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Topic Title: Minimum Correlation
Created On Fri Feb 04, 05 09:42 AM

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Minimum Correlation - jiantao - Fri Feb 04, 05 09:42 AM

RE:Minimum Correlation - alexandreC - Fri Feb 04, 05 11:02 AM

RE:Minimum Correlation - zerdna - Fri Feb 04, 05 03:30 PM

RE: Minimum Correlation - wstguru - Fri Feb 04, 05 04:33 PM

RE:Minimum Correlation - gjlipman - Fri Feb 04, 05 04:42 PM

RE:Minimum Correlation - zerdna - Fri Feb 04, 05 05:00 PM

RE: Minimum Correlation - Mandark - Fri Feb 04, 05 07:24 PM

RE:Minimum Correlation - Aaron - Sun Feb 06, 05 04:04 AM

RE:Minimum Correlation - jiantao - Sun Feb 06, 05 08:08 PM

RE:Minimum Correlation - freyzi - Fri Feb 11, 05 12:26 AM

RE: Minimum Correlation - exotiq - Sun Feb 13, 05 04:45 PM

RE:Minimum Correlation - freyzi - Mon Feb 14, 05 04:31 PM

RE:Minimum Correlation - Aaron - Sun Feb 13, 05 04:55 PM

RE:Minimum Correlation - energydude - Sun Feb 13, 05 11:17 PM

RE:Minimum Correlation - greenleaf - Sun Feb 13, 05 11:53 PM

RE:Minimum Correlation - Aaron - Mon Feb 14, 05 10:08 PM

RE:Minimum Correlation - Aaron - Mon Feb 14, 05 10:08 PM

RE:Minimum Correlation - Aaron - Mon Feb 14, 05 10:08 PM

RE:Minimum Correlation - Aaron - Mon Feb 14, 05 10:08 PM

RE:Minimum Correlation - energydude - Mon Feb 14, 05 11:55 PM

RE:Minimum Correlation - Aaron - Wed Feb 16, 05 04:57 PM

RE:Minimum Correlation - Chukchi - Sun Feb 13, 05 09:27 PM

RE:Minimum Correlation - JWD - Tue May 10, 05 04:45 AM

RE:Minimum Correlation - PaperCut - Sun May 15, 05 03:09 AM

RE:Minimum Correlation - bertstein - Sun May 15, 05 03:02 PM

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jiantao
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If X, Y, and Z are 3 random variables such that X and Y are 90% correlated, and Y and Z are 80% correlated, what is the minimum correlation that X and Z can have?

alexandreC
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wrong answer deleted.

i am bad when additions are involved...

Edited: Fri Feb 04, 05 at 05:34 PM by alexandreC

zerdna
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Quote
Minimum is 70%.

hmm, i got smth like 0.46, could have miscalculated -- i am bad when more than two additions are involved...

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wstguru
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gjlipman
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You can certainly get under 50%. For example
each of these scenarios happens with 10% probability:
scenario y x z
1 0.036723701 -0.381694905 0.05761223
2 0.139932364 -0.247475254 0.155952409
3 0.243140866 -0.113255576 0.254292605
4 0.386888018 0.090560882 0.341931464
5 0.554425706 0.335219473 0.423290214
6 0.533540724 0.256399114 0.554388144
7 0.630676507 0.380194717 0.654331331
8 0.727812984 0.503990069 0.754274393
9 0.824949091 0.627786067 0.854217606
10 0.956315469 0.27646133 2.505365596

Then there is a 46% correlation between x and z.

zerdna
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Geez, suppose you've got w1, w2, w3, independent and all distributed like standard normal. Set

X = w1
Y = 0.9 w1 + 0.4359 w2
Z = 0.47 w1 + 0.8649 w2 + 0.1762 w3

they will have correlations 0.9, 0.8, and 0.47, wouldn't you say?

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Mandark
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According to my calcs the roots of the equation w^2 - 2*.8*.9*w - 1 + .9^2 + .8^2 = 0 are the max and min allowable correlations assuming normal randoms.

Min = 0.45846606338756

Max = 0.98153393661244

Aaron
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Normal has nothing to do with it, but the answer is correct. Assume without loss of generality that X, Y and Z have mean zero and standard deviation one. Then we know E(XY) = 0.9 and E(YZ) = 0.8. That implies that X - 0.9Y has mean zero and standard deviation 0.19^0.5 and Z - 0.8Y has mean zero and standard deviation 0.36^0.5.

Multiply these together and you get XZ - 0.9YZ - 0.8XY + 0.72Y^2. The expected value of this is E(XZ) - 0.72. This must be between plus or minus (0.19*0.36)^0.5. Since E(XZ) is the correlation between X and Z, it must be between 0.72 - 0.2615 (0.4585) and 0.72 + 0.2615 (0.9815).

The correlation of X and Z depends on the correlation of the residuals when each of these is regressed on Y. If the correlation of the residuals is zero, the correlation of X and Z is 0.72 (0.9*0.8). If it is one, the correlation of X and Z is 0.98. If it is negative one, the correlation of X and Z is 0.46.

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Aaron Brown

Edited: Sun Feb 06, 05 at 04:06 AM by Aaron

jiantao
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Quote

The correlation of X and Z depends on the correlation of the residuals when each of these is regressed on Y. If the correlation of the residuals is zero, the correlation of X and Z is 0.72 (0.9*0.8). If it is one, the correlation of X and Z is 0.98. If it is negative one, the correlation of X and Z is 0.46.

This is a perfect answer and explanation.

freyzi
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As Aaron said "assume without loss of generality that X, Y and Z have mean zero and standard deviation one" then it can be easily shown that:

exotiq
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So using this or a similar method, can one solve this generalized minimum correlation problem:

What is the minimum possible value for the average of the non-diagonal values in an N x N correlation matrix? For example, for a 4x4 correlation matrix, what is the minimum possible average of the 6 numbers in the upper right or lower left triangle of the matrix?

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Aaron
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I don't know the answer to that (but I'll think about it). If all the off-diagonal elements are equal, I think the answer is -1/(n-1), so -1/3 for a 4x4 matrix. My guess is that is also the minimum average value.

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Aaron Brown

Chukchi
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Quote
Originally posted by: freyzi

Just a liitle explanation.

Suppose we have a triangle in volatility space, where the Cosines of angles are equal to correlation coefficients.

We know that a simple identity is valid for inner angles in any triangle ABC:

A + B + C = Pi

So, if we don't know Cos(C), but we know Cos(A) and Cos(B), then

Cos(C) = Cos(Pi - (A+B))= - Cos(A+B) = - Cos(A)*Cos(B)+ Sin(A)*Sin(B)

Sin(A) and Sin(B) could be both positive, both negative, or could have opposite signs.

We have two positive correlations Cos(A)=0.9 and Cos(B)=0.8

It gives us an estimate range for Abs(Cos(C))= {0.458466, 0.981534}.

Edited: Sun Feb 13, 05 at 10:21 PM by Chukchi

energydude
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Quote
Originally posted by: Aaron
I don't know the answer to that (but I'll think about it). If all the off-diagonal elements are equal, I think the answer is -1/(n-1), so -1/3 for a 4x4 matrix. My guess is that is also the minimum average value.

I think I can prove this at least for equal off-diagonal elements (Aaron probably has the proof already anyway

)

Suppose C is a correlation matrix with all off-diagonal elements = x. Then we can write

where D is an n x n democratic matrix (all elements = 1) and I is the usual identity matrix. But since

. But the eigenvalues of any matrix also satisfy the same equation that the matrix does. So,

. So there are only two distinct eigenvalues:

.

Since the matrix

is positive definite, all eigenvalues must be positive. The first value is always positive and the second is when

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greenleaf
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If X, Y, and Z are 3 random variables such that X and Y are 90% correlated, and Y and Z are 80% correlated, what is the minimum correlation that X and Z can have?

The answer is that the correlation matrix of (X,Y,Z) is semi-positive definite. Say: rho=

1 0.9 r
0.9 1 0.8
r 0.8 1

From Det(rho)>=0, one obtains an inequality quadratic in r. From there to the answer requires just a little more algebra.

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Edited: Sun Feb 13, 05 at 11:54 PM by greenleaf

freyzi
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Now, rescale volatility space to vector length one and the lower bound pops out:

-1/(N-1)

Next prove that the upper bound can only be bounded by 1?

What about extending the bounds put forward below from 3 dimensions to N? To do this we could think along the lines of Chukchi about angles in volatility space as correlation.

Two vectors define a plan and thus X,Y and Y,Z define two planes. Choose an angle from 0 to pi and we have defined a 3-dementional space. The correlation between X and Y depends on the chosen angel and is bounded as mentioned.

In N dimensions we have N(N-1)/2-1 different planes given. What restrictions do we have when we glue them together?

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Aaron Brown

Edited: Mon Feb 14, 05 at 10:09 PM by Aaron

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Aaron Brown

Edited: Mon Feb 14, 05 at 10:09 PM by Aaron

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Aaron Brown

Edited: Mon Feb 14, 05 at 10:09 PM by Aaron

Aaron
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Sorry for the repeats.

A clever proof, energydude, but I think there may be a flaw. You proved there are only two possible eigenvalues, but not (I think) that 1+x(n-1) has to be an eigenvalue.

I don't have a proof, but I was thinking along freyzi's lines. His argument lost me. It's probably obvious, but I don't see it.

I would instead assume all the variances are 1, for simplicity. If all the covariances among distinct variables are the same value x, then we get:

n + n*(n - 1)*x >= 0

or x >= -1/(n-1)

This also gives a proof for the average. The average correlation coefficient is the same as the average covariance if we set variance to 1. Therefore we have:

Sum of n*(n-1) off-diagonal correlations >= -n

Averay of off-diagonal correlations >= -1/(n-1)

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Aaron Brown

Edited: Mon Feb 14, 05 at 10:12 PM by Aaron

energydude
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Aaron: Your proof is far better...mine is more like using a hammer to crack a peanut...! However, I believe all the roots of the minimal polynomial of a matrix are its eigenvalues and all eigenvalues are its roots, so my proof though long is correct.

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Aaron
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Yes, you are correct.

I think your proof is a useful insight because it relates the minimum correlation question to decomposition of the correlation matrix.

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Aaron Brown

JWD
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The issue of possible physical values of correlations for general dimension was discussed on the technical forum, ?Correlations: Stressed matrices, PD approximations, the N-Sphere, SVD, Cholesky and all that?, Click Here .

For 3 variables, as freyzi and Chukchi indicated, we simply need to satisfy a triangle condition. However, it is most useful to introduce azimuthal angles. Call C12, C13, C23 the correlations between 12, 13, 23. Also call S12 = sqrt(1 ? C12^2) etc. Now calculate the azimuthal angle phi23 from
cos (phi23) = [C23-C12*C13]/[S12*S13] , using ordinary trigonometry. If this angle phi23 is a real number, then the correlations are physically possible. So if C12 = 0.9 and C13 = 0.8, then
cos (phi23)= -1 if C23 = 0.458, the minimum value as noted before by several people.

For larger matrices, more azimuthal angles are introduced, and their physical limits give the various extremal allowed values of the correlations. The independent angles turn out to live on an N-sphere, and the Cholesky decomposition takes a simple form when expressed in terms of these angles. My book has the details, Ch. 22.
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PaperCut
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JanDash -

Please, please - just once - I would love to see you make a post that doesn't end with, "for further elaboration, please refer to my book." Give me a break. I'm almost inclined to buy the damned thing if I thought it would get you to stop - but I think it won't matter. Maybe I will keep my money, and spend it on books written by Staunton, Collector and anyone else here who isn't full of baloney.

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bertstein
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Quote
Originally posted by: PaperCut
JanDash -

Please, please - just once - I would love to see you make a post that doesn't end with, "for further elaboration, please refer to my book." Give me a break. I'm almost inclined to buy the damned thing if I thought it would get you to stop - but I think it won't matter. Maybe I will keep my money, and spend it on books written by Staunton, Collector and anyone else here who isn't full of baloney.

i) Nursing a grudge over the dark matter thread, papercut?
ii) I'm having difficulty finding a post by jandash that could be described as "baloney". Could you give me an example?

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