About minimizing variance of a mimicking portfolio
Show older comments
I have been working on an assignment and encountered the following problem: I have one-factor model for returns (say CAPM). I regress the returns of fifteen companies on market returns (my factor) and obtain 15x1 vector of betas. My task is to create a 15x1 vector of weights thetaprime that is the mimicking portfolio of the factor. For this vector of weights to be the mimicking portfolios, I need the following conditions:
1. sum(thetaprime) = 1
2. thetaprime*beta' = 1
3. The variance of the resulting mimicking portfolio (constructed with the thetaprime components) to be the minimum variance possible and the variance is given by:
variance_mimicking = thetaprime*SIGMA*thetaprime' (where SIGMA is a 15x15 variance-covariance symmetric matrix that I have calculated)
I am doing the following steps:
1. N = null([ones(15,1) beta].').';
2.thetaprime = thetaa + randn(1,13)*N;
in this way I can generate infinitely many vectors of thetaprime that satisfy conditions 1) and 2) above. When I substitute thetaprime in the formula for the variance, I get the same variance every time independently of the variables in thetaprime which does not make sense for me. I expectted different result for variance for different values of thetaprime that I can minimize with an optimization routine in order to find one unique value for thetaprime. Any suggestions?
Thank you very much in advance!
Answers (2)
The only reason I can think of why that would happen is if
Aeq=[ones(15,1) beta]
is in the row space of sigma, or equivalently that null(SIGMA) contains null(Aeq). A simple test would be to check whether
SIGMA*N=0
Incidentally, though, you do know that QUADPROG is the more standard way to solve these kinds of problems, right? Do you not have the Optimization Toolbox?
12 Comments
N =B>=>2
on 3 Mar 2013
Matt J
on 3 Mar 2013
I didn't notice that you had transposed the output of null(Aeq). So, really what I meant was
N*SIGMA=0
N =B>=>2
on 3 Mar 2013
Matt J
on 3 Mar 2013
We probably need to see your SIGMA, beta, thetaa data so that we can try this ourselves.
N =B>=>2
on 3 Mar 2013
Matt J
on 3 Mar 2013
We need SIGMA as well. That's the most important part.
N =B>=>2
on 3 Mar 2013
I do not get the same variance for every thetaprime, although the difference is small and possibly due to the fact that the SIGMA data you've given is a 4 decimal place truncation of your actual data,
>> f=@(v)v*SIGMA*v';
>> v=thetaa + rand(1,13)*N;
>> f(v), f(thetaa)
ans =
19.7648
ans =
19.7642
The similarity between the 2 values can be explained by the fact that N is approximately in the null space of SIGMA, as I was theorizing before. N*SIGMA doesn't give a result that is precisely zero, but the numbers are small enough to be attributable to round-off
>> max(max(N*SIGMA))/norm(SIGMA)
ans =
5.6232e-06
If you repeat this, you might get an even smaller number than this since again, I'm working with a 4 decimal place approx to your data.
N =B>=>2
on 3 Mar 2013
N =B>=>2
on 3 Mar 2013
Matt J
on 3 Mar 2013
You should re-examine your SIGMA and how it was generated. It does not admit a unique solution.
Another thing I'll point out is that the problem is small enough to be solved analytically, and you've already done most of the work.
L=sqrtm(SIGMA);
dtheta = -thetaa*L/N*L;
thetaprime=thetaa+dtheta;
Categories
Find more on Quadratic Programming and Cone Programming in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
