Re: Using a Correlation Matrix to reduce risk

*To*: mathgroup at smc.vnet.net*Subject*: [mg114423] Re: Using a Correlation Matrix to reduce risk*From*: Andreas <aagas at ix.netcom.com>*Date*: Sat, 4 Dec 2010 06:12:10 -0500 (EST)*References*: <id7t4n$l8c$1@smc.vnet.net> <idag9a$jpe$1@smc.vnet.net>

Thank you all for your replies, Ray Koopman's idea to minimize variance may give me a complete solution, but I have some questions. In the following, the simple case I described in my original post, it gives me exactly what I expected and hoped for from a solution: m3 = {{1, 1, 0}, {1, 1, 0 }, { 0, 0 , 1}};=E2=80=A8p = Array[x, Length@m3]; Minimize[{p.m3.p, Tr@p == 1 && And @@ Thread[p >= 0]}, p] =E2=80=A8{1/2, {x[1] -> 1/4, x[2] -> 1/4, x[3] -> 1/2}} But in the case where I have 4 instruments with correlations of 1 and 1 with a correlation of 0 to all of the others it gives me the following: =E2=80=A8m5 = {{1, 1, 1, 1, 0}, {1 , 1, 1, 1, 0 }, {1 , 1, 1, 1, 0 }, {1 , 1, 1, 1, 0 }, { 0, 0 , 0, 0, 1}};=E2=80=A8p = Array[x, Length@m5]; Minimize[{p.m5.p, Tr@p == 1 && And @@ Thread[p >= 0]}, p] {1/2, {x[1] -> 1/16, x[2] -> 1/4, x[3] -> 1/8, x[4] -> 1/16, x[5] -> 1/2}} In this case I would think the more or most intuitive result would be: {1/2, {x[1] -> 1/8, x[2] -> 1/8, x[3] -> 1/8, x[4] -> 1/8, x[5] -> 1/2}} I do recognize that practically speaking the risk of these two compositions of instruments would be equal because of the correlations, but then so would: {1/2, {x[1] -> 1/2, x[2] -> 0, x[3] -> 0, x[4] -> 0, x[5] -> 1/2}} or a variety of others. How do I get to the more intuitive result: {1/2, {x[1] -> 1/8, x[2] -> 1/8, x[3] -> 1/8, x[4] -> 1/8, x[5] -> 1/2}} ? I think this is important beyond the kind of artificial cases described above. A more real world example such as cMatrix from my first post gives me the following: cMatrix = {{1., 0.635562, 0.698852, 0.404792, -0.32746}, {0.635562, 1., 0.410075, 0.314375, -0.0636438}, {0.698852, 0.410075, 1., 0.374416, -0.260137}, {0.404792, 0.314375, 0.374416, 1., 0.293135}, {-0.32746, -0.0636438, -0.260137, 0.293135, 1.}}; =E2=80=A8p = Array[x, Length@cMatrix];=E2=80=A8Minimize[{p.cMatrix.p, Tr@p == 1 && And @@ Thread[p >= 0]}, p] =E2=80=A8{0.306671, {x[1] -> 0.246238, x[2] -> 0.0909659, x[3] -> 0.214026, x[4] -> 0., x[5] -> 0.44877}} In this case x[1] and x[3] have the highest correlation, x[1] and x[2] the 2nd highest correlation, etc. But again the result doesn't look as intuitive as I would expect. Displayed differently with some comments and questions: x[1] -> 0.246238 Seems a bit heavy in this? x[2] -> 0.0909659 Why so little in this? x[3] -> 0.214026 x[4] -> 0. Why nothing in this? x[5] -> 0.44877 This makes perfect sense, the highest % in the least correlated instrument. This seems to give me something like the result for the m5 matrix rather than something closer to the intuitive solution for the m5 matrix ({1/2, {x[1] -> 1/8, x[2] -> 1/8, x[3] -> 1/8, x[4] -> 1/8, x[5] -> 1/2}}). I very much appreciate everyone's help in this. Can I get to something closer to the more intuitive type of answer for this?