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I would like to know which formula to use in order to optimize a portfolio based on highest Treynor and Jensens Alpha. I am aware that usually one optimize a portfolio by highest Sharpe ratio (the tangency portfolio) by following formula:

$$\textbf{w}_{tan} = \frac{1}{(\boldsymbol{\mu}^e)^\top\boldsymbol{\Sigma}^{-1}\textbf{1}}\boldsymbol{\Sigma}^{-1}\boldsymbol{\mu}^e. $$

Which formula can I use to maximize Treynor or Jensens Alpha, or do I need to create a maximization problem?

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1 Answer 1

This optimization is trivial

$$ w^{T,J}_i = \begin{cases} 1 \quad \text{if } i=\arg \max_i R^{T,J}(S_i) \\0 \quad \text{otherwise} \end{cases} $$

That is to say, when you optimize only one weight will be nonzero. That's because these ratios incorporate no notion of distributional width, and therefore do not reward diversification.

With no concentration penalty, they will simply put all their weight on the most "attractive" asset.

More generally, if you take any quantitative attribute $a$ of a stock, and apply it as a whole to the portfolio by summing it over the entire collection, you get exposure

$$ a_\bf{w}=\bf{w}^*\bf{a} $$

To achieve unit exposure to this attribute while minimizing variance, you optimize (using Lagrange multipliers), obtaining

$$ \bf{w}_a=\frac{\bf{\Sigma}^{-1}\bf{a}}{\bf{a}^*\bf{\Sigma}^{-1}\bf{a}} $$

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This is also what I've concluded. But is there no way to incorporate a reward for diversification? If not, how come mathematically, that it is not possible? –  Jens Jensen May 20 at 12:48
    
If you make your metric Sharpe-like, by including a measure of distributional width in the denominator, then there is of course a reward. If that measure is standard deviation, then the mathematics even works out to be nearly identical to Markowitz. –  Brian B May 20 at 13:20
    
I do not really understand how to solve this. Is it possible to edit your answer with a more detailed description of how to solve the question? –  Jens Jensen May 20 at 22:09

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