I am learning a bit more about CAPM, and wanted to know if there was a specific way that weightings of assets in the optimal mean-variance portfolio changed (for constant risk aversion, expected return, and risk free rate) if the covariance matrix was altered in different ways.

For example, for $n$ risky assets there is an $n \times n$ covariance matrix $\Sigma$ which holds the covariances of the assets in each of its indices. So I was wondering what would happen to the portfolio weights if 'suddenly' the covariance matrix was changed such that risky assets all became independent but their variances did not change. This would make a new covariance matrix, $\Sigma'$, a diagonal matrix, with the same diagonal as $\Sigma$.

So I know the weights of the risky assets should be $w = \lambda \Sigma^{-1}(\mu - r \mathbb 1)$. In our case let's just make the risk aversion, $\lambda = 1$, and the risk free rate $r = 0$. So the formula for each individual weight, $w_i, i \in [1,n]$ should be $w_i = \sum_{k = 1}^{n} \Sigma^{-1}_{ik}\cdot\mu_k$.

So in the case where we use the diagonal covariance matrix, $\Sigma'$, we would have that $w_i = \frac{\mu_i}{\text{Var}_i}$, and now I need to compare $\sum_{k = 1}^{n} \Sigma^{-1}_{ik}\cdot\mu_k$ and $\frac{\mu_i}{\text{Var}_i}$, but since the inverse of the covariance matrix is in the first equation, I cannot tell whether the coefficients of each $\mu_k$ will be lead to values less than or greater than $\frac{\mu_i}{\text{Var}_i}$.

So I was wondering if there is a definite answer to my question (i.e. do the weights necessarily change in a certain manner), or if it depends on the particular covariance matrix of the risky assets.


1 Answer 1


One general representation of the CAPM model is to suggest that it is the solution of:

$$ \min_w f(w) = \alpha \frac{1}{2}w^T(2\Sigma)w - (1-\alpha)\mu^t w $$ $$ s.t. \quad \delta^T w = 1$$

and different $\alpha$ determine different points on the efficient frontier.

Note that this form allows short selling and the solution is closed form. The reformulated KKT condition of stationarity reduce to:

$$ \begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix} \begin{bmatrix} w \\ \lambda \end{bmatrix} = \begin{bmatrix} (1-\alpha) \mu \\ 1 \end{bmatrix}$$

$$ \implies \begin{bmatrix} w \\ \lambda \end{bmatrix} = \begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix}^{-1}\begin{bmatrix} (1-\alpha) \mu \\ 1 \end{bmatrix}$$

So say you wanted to analyse the change in $w$ with respect to $\sigma_{11}$; $\frac{\partial w}{\partial \sigma_{11}}$ you can apply calculus directly:

$$ \partial_{\sigma_{11}} \begin{bmatrix} w \\ \lambda \end{bmatrix} = \partial_{\sigma_{11}} \begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix}^{-1}\begin{bmatrix} (1-\alpha) \mu \\ 1 \end{bmatrix}$$ $$ \partial_{\sigma_{11}} \begin{bmatrix} w \\ \lambda \end{bmatrix} = -\begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix}^{-1} \left ( \partial_{\sigma_{11}} \begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix} \right ) \begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha) \mu \\ 1 \end{bmatrix}$$ $$ \partial_{\sigma_{11}} \begin{bmatrix} w \\ \lambda \end{bmatrix} = -\begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix}^{-1} \begin{bmatrix} \begin{bmatrix} 2\alpha & 0 \\ 0 & 0 \end{bmatrix} & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 2 \alpha \Sigma & \delta \\ \delta^T & 0 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha) \mu \\ 1 \end{bmatrix}$$

For short selling not permitted the optimisation has no closed form so requires an optimisation solver and the derivatives must be calculated numerically as well, I expect.


To answer the specific question, let me re-formulate it:

Suppose you have a diagonal covariance matrix and returns vector, which gives rise to a specific set of weights as the solution of the mean-variance optimisation.

Now consider the scope of the set of all possible other matrices which have the same diagonal but other non-diagonal elements are allowed to change but still result in a valid new covariance matrix.

This set of new potential weight vectors (as a new solution) I expect is quite varied, so much so that the difference between the base weight vector and any new weight vector is completely dependent upon the change in the covariance matrix (as you state).

This is also clear from the calculus which asserts that the change in weight vector is completely dependent upon the make up of the covariance matrix, and also of the returns vector.

  • $\begingroup$ Thanks for the answer. Since in my case, the variances aren't changing but the covariances are, it seems like the weights will change in different manners dependent on the exact covariance matrix. Does that make sense? $\endgroup$
    – Slade
    Nov 6, 2019 at 19:38
  • $\begingroup$ edited in line.. $\endgroup$
    – Attack68
    Nov 6, 2019 at 19:49

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