How to find the correlation between two random numbers

Hi,

I want to give a specific correlation between two random numbers. I am using them to find the inverse lognormal distribution, so they have to remain between 0 and 1. How can I do this?

script:

N = 10;
ro = 0.75;         %correlation between random numbers
f = rand(N,2);
mu7 = 0.84208;
su7 = 0.3852;
my7 = 0.688;
sy7 = 0.09975;
j = f(:,1);
g = f(:,2);

epy = logninv(j,my7,sy7);                                    
epu_y = logninv(g,mu7,su7);                                   
epu = epu_y + epy;

Thanks!

Answer 1

[Greig]

If you have two uncorrelated normally distributed random numbers, given by x, you can use the following to determine Y, which will be correlated with x(:,1)....

x=randn(1e3,2);
R0 = 0.75;
corr(x(:,1), x(:,2))
Y=R0*x(:,1)+(sqrt(1-R0)*x(:,2));
corr(x(:,1), Y)

It also appears to work well if x is from a uniform distribution, but the returned results are no longer uniformly distributed.

x=rand(1e3,2);
R0 = 0.75;
corr(x(:,1), x(:,2))
Y=R0*x(:,1)+(sqrt(1-R0)*x(:,2));
corr(x(:,1), Y)

It should be noted that the real correlation between x(:,1) and Y can vary significantly from R0 if N is small, also it works less well if x(:,1) and x(:,2) are coincidently correlated.

There are ways of doing this with Cholesky and eignevector decomposition, but I can't remember them off the top of my head.

UPDATED:

As John pointed out, the about solution does not maintain the uniformity of the random numbers after adjusted for correlations. I have attached a function that I sometimes use for this purpose, which does return uniform results

This is based on an example script provided here http://comisef.wikidot.com/tutorial:correlateduniformvariates

[John D'Errico]

Except that the result fails for uniform variates in the sense that they will not remain bounded in the interval [0,1].

[Greig]

That is true, so the approach is only valid for random samples from unbounded distributions. Given that the problem is dealing with logninv, this might throw up some errors.

I have some scripts for this sort of thing that I can check in when I'm back at work in the morning.

[Greig]

The answer has been updated with a script that returns uniformly distributed results