Can anyone help me calculating the average covariance of return?

• Hi everyone. I'm trying to convert this formula into MatLab code:

where :

Fcvc is a target matrix used for shrinking the sample covariance matrix of return towards their average variance and average covariance, in order to obtain a better estimator. The topic is fromt De Nard paper "Oops! I Shrunk the Sample Covariance Matrix Again: Blockbuster Meets Shrinkage"

I don't know how to compute the average covariance (average of off-diagonal entries of a simmetric matrix), can anyone help me?

Answer 1

[William Rose]

@Silvia Grasso,

[Edit: change the name of the initial matrix; it should not be Fcvc, it should be S. Add code to construct Fcvc using the which we calculate.]

You can just write a nested for loop (i,j) that does the calculation:

%Make a 4x4 covariance matrix

N=4; S=eye(N);  %identity matrix
for i=1:N, 
    for j=i+1:N
        S(i,j)=-1+2*rand(1,1);
        S(j,i)=S(i,j);
    end
end
S=10*S   %multiply by some arbitray constant

S = 4×4

10.0000 -7.0228 -1.6309 6.1487 -7.0228 10.0000 -7.4054 5.1093 -1.6309 -7.4054 10.0000 0.3682 6.1487 5.1093 0.3682 10.0000

That looks like a reasonable covariance matrix: it is positive along the diagonal, and the absolute value of off-diagonal element are less than the diagoonal, and it is symmetric.

Now compute and :

siibar=sum(sum(S))/N^2;
sijbar=0;
for i=1:N
    for j=i+1:N
        sijbar=sijbar+S(i,j);
    end
end
sijbar=sijbar/((N^2-N)/2);
fprintf('siibar=%.4f, sijbar=%.4f\n',siibar,sijbar)

siibar=1.9459, sijbar=-0.7388

Those values look reasonable.

Use siibar and sijbar to make Fcvc

Fcvc=sijbar*(ones(N)-eye(N))+siibar*eye(N)

Fcvc = 4×4

1.9459 -0.7388 -0.7388 -0.7388 -0.7388 1.9459 -0.7388 -0.7388 -0.7388 -0.7388 1.9459 -0.7388 -0.7388 -0.7388 -0.7388 1.9459

That looks good.

[William Rose]

@Silvia Grasso

You may note that when calculating sijbar = , I added up the values on one side of the diagonal, then divided by (N^2-N)/2. In the formula you quoted, they added up the values on both sides of the diagonal and divided by (N^2-N). Since the matrix is symmetric (covariance matrices are alway ssymmetric), the result is the same for these two different ways of doing it.

[Silvia Grasso]

I tried to do this:

meanvar=mean(diag(sample));

meancov=sum(sum(sample(~eye(p))))/p*(p-1); %

target=meanvar*eye(p)+meancov*(~eye(p));

Could have sense? @William Rose