- #1
LocalVol
- 1
- 0
Determinant of "enlarged" Correlation Matrix
Hi guys,
I am not a physicist but saw that you guys are actively discussing math problems in this forum. I have the following problem that I've been fighting with for some time now: I have a n-dimensional (n >= 3) correlation matrix with the following special structure
C(n, \rho, x) = \left( \begin{array}{c c c c c} 1 & \rho & x & \dots & \rho\\ \rho & 1 & \rho & \dots & \rho\\ x & \rho & 1 & \dots & \rho\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \rho & \rho & \rho & \dots & 1 \end{array} \right).
I.e. all off-diagonal entries are equal to \rho except for the correlation pair (1, 3). Here is my problem: Let's say I know that the determinant for dimension n is given by \psi(n). What is the determinant for dimension n + 1? I played around with Mathematica and found that for n = 3
\psi(3) = 1 - x^2 - 2 \rho^2 + 2 x \rho^2
(this was the easy part) and for n >= 4, we have the relationship
\psi(n) = (\rho - 1) \left( -\psi(n - 1) + (-1)^n (x - 1) (\rho - 1)^{n - 4} \left( \rho (1 + x) - 2 \rho^2 \right) \right)
(this is my problem - how to show this?). Equivalently, we can directly state the formula for the n-th determinant (n >= 3) as
\psi(n) = (-1)^n (x - 1) (\rho - 1)^{n - 3} \left( 1 + x + (n - 3) \rho (1 + x) - 2 (n - 2) \rho^2 \right)
but I have no clue of how to get there. Note that by increasing the dimension by one, we basically set
C(n, \rho, x) = \left( \begin{array}{c c} C(n - 1, \rho, x) & \rho 1_{\{ (n - 1) \times 1 \}}\\ \rho 1_{\{ 1 \times (n - 1) \}} & 1 \end{array} \right)
Any help is appreciated!
Hi guys,
I am not a physicist but saw that you guys are actively discussing math problems in this forum. I have the following problem that I've been fighting with for some time now: I have a n-dimensional (n >= 3) correlation matrix with the following special structure
C(n, \rho, x) = \left( \begin{array}{c c c c c} 1 & \rho & x & \dots & \rho\\ \rho & 1 & \rho & \dots & \rho\\ x & \rho & 1 & \dots & \rho\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \rho & \rho & \rho & \dots & 1 \end{array} \right).
I.e. all off-diagonal entries are equal to \rho except for the correlation pair (1, 3). Here is my problem: Let's say I know that the determinant for dimension n is given by \psi(n). What is the determinant for dimension n + 1? I played around with Mathematica and found that for n = 3
\psi(3) = 1 - x^2 - 2 \rho^2 + 2 x \rho^2
(this was the easy part) and for n >= 4, we have the relationship
\psi(n) = (\rho - 1) \left( -\psi(n - 1) + (-1)^n (x - 1) (\rho - 1)^{n - 4} \left( \rho (1 + x) - 2 \rho^2 \right) \right)
(this is my problem - how to show this?). Equivalently, we can directly state the formula for the n-th determinant (n >= 3) as
\psi(n) = (-1)^n (x - 1) (\rho - 1)^{n - 3} \left( 1 + x + (n - 3) \rho (1 + x) - 2 (n - 2) \rho^2 \right)
but I have no clue of how to get there. Note that by increasing the dimension by one, we basically set
C(n, \rho, x) = \left( \begin{array}{c c} C(n - 1, \rho, x) & \rho 1_{\{ (n - 1) \times 1 \}}\\ \rho 1_{\{ 1 \times (n - 1) \}} & 1 \end{array} \right)
Any help is appreciated!