About minimizing variance of a mimicking portfolio

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!

[Matt J]

Don't you mean

thetaa + N*randn(13,1)

You want a random linear combination of the columns of N, I think.

[N =B>=>2]

Even if I do it in this way, I get the same problem in the end. Values of thetaprime do vary,but the variance is still constant.

Answer 1

[Matt J]

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?

[N =B>=>2]

Thank you for your answer. N is 13x15, so SIGMA*N is not a possible operation, since SIGMA is 15x15. I am a Matlab rookie, so I haven't used QUADPROG yet, but I do have the Optimization Toolbox

[Matt J]

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]

N*SIGMA yields a result which is different from zero

[Matt J]

We probably need to see your SIGMA, beta, thetaa data so that we can try this ourselves.

[N =B>=>2]

beta is:

0.4829
  0.7327
  1.1341
  1.2864
  1.1280
  1.3539
  1.2257
  0.9682
  1.5092
  1.4600
  0.4778
  0.7881
  0.5662
  0.6209
  0.4340

thetaa' is:

0
       0
       0
       0
       0
       0
       0
       0
  0.5264
       0
       0
       0
       0
       0
  0.4736

[Matt J]

We need SIGMA as well. That's the most important part.

[N =B>=>2]

   SIGMA Columns 1 to 5
    4.6089    6.9931   10.8237   12.2776   10.7660
    6.9931   10.6107   16.4230   18.6291   16.3354
   10.8237   16.4230   25.4189   28.8335   25.2833
   12.2776   18.6291   28.8335   32.7067   28.6797
   10.7660   16.3354   25.2833   28.6797   25.1485
   12.9218   19.6065   30.3463   34.4228   30.1845
   11.6980   17.7495   27.4721   31.1625   27.3256
    9.2408   14.0213   21.7017   24.6169   21.5859
   14.4046   21.8563   33.8285   38.3727   33.6481
   13.9344   21.1429   32.7243   37.1202   32.5498
    4.5607    6.9200   10.7105   12.1492   10.6534
    7.5214   11.4123   17.6636   20.0364   17.5694
    5.4043    8.2001   12.6918   14.3967   12.6241
    5.9263    8.9920   13.9175   15.7871   13.8433
    4.1418    6.2845    9.7269   11.0335    9.6750

Columns 6 to 10

   12.9218   11.6980    9.2408   14.4046   13.9344
   19.6065   17.7495   14.0213   21.8563   21.1429
   30.3463   27.4721   21.7017   33.8285   32.7243
   34.4228   31.1625   24.6169   38.3727   37.1202
   30.1845   27.3256   21.5859   33.6481   32.5498
   36.2289   32.7975   25.9085   40.3861   39.0679
   32.7975   29.6912   23.4546   36.5610   35.3676
   25.9085   23.4546   18.5281   28.8815   27.9388
   40.3861   36.5610   28.8815   45.0203   43.5509
   39.0679   35.3676   27.9388   43.5509   42.1294
   12.7867   11.5756    9.1442   14.2539   13.7887
   21.0876   19.0904   15.0805   23.5074   22.7401
   15.1521   13.7170   10.8358   16.8908   16.3395
   16.6154   15.0417   11.8822   18.5220   17.9174
   11.6124   10.5126    8.3044   12.9449   12.5224

Columns 11 to 15

    4.5607    7.5214    5.4043    5.9263    4.1418
    6.9200   11.4123    8.2001    8.9920    6.2845
   10.7105   17.6636   12.6918   13.9175    9.7269
   12.1492   20.0364   14.3967   15.7871   11.0335
   10.6534   17.5694   12.6241   13.8433    9.6750
   12.7867   21.0876   15.1521   16.6154   11.6124
   11.5756   19.0904   13.7170   15.0417   10.5126
    9.1442   15.0805   10.8358   11.8822    8.3044
   14.2539   23.5074   16.8908   18.5220   12.9449
   13.7887   22.7401   16.3395   17.9174   12.5224
    4.5130    7.4427    5.3478    5.8643    4.0985
    7.4427   12.2744    8.8195    9.6713    6.7592
    5.3478    8.8195    6.3371    6.9491    4.8567
    5.8643    9.6713    6.9491    7.6202    5.3257
    4.0985    6.7592    4.8567    5.3257    3.7221

Thanks for the help!

[Matt J]

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]

You are absolutly right! Repeating

>> max(max(N*SIGMA))/norm(SIGMA)

yields:

 2.9163e-17

I also realized that variances are slightly different. So an optimization routine for the variance now should help me get my unique answer for thetaprime?

[Matt J]

No. The computation that you've just done confirms that there is no unique answer for thetaprime. In fact, all thetaprime satisfying constraints (1) and (2) minimize the variance to within numerical precision.

[N =B>=>2]

This is a bit strange. The assignment asks for an uniqie solution for thetaprime and economic interpretation of its components (the magnitude of loadings on different stocks and their relative importance in the portfolio).

[Matt J]

You should re-examine your SIGMA and how it was generated. It does not admit a unique solution.

Answer 2

[Matt J]

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;