Prove that a determinant in markowitz method derivation is greater than zero

Question

I want to prove that the following determinant, that appears in the markowitz method of portfolio allocation is greater than zero. ($\mu$ is the vector of returns and $\sum$ is the covariance matrix)

Answer 1

The comments above re all the entries of $\mu$ not being the same is true, but can be removed if you make the 2x2 determinant in question $\ge 0$ instead of $> 0$. The commenters know this of course.

The answer to your question can be obtained by an application of the Cauchy-Schwartz inequality along with knowledge that a symmetric positive definite matrix has a square root.

Since $\Sigma^{-1}$ is positive definite, there exists a symmetric matrix $A$ such that $A^2=\Sigma^{-1}$. One might say that $A=\Sigma^{-1/2}$. The existence of $A$ can be seen by noting that $\Sigma^{-1}$ is diagonalizable. Look that up.

Let's call your 2x2 determinant $D$. Note that $D$ can be expressed as a bunch of inner products as follows:

$$<\Sigma^{-1}\mu, \mu><\Sigma^{-1}1_n,1_n>-<\Sigma^{-1}\mu,1_n><\Sigma^{-1}1_n, \mu>$$

Since $\Sigma^{-1}$ is symmetric and real, it is self-adjoint, which means that the product in the second term is of equal numbers (the second equality is because we are in a real vector space - in a complex vector space, we would need to take the complex conjugate to retain equality): $$<\Sigma^{-1}\mu,1_n>=<\mu,\Sigma^{-1}1_n>=<\Sigma^{-1}1_n,\mu>$$

Let's rewrite in terms of $A$:

$$<A^2\mu,\mu><A^21_n,1_n>-<A^2\mu, 1_n>^2$$

Again, A is symmetric and real, so it is also self-adjoint and this becomes:

$$<A\mu,A\mu><A1_n,A1_n>-<A\mu,A1_n>^2$$

The Cauchy-Schwarz inequality finishes us off. As a reminder, the Cauchy-Schwarz inequality states that using the usual inner product in $R^n$ or $C^n$, we get that:

$$|<x,y>| \le <x,x>^{1/2}<y,y>^{1/2}$$

So we get that:

$$<A\mu,A1_n>^2 \le <A\mu,A\mu><A1_n,A1_n>$$

Then subtracting the left handside on both sides of this inequality gives us $D \ge 0$.