       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 -> 1/4, x -> 1/4, x -> 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 -> 1/16, x -> 1/4, x -> 1/8, x -> 1/16, x ->
1/2}}
In this case I would think the more or most intuitive result would be:

{1/2, {x -> 1/8, x -> 1/8, x -> 1/8, x -> 1/8, x ->
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 -> 1/2, x -> 0, x -> 0, x -> 0, x -> 1/2}} or
a variety of others.

How do I get to the more intuitive result: {1/2, {x -> 1/8, x ->
1/8, x -> 1/8, x -> 1/8, x -> 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
=E2=80=A8{0.306671, {x -> 0.246238, x -> 0.0909659, x -> 0.214026,
x -> 0., x -> 0.44877}}

In this case x and x have the highest correlation, x and x
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 -> 0.246238      Seems a bit heavy in this?
x -> 0.0909659    Why so little in this?
x -> 0.214026
x -> 0.		       Why nothing in this?
x -> 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 -> 1/8, x -> 1/8, x -> 1/8, x -> 1/8,
x -> 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?

```

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