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Topic Title: How can I simulate correlated random numbers?
Created On Sun Jan 12, 03 07:30 PM

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Paul
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Sun Jan 12, 03 07:30 PM

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Alan
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Mon Jan 13, 03 03:25 AM

For standard normals, make two draws Z1 and Z2 from your
favorite (random normal) generator.
Then, use Z1 and Z3 = rho Z1 + Sqrt ( 1 - rho^2 ) Z2,
where rho is the desired correlation.

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Mon Jan 13, 03 07:53 AM

If you want n random normals with correlation matrix C then find a matrix A such that

C = AA^{t}.

Now draw n independent normals W= (W_1,... W_n) and then put

Z = AW.

NB there lots of such A s.

MJ

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Tue Feb 04, 03 10:36 AM

most widely used being the cholesky decomposition where A is a lower triangular matrix with positive diagonal elements.

someone on the forum mentioned a link which explicitly shows how to perform cholesky decomposition: http://ikpe1101.ikp.kfa-juelich.de/briefbook_data_analysis/node33.html

d.

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reza
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Tue Feb 04, 03 12:05 PM

Cholesky is indeed one popular method using triangular decomposition
however many use a spectral method using matrix Diagonalization and argue the results are better
PJ is the expert ...

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James
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Tue Feb 04, 03 03:19 PM

I was taught Cholesky in grad school, but Reza wrote "spectral method using matrix Diagonalization and argue the results are better"
I have no idea what this is, PJ or Reza, please explain....please?

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reza
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Tue Feb 04, 03 03:30 PM

again the expert id PJ
but if you just replace the decmposition of the covariance matrix into triangular matrices with
diagonalization you could do the same thing, no?
instead of
x = X
y= rX + sqrt(1-r^2) Y
you would have something like
x= aX+bY
y= cX+dY

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Tue Feb 04, 03 08:33 PM

James -
I'm guessing the spectral approach is as follows: Write M = ADA^-1. Take sqrts of the diagonal elements and call that S i.e. D = SS. Then M = ASA^-1 . ASA^-1. If A is orthonormal by construction (i.e. it is orthogonal by construction and we took the 'ovals' out by using D), then A^-1 = A^T. S=S^T also, so if C = ASA^-1 then M=CC^T as well.

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Tue Feb 04, 03 08:36 PM

i am guessing that the performance is improved because the A's are orthonormal and hence behave well when multiplied by a vector of normal random variates... all the magnification / compression is inside S, which is safely diagonal.

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James
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Wed Feb 05, 03 09:21 AM

Thanks Reza & kr, I am going to try this, and will be back for more advice if i screw it up (p=100%).

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"The good Christian should beware of mathematicians and all those who make empty prophecies. The danger already exists that mathematicians have made a covenant with the devil to darken the spirit and confine man in the bonds of Hell." -- St. Augustine

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Joined: Dec 2001

Thu Feb 06, 03 02:18 PM

One issue is the performance of low-discrepancy numbers.

If you take low discrepancy numbers that work better in low dimensions than in high ones (this is true of all low-discrepancy numbers of which I am aware,) then you want the low dimensions to do more work.

Spectral pseudo-square rooting is one way to achieve this. For ordinary pseudo-randoms it shouldn't make any difference to the convergence rate.

Additionally, some people prefer the spectral method for numerical stability reasons.

The method is write

C = PDP^{t}

with P orthogonal and D diagonal. Let D^{1/2} be D with the elements square-rooted.

Put A = PD^{1/2}.

MJ

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WaaghBakri
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Sat Feb 08, 03 06:32 AM

A naive question ...... to create n correlated or independent RV's, must we worry about non-sequential drawing that results when we use the same generator? Or equivalently, if I choose every nth. sample from a single stream output from a random number generator, will I ruin the random number generators properties?

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sloth
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Thu Feb 13, 03 03:46 PM

The whole discussion on this topic so far has been concerned with normal distributions.

More generally, you might want to simulate correlated numbers from an arbitrary collection of marginal distributions.

This is the basic problem that insurance companies face all the time: estimating their total p+l distribution, when the individual parts of their business (different classes of insurance, investments, and so on) are correlated, and have weird and wonderful marginal distributions.

The only 'derivatives' application I know is for modelling portfolios of weather derivatives, some of which may have very non-normal distributions (e.g. the number of days it snows in winter).

Even given the correlations, there is no unique answer, unlike for the normal case.

The standard method used in practice is due to Iman and Connover.

My understanding of it is that it works as follows:
1)calculate rank correlations
2)convert them to linear correlations using a cunning formula
3)simulate (using Choleski or SVD) from a multivariate normal with these correlations
4)convert the normally distributed simulations to the right marginal distributions using the CDF of the normal distribution and the inverse CDF of the marginal distribution.

To go further than this, you could start looking at copulas.

Sloth

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Wed Feb 26, 03 12:18 PM

I think that to generate correlated multivariate normal random variables, one would do the Cholesky or Spectral method on the CORRELATION matrix. Some people here talk about the COVARIANCE matrix.

Will using the Covariance matrix (X*sqrt(t), where X is N(0,1)) be same as using the correlation matrix and then multiplying by the individual volatilities (sigma*Z*sqrt(t))

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Wed Feb 26, 03 04:08 PM

why do two mults when you can get away with one.

MJ

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The second edition of Quant Job Interview Questions and Answers is now out.

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challenged
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Thu Feb 27, 03 12:00 AM

There is a VBA example for use in excel at?

http://www.geocities.com/WallStreet/9245/vba11.htm

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bluesea
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Sun Nov 09, 03 05:17 AM

searching choleskey decomposition in google lead me to this page.
I have such topic this quarter and what my instructor`s equation is X=chol(cov())*randn()+Mu()
I have no idea why it is the product of chol, not directly of covariance? Cholesky decoposition remain the properties of the original covariance?
and what is the second term in your folks description(Sqrt ( 1 - rho^2 ) Z2), it`s the same to the one of mine(Mu)? I also wonder if it converts the mean properties to the X with such addition.

I appreciate your all inputs.

Hai

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nastradamus
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Wed Nov 12, 03 06:12 AM

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Originally posted by: bluesea
searching choleskey decomposition in google lead me to this page.
I have such topic this quarter and what my instructor`s equation is X=chol(cov())*randn()+Mu()
I have no idea why it is the product of chol, not directly of covariance? Cholesky decoposition remain the properties of the original covariance?
and what is the second term in your folks description(Sqrt ( 1 - rho^2 ) Z2), it`s the same to the one of mine(Mu)? I also wonder if it converts the mean properties to the X with such addition.

I appreciate your all inputs.

Hai

You can do the math ... If x = sqrt(var)*randn(), then x~N(0,var). Likewise, if X=chol(cov())*randn(), then X ~N(0, Cov()). You can think of chol as a square root function for matrix.