How to find the correlation between two random numbers
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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!
采纳的回答
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
3 个评论
John D'Errico
2015-2-24
Except that the result fails for uniform variates in the sense that they will not remain bounded in the interval [0,1].
Greig
2015-2-24
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
2015-2-25
The answer has been updated with a script that returns uniformly distributed results
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