# Maximum Diversification Strategy without risk free asset

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

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.

• 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. – Dirty Dan Apr 28 at 7:24
• 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. – James Spencer-Lavan Apr 28 at 10:51