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)

enter image description here

  • 1
    $\begingroup$ Where does this question come from? Since you took a picture of the inequality - could you also please provide the reference to give the context? $\endgroup$ Mar 17, 2017 at 22:40
  • $\begingroup$ For it to be true you probably need that the vector $\mu$ is not proportional to $1_n$, i.e the the components of $\mu$ are not all the same (asset returns are not identical). Also you need that $\Sigma$ is invertible (duh). $\endgroup$
    – Alex C
    Mar 18, 2017 at 0:10
  • $\begingroup$ @AlexC could you please come up with the proof supposing that the entries of $\mu$ are not all the same and that $\sum$ is invertible $\endgroup$
    – Joanna
    Mar 18, 2017 at 2:45
  • $\begingroup$ @LocalVolatility the context is the derivation of Markowitz mean-variance optimization. It is in one of my classes study notes. $\endgroup$
    – Joanna
    Mar 18, 2017 at 2:50
  • 1
    $\begingroup$ @AlexC But it does mean that the top left and bottom right entries to the matrix have to be positive. (note: I'm not the same John as above) $\endgroup$
    – John
    Mar 20, 2017 at 16:28

1 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:


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 $D \ge 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.