an illustration

where the entry n in rmvnorm(n,mean,sigma) is the number of elements in the random sample, while mean is the mean vector and sigma is the variance-covariance matrix.

Answer 2

Are you allowed to use a package that provides you with independent normal distributed RVs? If not, the Box-Muller method (https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform) could be used for that.

Given that you have n independent normal distributed random variables, using the cholesky decomposition $L^\top L = \Sigma$ (for the wanted covariance matrix $\Sigma$) and multiplying $L$ by the vector of independent normal RVs gives you a multivariate normal distributed random variable with covariance $\Sigma$.