# Covariance Matrix and minimizing variance

Let $$X = (X_1 X_2 \cdots X_n)^T$$ be $$n$$ not necessarily independent investment vehicles. Assume each has expected return $$\mu$$ with covariance matrix $$\operatorname{Var}(X)=(\operatorname{Cov}(X_i,X_j))_{i,j}$$. What values of $$0\leq c_1, \cdots, c_n \leq 1, c_1+\cdots +c_n=1$$ minimizes the variance of the variable $$Z=c_1X_1+\cdots +c_nX_n$$?

Let $$c=(c_1 \cdots c_n)$$, I know that $$Z=cX$$ and that $$\operatorname{Var}(Z)=c\operatorname{X}c^T=\sum_{i=1}^n\sum_{j=1}^nc_ic_j\operatorname{Cov}(X_i,X_j).$$

Let $$f(c_1, \cdots, c_n)=\sum_{i=1}^n\sum_{j=1}^nc_ic_j\operatorname{Cov}(X_i,X_j),$$ then it suffices to find the partial derivatives of $$f$$:

$$\frac{\partial f}{\partial c_j}=\sum_{k=1}^n2\operatorname{Cov}(X_j,X_k)c_k, \ j=1,\cdots, n$$

But I had troubles finding the minimum of $$f$$ subjected to the constraint $$c_1+\cdots+c_n=1$$. I tried using Largange multiplier but it didn't seem to work. Thanks in advance.

• It sounds like you're trying to minimize something of the form $c^T M c$ over a vector $c$ subject to $c^T c$. You can indeed do that with Lagrange multipliers (and I've made two answers over the last couple days on that) but the requirement that $0 \le c_i \le 1$ seems to turn it into a linear programming problem, which would need nastier methods. Also, I assume $c_i c_k$ should be $c_i c_j$ and $X_1$ should be $X_i$. Aug 2, 2021 at 18:37
• @David yes those were typos, updated just now thank you. Aug 2, 2021 at 18:45
• You might want to look into convex optimization Aug 10, 2021 at 18:53

Setting up the Lagrange function (For $$C_{ij} = Cov(X_i,X_j)$$)

$$\sum_{i=1}^n \sum_{j=1}^n c_i c_j C_{ij} - 2 \lambda \sum_{i=1}^n c_i$$

Taking the derivative w.r.t. $$c_i$$ and setting it to zero

$$2 c_{i} C_{ii} + 2 \sum_{j=1,i \neq j }^n c_j C_{ij} = 2 \lambda$$

Defining the matrix $$C$$, we can restate the system as $$c \cdot C = \lambda \cdot 1_n$$ where $$1_n$$ is the $$n$$-vector of ones.

Substituting back in the constraint, $$c \cdot 1_n = \lambda \, 1_n \cdot C^{-1} \cdot 1_n = 1$$ we arrive at the solution

$$c = \frac{1}{1_n \cdot C^{-1} \cdot 1_n} \times 1_n \cdot C^{-1}$$

So, you can just sum columns of $$C^{-1}$$ to get the coefficients and normalize them.

• Shouldn't the Lagrange equation be $\sum_{i=1}^n \sum_{j=1}^n c_ic_jC_{ij}-\lambda ((\sum_{i=1}^nc_i)-1)$? Since the constraint equation is $c_1+\cdots+c_n=1.$ Aug 29, 2021 at 7:34
• Yes, but the 1 doesn’t matter since we care about the derivative of the Lagrange function. Aug 29, 2021 at 7:36
• Thank you, but how did the $2$ show up in $2\lambda (c_1+\cdots+c_n)$? Aug 29, 2021 at 7:43
• Lambda is an arbitrary constant. I put 2 there to make calculations slightly easier. The result won’t change with or without 2. Aug 29, 2021 at 7:45
• Yes, that was a typo (the 2 is the reason I used 2 lambda). Corrected. Aug 29, 2021 at 8:20