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












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you don't get good responses here, you could try Stats SE or SO.
    $endgroup$
    – user21820
    Jan 12 at 9:54
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you don't get good responses here, you could try Stats SE or SO.
    $endgroup$
    – user21820
    Jan 12 at 9:54














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

In Mathematica:



n = 4;

RandomVariate[

 MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]





share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

In Mathematica:



n = 4;

RandomVariate[

 MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]





share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

In Mathematica:



n = 4;

RandomVariate[

 MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]





share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

In Mathematica:



n = 4;

RandomVariate[

 MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]





share|cite|improve this answer









$endgroup$



In Mathematica:



n = 4;

RandomVariate[

 MultinormalDistribution[ConstantArray[0, n], IdentityMatrix[n]], 3]






share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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