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In the Black-Litterman framework views of inverstors on the market are modelled. These views have a covariance-matrix $\Omega$.

I always found it quite natural to model $\Omega$ proportional to the market covaraince $\Sigma$. But in the literature by Litterman himself I find that $\Omega$ should be diagonal. Why should this be useful?

I take this from Appendix B of The Intuition Behind Black-Litterman Model Portfolios and Global Asset Allocation With Equities, Bonds, and Currencies p.38

There is no proof in the papers mentioned. I don't think that $\Omega$ has to be diagonal for the framework to work, but why do they assume it? Mabye the proof is in Black, Fischer, and Robert Litterman, Asset Allocation: Combining Investor Views With Market Equilibrium, Goldman, Sachs & Co.,September 1990. But I can't find it on the web.


2 Answers 2


It depends on your investment process: more specifically, on how you generate views. Here are three practical cases which lead to different choices for $\Omega$:

  1. Let's assume you are an investor who acts on (more or less) arbitrary bits of opinion: e.g. you like Italian equities because you like Italy, and German equities because you find Angela Merkel's economic policies convincing. Then your views are uncorrelated and the off-diagonal elements of $\Omega$ are zero.
  2. Your views come from a systematic approach, say a regression model based on European credit spreads. Then you can estimate $\Omega$ directly from the residuals of your forecast. The off-diagonal elements will not be zero because they are based on a common factor.
  3. You talk to two different brokers. Both are bullish on equities, and broker A recommends Italian, broker B German equities. In this case you probably want to use the covariance structure of the market in equilibrium, $\Omega \propto P \Sigma P^T$. Again, the off-diagonal elements will be non-zero.
  • $\begingroup$ Thanks for your nice examples-I really like the ideas behind them. I just still wonder why B and L themselves use diagonal in the papers. Usually whenever possible one works as general as it makes sense. Non-diagonal absolutely makes sense... $\endgroup$
    – Richi Wa
    Commented Feb 6, 2015 at 12:13

In The Black-Litterman Model In Detail Jay Walters says the following on p. 13 (top paragraph):

First, by construction we will require each view to be unique and uncorrelated with the other views. This will give the conditional distribution the property that the covariance matrix will be diagonal, with all offdiagonal entries equal to 0. We constrain the problem this way in order to improve the stability of the results and to simplify the problem. Estimating the covariances between views would be even more complicated and error prone than estimating the view variances.

On p. 14 there is a whole section on how to specify $\Omega$. There he also says that Meucci (2006) uses a non-diagonal matrix $\Omega$.

Maybe this is helpful.

  • 1
    $\begingroup$ Yes, Jay Walters writes this on top of page 13 and starting with page 14 he also states the specification $\Omega = P \Sigma P^T$. So he speaks about both choices ... looking at his survey paper diagonal and full matrices are possible. My question is, why does Litterman himself use a diagonal matrix? $\endgroup$
    – Richi Wa
    Commented Jan 27, 2015 at 11:36
  • $\begingroup$ I didn't understand your question in such a way that you are only interested in what Litterman himself thought. The last two sentences in the above quote from Walters give some reasoning. I think it all depends on your model or reasoning on how to model/estimate $\Omega$. Specifying all the covariances manually is not really an option, because $\Omega$ should be pos. semi-definit. In an Bayesian framework $\Omega$ are model parameters (about some uncertainty). The result does not change much if you specify all covariances, but you have $n(n-1)/2$ parameters more. $\endgroup$ Commented Jan 27, 2015 at 13:03
  • $\begingroup$ Just imagine you do not specify single investor uncertainty but factors (i.e. eigenvalues) of it. It is only a slight model difference but much more parsimonious. $\endgroup$ Commented Jan 27, 2015 at 13:07

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