Solved – Generating Multivariate Uniform Distribution in R

multivariate distributionrrandom-generation

Let $d$ a positive integer How to generate a sample of $n$ random variables with a multivariate uniform distribution on the cube $[a,b]^d$ in R?

I don't know what to do. I know that I need a covariance matrix for the random vectors that I will generate…

Best Answer

It depends a little bit on the terminology, but usually multivariate uniform refers to a distribution where every point in $[a,b]^d$ is equally likely. Hence, the dimensions are independent, and you can draw uniformly between $[a,b]$ d times individually to get a sample from the multivariate uniform.

If you don't want the dimensions to be independent, it might be worth looking into Copulas

Related Solutions

Multivariate Normal Distribution – Generating Values from It

If $X \sim \mathcal{N}(0,I)$ is a column vector of standard normal RV's, then if you set $Y = L X$, the covariance of $Y$ is $L L^T$.

I think the problem you're having may arise from the fact that matlab's mvnrnd function returns row vectors as samples, even if you specify the mean as a column vector. e.g.,

 > size(mvnrnd(ones(10,1),eye(10))  
 > ans =
 >      1    10

And note that transforming a row vector gives you the opposite formula. if $X$ is a row vector, then $Z = X L^T$ is also a row vector, so $Z^T = L X^T$ is a column vector, and the covariance of $Z^T$ can be written $E[Z^TZ] = LL^T$.

Based on what you wrote though, the Wikipedia formula is correct: if $\Phi^{-1}(U)$ were a row vector returned by matlab, you can't left-multiply it by $L^T$. (But right-multiplying by $L^T$ would give you a sample with the same covariance of $LL^T$).

Solved – Constraints on generating valid covariance matrices for multivariate normal distributions

Yes, your constraints are correct (though I wouldn't really call the first one a constraint, necessarily...).

One way to generate symmetric positive semi-definite matrices is as follows: if $X \in \mathbb R^{n \times p}$, then $X X^T \in \mathbb R^{n \times n}$ is psd: $(X X^T)^T = X X^T$ so that it's symmetric, and $z^T X X^T z = \lVert X^T z \rVert^2 \ge 0$. So any way that you choose $X$ will give you a valid covariance $X X^T$, and indeed any psd matrix can be decomposed in such a way (e.g. the Cholesky decomposition).

If you choose the entries of $X$ to be iid normal, for example, you get a Wishart distribution.

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Related Solutions

Multivariate Normal Distribution – Generating Values from It

Solved – Constraints on generating valid covariance matrices for multivariate normal distributions

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