Ted,
Thanks for your help, it is right on the money!
for your comments:
1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1.
2. The correlation is the only free parameter.
Michael
On Sat, Nov 6, 2010 at 7:07 PM, Ted Harding <ted.hard...@wlandres.net>wrote:
> On 06-Nov-10 21:54:41, michael wrote:
> > I wish to generate 100 by 1 vector of x1 and x2 both are uniform
> > distributed with covariance matrix \Sigma.
> >
> > Thanks,
> > Michael
>
> First, some comments.
>
> 1. I don't think you mean a "100 by 1 vector of x1 and x2" since
> you have two variables. "100 by 2" perhaps?
>
> 2. Given the range over which x1 is to be uniformly distributed
> (say of length A) and the range for x2 (say of length B),
> then by virtue of the uniform distribution the variance of x1
> will be (A^2)/12 and the variance of x2 will be (B^2)/12,
> so you don't have a choice about these. So your only "free
> parameter" is the covariance (or the correlation) between
> x1 and x2.
>
> That said, it is a curious little problem. Here is one simple
> solution (though it may have properties that you do not want).
> Using X and Y for x1 and x2 (for simplicity), and using ranges
> (-1/2,1/2) for the ranges of X and Y (you could rescale later):
>
> 1. Generate X uniformly distributed on (-1/2,1/2).
> 2. With probability p (0 < p < 1) let Y=X.
> Otherwise, let Y be independently uniform on (-1/2,1/2).
>
> This can be implemented by the following code:
>
> n <- 100
> p <- 0.6 # (say)
> X <- runif(n)-1/2
> Z <- runif(n)-1/2
> U <- rbinom(n,1,p) # =1 where Y=X, = 0 when Y independent of X
> Y <- X*U + Z*(1-U)
> XY <- cbind(X,Y) # your n by 2 matrix of bivariate (X,Y)
>
> Now the marginal distributions of X and Y are both uniform.
> var(X) = 1/12 and var(Y) = 1/12. Since the means are 0,
> cov(X,Y) = Exp(X*Y) = p*Exp(X^2) = p/12
> cor(X,Y) = cov(X,Y)/sqrt(var(X)*var(Y)) = (p/12)/(1/12) = p.
>
> So all you need to do to get a desired correlation rho between
> X and Y is to set p = rho.
>
> The one thing you may not like about this is the diagonal line
> you will get for the cases where X=Y:
>
> plot(X,Y)
>
> Test:
>
> n <- 100000
> p <- 0.6 # (say)
> X <- runif(n)-1/2
> Z <- runif(n)-1/2
> U <- rbinom(n,1,p)
> Y <- X*U + Z*(1-U)
> var(X)
> # [1] 0.08340525 # theory: var = 1/12 = 0.08333333
> > var(Y)
> [1] 0.08318914 # theory: var = 1/12 = 0.08333333
> > cov(X,Y)
> [1] 0.04953733 # theory: cov = p/12 = 0.6/12 = 0.05
> > cor(X,Y)
> [1] 0.5947063 # theory: cor = p = 0.6
>
> It would be interesting to see a solution which did not involve
> having cases with X=Y!
>
> Hoping this helps,
> Ted.
>
> --------------------------------------------------------------------
> E-Mail: (Ted Harding) <ted.hard...@wlandres.net>
> Fax-to-email: +44 (0)870 094 0861
> Date: 06-Nov-10 Time: 23:07:26
> ------------------------------ XFMail ------------------------------
>
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