I am currently dealing with the Maximum Diversification Strategy and I am trying to understand the synthetic universe approach from Choueifaty et al.(https://www.tobam.fr/wp-content/uploads/2014/12/TOBAM-JoPM-Maximum-Div-2008.pdf)

Specifically I would like to know if you guys think that it is still mathematically correct to omit the risk free asset (since other investment strategies which I am comparing it to, don't have the possibility to invest in a risk free asset) and rescale the portfolio weights at the end of the optimaziation simply to 1, without changing his fundamental approach?

Best regards


1 Answer 1


Yes, it is easy to find the MDP of N risky assets if you have the covariance matrix V (assumed non-singular) if there are no constraints:

Step 1. Compute the inverse of the covariance matrix: $CINV = V^{-1}$

Step 2. Find the standard deviations $\sigma$ by taking the square roots of the diagonal elemnts of $V$

Step 3. Find $X = CINV \times\sigma$ i.e. multiply the matrix CINV by the column vector $\sigma$. So $X$ consists of weighted row sums of CINV, with the $\sigma_i$s as weights

Step 4. Normalize X so the elements sum to 1, i.e take $W=\frac{X}{\sum_i x_i}$. These are the MDP portfolio weights.

It is quite simple, the clearest explanation I found is in Theron and Van Vuuren: The maximum diversification investment strategy: A portfolio performance comparison (link), it is also equivalent to what is described in the Chouefaty article.

  • $\begingroup$ Thank you for the answer! But could you explain why it is allowed to normalize X, so that it sums to 1? I am writing a seminar paper and my supervisor insists on an explanation why this is allowed, since Chouefaty uses risk free assets in his investment universe and does not normalize the elements. $\endgroup$
    – Dirty Dan
    Commented Apr 28, 2019 at 7:24
  • 1
    $\begingroup$ Assume you had $100 allocated according to MDP, following the above. If you had another $100 allocated similarly, the sum of the two portfolios the $200 is also MDP. It is scale-invariant, so you can normalise. $\endgroup$ Commented Apr 28, 2019 at 10:51

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