Date: 2 Apr 1987 02:48:54 EST (Thu) From: Dan Hoey Subject: Re: Randomness To: "Keith F. Lynch" Keith, From: "Keith F. Lynch" Thanks. Lambert Marteens came up with a simpler formula for the constraints. If A,B, and C are three correlations (for instance C12, C23, C13), then A^2 + B^2 + C^2 <= 1 + 2ABC. From this it follows that given an A and a B, C must be at least AB - SQRT((A^2 - 1) * (B^2 - 1)) and can be at most AB + SQRT((A^2 - 1) * (B^2 - 1)). Interesting. His inequalities are weaker than mine, but it may be he hasn't made the assumption (as I have) that the RV's can take on only the values -1 and 1. So I get AB - MIN((1-A)(1-B), (1+A)(1+B)) <= C <= AB + MIN((1-A)(1+B), (1+A)(1-B)) whereas he uses the geometric mean instead of MIN, yielding a larger interval. Is his proof, or an indication of his method, online? If communicating electronically, I'd appreciate copies. I do not know what it would be for the six correlations between four random variables. I've been working on this, and I may in fact call upon poor MX for some help. I considered considering tetrahedra and higher dimensional simplexes (simplices?). But I couldn't even figure out what the side length rules were, i.e. given six side lenghts, could those sides be assembled into a tetrahedron? I did some work on this a while back. Basically, you solve for the altitude, and if that's imaginary you haven't got a tetrahedron. If I can find the equation, I'll send it along, but I don't know what this has to do with random variables and correlations. What I am really trying to do is generate N random variables with a specified set of correlations. Except for N < 3, I have had no success as yet. Well I gave it to you for the restricted range of correlations I showed: 0 <= 1 - C12 + C13 - C23 = 4 Prob(X2 NE X3 = X1) 0 <= 1 + C12 - C13 - C23 = 4 Prob(X1 = X2 NE X3) 0 <= 1 - C12 - C13 + C23 = 4 Prob(X1 NE X2 = X3) 0 <= 1 + C12 + C13 + C23 = 4 Prob(X1 = X2 = X3) You take the sums of correlations, divide by four, and get four probabilities summing to one. Pick X1 equiprobable from {-1,1}, and pick one of the four cases according to their probabilities. Then X2 and X3 are either X1 or -X1, according to which case you picked. Either this is tougher than I thought, or I am dumber than I thought. Well, it doesn't seem too easy, and I think we're all missing something--that is, unless Marteens has figured it all out. I'd really like to know if he can demonstrate RV's with the correlations he gives. One curious spinoff is I now have an exact solution for the square root of a two by two matrix. What a pity I have no use for it now. It doesn't seem to generalize to 3 by 3 or larger. Seems to me I saw a method once, but I can't say where offhand. What is the square root of a 2x2?