How to generate bivariate random normally distributed 3d array?
5 views (last 30 days)
Show older comments
Nainsi Gupta
on 31 May 2023
Commented: Nainsi Gupta
on 1 Jun 2023
mu= [0 0]
sigma= [1 0.25; .25 1]
mvnrnd(mu,sigma,100)
I want to generate a 3-by-2-by-100 array which is normally distributed with given mu and sigma. Please help me how can I do this?
5 Comments
John D'Errico
on 31 May 2023
Since the realizations of such a random variable are i.i.d., it is sort of irrelevant what the 1st and third diemsnions mean!
Accepted Answer
John D'Errico
on 31 May 2023
Edited: John D'Errico
on 31 May 2023
Um, trivial?
You apparently want to generate 300 samples of a bivariate normal. So generate them as a 300x2 array, Then reshape and permute them into the desired 3x2x100 array.
mu= [0 0];
sigma= [1 0.25; .25 1];
X = mvnrnd(mu,sigma,300);
X = reshape(X,[3,100,2]);
X = permute(X,[1 3 2]);
size(X)
3 Comments
John D'Errico
on 31 May 2023
Permute is like transpose. But it applies to arrays of multiple dimensions.
Look at what I did.
mu= [0 0];
sigma= [1 0.25; .25 1];
X = mvnrnd(mu,sigma,300);
What is the initial size of X?
size(X)
I generated 300 samples of bivariate random numbers with the desired distribution. What did reshape do?
X = reshape(X,[3,100,2]);
size(X)
So I simply reshaped that into an array of 3x100 instances of the same random numbers. FInally, the call to permute does nothing more than transpose the second and third dimensions.
X = permute(X,[1 3 2]);
size(X)
Again, permute is just like transpose. It allows you to transpose any dimensions you wish. It applies to 3-d and higher dimension arrays.
You will do exactly the same thing, IF you wanted to generate trivariate random numbers. But now you will have a 3 instead of a 2 in those places in the reshape and permute.
More Answers (0)
See Also
Categories
Find more on Random Number Generation in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!