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I have a correlation matrix A given below. Here A should be a positive-definite matrix so that we can perform Cholesky decomposition of A.

1.0,  0.5,  0.6,  0.7

0.5,  1.0,  ??,   ??

0.6,  ??,   1.0,  ??

0.7,  ??,   ??,   1.0

How to find the unknown correlation coefficients in A such that A will remain as positive-definite matrix without changing the given coefficients?

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The missing entries are not uniquely determined, unless additional information is known.

The only requirements on the missing entries are that they be symmetric, i.e., A(2,3) = A(3,2), A(2,4) = A(4,2), A(3,4) = A(4,3), and that A be positive semi-definite.

I used CVX to maximize the sum of the unknown entries, then separately ran it to minimize the sum of the unknown entries, then separately ran it to minimize the sum of absolute values of the unknown entries. This produces three different solutions, among an infinite number of possibilities.

Maximum sum:

 A =
    1.0000    0.5000    0.6000    0.7000
    0.5000    1.0000    0.9928    0.9685
    0.6000    0.9928    1.0000    0.9913
    0.7000    0.9685    0.9913    1.0000

Minimum sum:

 A =
    1.0000    0.5000    0.6000    0.7000
    0.5000    1.0000   -0.1400    0.0400
    0.6000   -0.1400    1.0000    0.2200
    0.7000    0.0400    0.2200    1.0000

Minimum sum of absolute value:

A=
    1.0000    0.5000    0.6000    0.7000
    0.5000    1.0000    0.0000    0.0000
    0.6000    0.0000    1.0000    0.1354
    0.7000    0.0000    0.1354    1.0000
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  • $\begingroup$ +1. Mathematica's solution is $$\frac{1}{10} \left(3-4 \sqrt{3}\right)<x<\frac{1}{10} \left(3+4 \sqrt{3}\right)\land \\\frac{1}{20} \left(7-3 \sqrt{17}\right)<y<\frac{1}{20} \left(7+3 \sqrt{17}\right)\land \\ \frac{1}{75} (100 x y-35 x-30 y+42) \\ -\frac{1}{75} \sqrt{2} \sqrt{5000 x^2 y^2-3500 x^2 y-1300 x^2-3000 x y^2+2100 x y+780 x-1950 y^2+1365 y+507}\\ <z<\frac{1}{75} \sqrt{2} \sqrt{5000 x^2 y^2-3500 x^2 y-1300 x^2-3000 x y^2+2100 x y+780 x-1950 y^2+1365 y+507}\\ +\frac{1}{75} (100 x y-35 x-30 y+42)$$ where $x,y,z$ are the (2,3), (2,4), and (3,4) entries in $A$. $\endgroup$
    – whuber ♦
    Commented Jun 19, 2015 at 13:54
  • $\begingroup$ Thanks for the solution. Here correlation matrix A should be positive definite so that the Cholesky Decomposition of A is possible. Is it possible? $\endgroup$
    – Jameesh Hussain
    Commented Jun 22, 2015 at 4:24
  • $\begingroup$ Yes, it's possible. But there are an infinite number of possible solutions. The extreme solutions (all 3 I showed) all produce singular A, but all other solutions have positive definite A. But the point is that positive definite A, and therefore Cholesky Decomposition of A are not unique. There are an infinite number of them. Your problem is underdetermined. $\endgroup$
    – Mark L. Stone
    Commented Jun 22, 2015 at 5:04
  • $\begingroup$ Any convex combination of the 3 matrices I provided with non-zero-weight on at least one of them will be a positive definite solution to your problem. For example, the average of any 2 or 3 of them. $\endgroup$
    – Mark L. Stone
    Commented Jun 22, 2015 at 5:12

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