# 11.9 Nearest Correlation Matrix Problem¶

A correlation matrix is a symmetric positive definite matrix with unit diagonal. This term has origins in statistics, since the matrix whose entries are the correlation coefficients of a sequence of random variables has all these properties.

In this section we study variants of the problem of approximating a given symmetric matrix $$A$$ with correlation matrices:

• find the correlation matrix $$X$$ nearest to $$A$$ in the Frobenius norm,

• find an approximation of the form $$D+X$$ where $$D$$ is a diagonal matrix with positive diagonal and $$X$$ is a positive semidefinite matrix of low rank, using the combination of Frobenius and nuclear norm.

Both problems are related to portfolio optimization, where one can often have a matrix $$A$$ that only approximates the correlations of stocks. For subsequent optimizations one would like to approximate $$A$$ with a correlation matrix or, in the factor model, with $$D+VV^T$$ with $$VV^T$$ of small rank.

## 11.9.1 Nearest correlation with the Frobenius norm¶

The Frobenius norm of a real matrix $$M$$ is defined as

$\|M\|_F = \left(\sum_{i,j}M_{i,j}^2\right)^{1/2}$

and with respect to this norm our optimization problem can be expressed simply as:

(11.36)$\begin{split}\begin{array}{ll} \minimize & \|A-X\|_F\\ \st & \mathbf{diag}(X) = e,\\ & X \succeq 0.\\ \end{array}\end{split}$

We can exploit the symmetry of $$A$$ and $$X$$ to get a compact vector representation. To this end we make use of the following mapping from a symmetric matrix to a flattened vector containing the (scaled) lower triangular part of the matrix:

(11.37)$\begin{split}\begin{array}{ll} \mbox{vec}: & \R^{n\times n} \rightarrow \real^{n(n+1)/2} \\ \mbox{vec}(M) = & (\alpha_{11}M_{11},\alpha_{21}M_{21},\alpha_{22}M_{22},\ldots,\alpha_{n1}M_{n1},\ldots,\alpha_{nn}M_{nn}) \\ \alpha_{ij}=&\begin{cases}1 & j=i\\ \sqrt{2} & j<i\end{cases} \end{array}\end{split}$

Note that $$\|M\|_F=\|\mbox{vec}(M)\|_2$$. The Fusion implementation of $$\mbox{vec}$$ is as follows:

Listing 11.17 Implementation of function $$vec$$ in (11.37). Click here to download.
def vec(e):
N = e.getShape()

msubi = range(N * (N + 1) // 2)
msubj = [i * N + j for i in range(N) for j in range(i + 1)]
mcof  = [2.0**0.5 if i !=
j else 1.0 for i in range(N) for j in range(i + 1)]

S = Matrix.sparse(N * (N + 1) // 2, N * N, msubi, msubj, mcof)
return Expr.mul(S, Expr.flatten(e))


That leads to an optimization problem with both conic quadratic and semidefinite constraints:

(11.38)$\begin{split}\begin{array}{ll} \minimize & t\\ \st & (t, \mbox{vec} (A-X)) \in \Q,\\ & \mathbf{diag}(X) = e,\\ & X \succeq 0.\\ \end{array}\end{split}$

Code example

Listing 11.18 Implementation of problem (11.38). Click here to download.
def nearestcorr(A):
N = A.numRows()

# Create a model
with Model("NearestCorrelation") as M:
# Setting up the variables
X = M.variable("X", Domain.inPSDCone(N))
t = M.variable("t", 1, Domain.unbounded())

# (t, vec (A-X)) \in Q
v = vec(Expr.sub(A, X))
M.constraint("C1", Expr.vstack(t, v), Domain.inQCone())

# diag(X) = e
M.constraint("C2", X.diag(), Domain.equalsTo(1.0))

# Objective: Minimize t
M.objective(ObjectiveSense.Minimize, t)
M.solve()

return X.level(), t.level()


We use the following input

Listing 11.19 Input for the nearest correlation problem.
    N = 5
A = Matrix.dense(N, N, [0.0, 0.5, -0.1, -0.2, 0.5,
0.5, 1.25, -0.05, -0.1, 0.25,
-0.1, -0.05, 0.51, 0.02, -0.05,
-0.2, -0.1, 0.02, 0.54, -0.1,
0.5, 0.25, -0.05, -0.1, 1.25])


The expected output is the following (small differences may apply):

X =
[[ 1.          0.50001941 -0.09999994 -0.20000084  0.50001941]
[ 0.50001941  1.         -0.04999551 -0.09999154  0.24999101]
[-0.09999994 -0.04999551  1.          0.01999746 -0.04999551]
[-0.20000084 -0.09999154  0.01999746  1.         -0.09999154]
[ 0.50001941  0.24999101 -0.04999551 -0.09999154  1.        ]]


## 11.9.2 Nearest Correlation with Nuclear-norm Penalty¶

Next, we consider the approximation of $$A$$ of the form $$D+X$$ where $$D=\diag(w),\ w\geq 0$$ and $$X\succeq 0$$. We will also aim at minimizing the rank of $$X$$. This can be approximated by a relaxed linear objective penalizing the trace $$\trace(X)$$ (which in this case is the nuclear norm of $$X$$ and happens to be the sum of its eigenvalues).

The combination of these constraints leads to a problem:

$\begin{split}\begin{array}{ll} \minimize & \left\|X+\diag(w)-A\right\|_F + \gamma \trace(X),\\ \st & X \succeq 0, w \geq 0, \end{array}\end{split}$

where the parameter $$\gamma$$ controls the tradeoff between the quality of approximation and the rank of $$X$$.

Exploit the mapping $$\mbox{vec}$$ defined in (11.37) we can express this problem as:

(11.39)$\begin{split}\begin{array}{ll} \minimize & t + \gamma\trace(X) \\ \st & (t, \mbox{vec} (X + \diag(w) - A) ) \in \Q, \\ & X \succeq 0 , w \geq 0. \end{array}\end{split}$

Code example

Listing 11.20 Implementation of problem (11.39). Click here to download.
def nearestcorr_nucnorm(A, gammas):
N = A.numRows()
with Model("NucNorm") as M:
# Setup variables
t = M.variable("t", 1, Domain.unbounded())
X = M.variable("X", Domain.inPSDCone(N))
w = M.variable("w", N, Domain.greaterThan(0.0))

# D = diag(w)
D = Expr.mulElm(Matrix.eye(N), Var.repeat(w, 1, N))
# (t, vec (X + D - A)) in Q
Domain.inQCone())

result = []
for g in gammas:
# Objective: Minimize t + gamma*Tr(X)
t, Expr.mul(g, Expr.sum(X.diag()))))
M.solve()

# Find eigenvalues of X and compute its rank
d = [0.0] * int(N)
LinAlg.syeig(mosek.uplo.lo, N, X.level(), d)
result.append(
(g, t.level(), sum([d[i] > 1e-6 for i in range(N)]), X.level()))

return result


We feed MOSEK with the same input as in Sec. 11.9.1 (Nearest correlation with the Frobenius norm). The problem is solved for a range of values $$\gamma$$ values, to demonstrate how the penalty term helps achieve a low rank solution. To this extent we report both the rank of $$X$$ and the residual norm $$\left\|X+\diag(w)-A\right\|_F$$.

--- Nearest Correlation with Nuclear Norm---
gamma=0.000000, res=3.076163e-01, rank=4
gamma=0.100000, res=4.251692e-01, rank=2
gamma=0.200000, res=5.112082e-01, rank=1
gamma=0.300000, res=5.298432e-01, rank=1
gamma=0.400000, res=5.592686e-01, rank=1
gamma=0.500000, res=6.045702e-01, rank=1
gamma=0.600000, res=6.764402e-01, rank=1
gamma=0.700000, res=8.009913e-01, rank=1
gamma=0.800000, res=1.062385e+00, rank=1
gamma=0.900000, res=1.129513e+00, rank=0
gamma=1.000000, res=1.129513e+00, rank=0