correlation range

[iniesta]

if cor(a,b)=.5 and cor(a,c)=0, what is the range for cor(b,c)

 

[Keith Tan]

-1 to 1

 

[Amir Yousefi]

it is not 1,-1 because in that case for(a,c) should be 0.5 too! but about the exact range, I am still thinking!

 

[diegosanaz]

Think of them as vectors. The correlation could be -0.5 also.

 

[crazybear]

Write down the correlation matrix. It should be positive definite.
Thus the determinant of the matrix should be positive. Then solve for cor(b.c)
It should be [-1,1)

 

[debkumar]

|1.0 0.5 0.0|
|0.5 1.0 r | = 1 -r*r -0.25 > 0 => - sqrt(0.75) <r < sqrt(0.75)
|0.0 r 1.0|
what am i doing wrong?

 

[diegosanaz]

debkumar, I believe you are right. In the centered but reduced variables space, the cos of the angle between the vector (variables) is their correlation. For A and B to have a 0.5 cor, their angle must be 60°, also, the angle for A and C must be 90°, so in a space where there are only two observations (the example picture) the C can only have 30° cos(30°)=sqrt(0.75) ) or 150° with B. In more dimentions, C is the orthogonal hyperplane to A, so there can be found values in the middle.

 

[Amir Yousefi]

|1.0 0.5 0.0|
|0.5 1.0 r | = 1 -r*r -0.25 > 0 => - sqrt(0.75) <r < sqrt(0.75)
|0.0 r 1.0|
what am i doing wrong?

What is the constraint for determinant of Correlation Matrix?
I thought it should be less than 1. how did you come up with zero?

 

[Shantanu Kumar]

What is the constraint for determinant of Correlation Matrix?
I thought it should be less than 1. how did you come up with zero?

Correlation Matrix properties are determined by Covariance Matrix properties.
Hence it should be positive-semidefinite and symmetric.

 

[Amir Yousefi]

Correlation Matrix properties are determined by Covariance Matrix properties.
Hence it should be positive-semidefinite and symmetric.

I see, But I also think that Determinant<1 (does not make any change here) should also hold.
Because variance of a portfolio can never be more than the variance of single assets, so if you think of Det as the magnitude ( I do not know what is the exact math terminology) of the matrix so it cannot be more than 1.

 

[Shantanu Kumar]

I see, But I also think that Determinant<1 (does not make any change here) should also hold.
Because variance of a portfolio can never be more than the variance of single assets, so if you think of Det as the magnitude ( I do not know what is the exact math terminology) of the matrix so it cannot be more than 1.

This reasoning is not correct. First of all these are variance/co-variance matrix not the variance/co-variance themselves. You need to multiply the matrix with the weight vector for calculating the variance/co-variance of the portfolio.

 

[Amir Yousefi]

This reasoning is not correct. First of all these are variance/co-variance matrix not the variance/co-variance themselves. You need to multiply the matrix with the weight vector for calculating the variance/co-variance of the portfolio.

I hardly understand what do you mean by "First of all these are variance/co-variance matrix not the variance/co-variance themselves. ".
If the determinant be more than 1, it means that you can pick weights in the way that the resulting portfolio would have a variance more than the weighted average variance of single assets, and we know this should not be possible!

An ugly proof which has little intuition behind could be the fact that product of eigenvalues cannot be more than one which is equal to determinant!

 

[Shantanu Kumar]

I hardly understand what do you mean by "First of all these are variance/co-variance matrix not the variance/co-variance themselves. ".
If the determinant be more than 1, it means that you can pick weights in the way that the resulting portfolio would have a variance more than the weighted average variance of single assets, and we know this should not be possible!

An ugly proof which has little intuition behind could be the fact that product of eigenvalues cannot be more than one which is equal to determinant!

Determinant of correlation matrix has maximum possible value of 1 but not for the variance-covariance matrix. And there is no way

"If the determinant be more than 1, it means that you can pick weights in the way that the resulting portfolio would have a variance more than the weighted average variance of single assets, and we know this should not be possible"

Verify yourself!

 

[Amir Yousefi]

I guess we are talking about correlation matrix in this thread! do not we?

 

[Amir Yousefi]

Determinant of correlation matrix has maximum possible value of 1 but not for the variance-covariance matrix. And there is no way

Verify yourself!

The determinant of a correlation matrix cannot be more than 1 because otherwise you can pick the weights in the way that the resulting portfolio would have a variance more than the weighted average variance of single assets, and we know this should not be possible"

 

[Shantanu Kumar]

The determinant of a correlation matrix cannot be more than 1 because otherwise you can pick the weights in the way that the resulting portfolio would have a variance more than the weighted average variance of single assets, and we know this should not be possible"

If we are talking about Correlation Matrix please use CORRELATION not VARIANCE. Since weights always belong in the interval [0,1], variance of portfolio can never be more than the weighted average of variance of single assets. If you disagree with that then give me just one example where you can make it true. I don't want any mathematical rigor.