Mixing
I want to create a random sample of length n from a normal multivariate distribution
$begingroup$
This may sound like a stupid question, but 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)}
using_rmvnorm(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.
statistics
asked Jan 8 at 15:36
qcc101qcc101
597213
$endgroup$
add a comment |
$begingroup$
This may sound like a stupid question, but 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)}
using_rmvnorm(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.
statistics
asked Jan 8 at 15:36
qcc101qcc101
597213
$endgroup$
$begingroup$
If you don't get good responses here, you could try Stats SE or SO.
$endgroup$
– user21820
Jan 12 at 9:54
add a comment |
$begingroup$
This may sound like a stupid question, but 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)}
using_rmvnorm(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.
statistics
asked Jan 8 at 15:36
qcc101qcc101
597213
$endgroup$
This may sound like a stupid question, but 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)}
using_rmvnorm(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.
statistics
statistics
asked Jan 8 at 15:36
qcc101qcc101
597213
asked Jan 8 at 15:36
qcc101qcc101
597213
asked Jan 8 at 15:36
qcc101qcc101
597213
asked Jan 8 at 15:36
qcc101qcc101
597213
asked Jan 8 at 15:36
qcc101qcc101
597213
597213
$begingroup$
If you don't get good responses here, you could try Stats SE or SO.
$endgroup$
– user21820
Jan 12 at 9:54
add a comment |
$begingroup$
If you don't get good responses here, you could try Stats SE or SO.
$endgroup$
– user21820
Jan 12 at 9:54
$begingroup$
If you don't get good responses here, you could try Stats SE or SO.
$endgroup$
– user21820
Jan 12 at 9:54
$begingroup$
If you don't get good responses here, you could try Stats SE or SO.
$endgroup$
– user21820
Jan 12 at 9:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In Mathematica:
n = 4;
RandomVariate[
MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In Mathematica:
n = 4;
RandomVariate[
MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
add a comment |
$begingroup$
In Mathematica:
n = 4;
RandomVariate[
MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
add a comment |
$begingroup$
In Mathematica:
n = 4;
RandomVariate[
MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
In Mathematica:
n = 4;
RandomVariate[
MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
answered Jan 8 at 15:41
David G. StorkDavid G. Stork
10.9k31432
10.9k31432
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
add a comment |
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
$begingroup$
So in your case n = 4 is the length of the sample and 3 is the number of samples you wanted to generate?
$endgroup$
– qcc101
Jan 8 at 16:02
add a comment |
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$begingroup$
If you don't get good responses here, you could try Stats SE or SO.
$endgroup$
– user21820
Jan 12 at 9:54