0
$\begingroup$

I am taking an introductory course to finance in my Master's, and wanted to go further in the topic of portfolio theory (I am an engineering bachelor graduate, but as I just hinted, I am new to advanced maths for statistics). My question might look dumb to some of you, but I think my main problem is simply understanding the notation of the variance-covariance matrix. I watched a video saying that when calculating the optimal vector of weights to obtain the minimum-variance portfolio, in the case of many assets, we have: $$ \sigma_{portfolio}^2 = w^T\Sigma w$$ $$w^T\textbf{1}=1 $$ , where $w$ is a column vector of weights, and $\textbf{1}$ is (I suppose?) a column vector filled with 1s with the same dimension as $w$.

He then goes on to write the Lagrangian as follows: $$ \operatorname{L}=w^T\Sigma w + \lambda(w^T -1) $$ And, by deriving the Lagrangian with respect to $w$ and setting the derivative to 0, and after a couple of other steps, he ends up with: $$ w=\frac{\Sigma^{-1}\textbf{1}}{\textbf{1}^T\Sigma^{-1}\textbf{1}} $$ (If you'd like to see the intermediary steps, please ask).

The reasoning seems pretty clear to me, however I have no idea how to interpret the notation of the variance-covariance matrix in this final equation: does $\Sigma^{-1}\textbf{1}$ mean the inverse variance-covariance matrix of vector $\textbf{1}$ (which wouldn't make much sense according to the calculations I did for that case), or is it implied that $\Sigma^{-1}$ is the inverse variance-covariance matrix of the expected returns of the respective stocks? And, therefore, $\textbf{1}^T\Sigma^{-1}\textbf{1}$ would be the elements of the inverse var-covar matrix of expected returns, but as scalars (which would make sense because we can only divide scalars together and not matrices?).

Am I right in this assumption, or am I still missing the point? Again, maybe this question will look awkward to some of you, but I'm new to the topic, and after some research on the web I couldn't find the var-covar notations well explained anywhere.

Thank you!

$\endgroup$
1
  • 2
    $\begingroup$ $\Sigma^{-1}\textbf{1}$ really is just multiplication of the matrix $\Sigma^{-1}$ with the column vector $\textbf{1}$, so standard multiplication rules apply. When multiplying with a column vector of ones, you get the row sums. $\endgroup$ Mar 19, 2021 at 14:42

1 Answer 1

4
$\begingroup$

I think what is going on here is the matrix-vector product $\Sigma^{-1} \mathbf{1}$ and quadratic form $\mathbf{1}^T\Sigma^{-1} \mathbf{1}$ are being used as a convenient way of specifying summation. That is, they bridge matrix-vector notation and summation notation.

Assuming $\Sigma^{-1} \in \mathbb{R}^{d \times d}$, then $\Sigma^{-1} \mathbf{1}$ is the product of the matrix $\Sigma^{-1}$ with the column vector $\mathbf{1} \in \mathbb{R}^d$. Hence the matrix-vector $\Sigma^{-1} \mathbf{1}$ product is just a column vector in $\mathbb{R}^d$ where each element is the sum of the rows of $\Sigma^{-1}$.

Without any further context, $\Sigma^{-1} \mathbf{1}$ means take your inverse variance-covariance matrix $\Sigma^{-1}$ and take the sum of each row, to give a column vector.

Further, the quadratic form $\mathbf{1}^T \Sigma^{-1} \mathbf{1}$ is a product of the row vector $\mathbf{1}^T \in \mathbb{R}^{1 \times d}$, the matrix $\Sigma^{-1}$, and the column vector $\mathbf{1} \in \mathbb{R}^{d \times 1}$. Computing this will yield the sum of all elements in $\Sigma^{-1}$.

Similar to above, $\mathbf{1}^T \Sigma^{-1} \mathbf{1}$ means take your inverse variance covariance matrix $\Sigma^{-1}$ and sum all the elements, yielding a scalar.

I guess then what the expression for $w$ means is to take the column vector $\Sigma^{-1} \mathbf{1}$ and divide each element by the scalar $(\mathbf{1}^T \Sigma^{-1} \mathbf{1})$ (or more precisely multiply each element in the column vector by the scalar $(\mathbf{1}^T \Sigma^{-1} \mathbf{1})^{-1}$).

You might wonder why do this when we can just use summation notation - you would be surprised just how useful this is when writing vectorised mathematical pseudocode.

$\endgroup$
3
  • $\begingroup$ I see now, things are becoming clearer. However one last thing, you're talking about $\Sigma^{-1}$, but $\Sigma^{-1}$ of what vector quantity exactly? I understand that the diagonal of $\Sigma$ contains the variances and the rest of cells the covariances, but of what vector quantity here exactly? How I see things is that $\Sigma$ alone doesn't calculate the (co)variances of anything unless it's implied somehow $\endgroup$
    – it'syasper
    Mar 19, 2021 at 15:08
  • $\begingroup$ My response only addresses the problematic interpretations of the matrix-vector product and quadratic form in the numerator and denominator respectively of $w$ from the perspective of linear algebra. If you want context specific interpretations concerning $\Sigma$, which I cannot guarantee, but others on the forum may be able to assist with , please supply more info on what is contained in $\Sigma$. $\endgroup$
    – microhaus
    Mar 19, 2021 at 15:27
  • $\begingroup$ Alright thanks for the explanations! All clear now $\endgroup$
    – it'syasper
    Mar 19, 2021 at 16:53

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.