Ok, on the advice of administration I open a new question, hoping that in this way it becomes clearer.

Like I said before, I am trying to understand how the authors of this (page 76) and this (page 31) works arrive to the condition $\left \| C(\widetilde{R}_1- \check{R}_1) \right \|\leq \Gamma$, with:

  • $\Gamma \in \mathbb{R}^+$ is a predefined parameter whose value reflects the degree of risk-tolerance of the investor, but for purposes of question it doesn't matter knowing what it means.

  • $\widetilde{R}_1=\begin{bmatrix} \frac{\widetilde{S}_1^1}{S_0^1} & \frac{\widetilde{S}_1^2}{S_0^2} & \cdots & \frac{\widetilde{S}_1^M}{S_0^M} \end{bmatrix}^T$ represents the vector of random returns for the first period (i.e. $t=1$) given the current price of $m$-th stock included in the underlying basket of multiasset option, for $m=1,...,M$.

  • $\check{R}_1=\begin{bmatrix} \frac{\check{S}_1^1}{S_0^1} & \frac{\check{S}_1^2}{S_0^2} & \cdots & \frac{\check{S}_1^M}{S_0^M} \end{bmatrix}^T$ represents the vector of expected returns for the first period (i.e. $t=1$) given the current price of $m$-th stock included in the underlying basket of multiasset option, for $m=1,...,M$.

  • $C=\sum^{-\frac{1}{2}}$ is a matrix obtained (I still got to figure out) applying Cholesky decomposition. Text only says that $C^TC=\sum^{-1}$.

  • $\left \| x \right \|$ is (I quote from page 31 second link that I inserted) a general norm of a factor, depending on the modeler's preference, it can be $L_1,L_2,L_{\infty}$ or $D$-norm. It's the only thing that authors say about that, so I can't be more specific. But I earlier found this paper (Robust Linear Optimization under General Norms - Bertsimas,Pachamanova,Sim) where, at page 513, it seems to be explained the meaning of $\left \| x \right \|$. In particular they consider (I don't know on what basis) uncertainty sets given by $\left \| M(vec(\widetilde{A})-vec(\check{A})) \right \|_{p}\leq \Delta$, and for the Proposition 1. page 512 they say that $\left \| x \right \|_p\doteq (\sum_{j=1}^{n}\left | x_j \right |^p)^{\frac{1}{p}}$, with $M$ a diagonal matrix that contains the inverses of the ranges of coefficient variation.

Now, in those circumstances, I would like to know if my reasoning below is correct in your opinion. In fact it doesn't deal with the mathematical meaning of condition above but with the reason that has led the authors to formalize it. This is what I think.

They suggest, in order to consider the correlation between the asset prices included in the basket, to study the correlation between the asset returns. Clearly, correlation must be quantized between each $m$-th asset included in the basket (or rather for each couple of asset on a set of $m$ elements, for a total of $\binom{m}{2}$ combinations without repetitions) and for each $t$-th instant of time. It's not the case to have such tremendous amount of information, moreover subject to uncertainty, so instead to constructing the variance-covariance matrix for each $t$ [i.e. to say considering, for each $t$, one by one, all the possible couples of assets and for any $j$-th possible realization of assets returns (for $j=1,...,n$ and $n$ the sample size) calculating the sum of cross products of deviations from the estimations of these realizations] they use the properties of $\sum$. We know that $∑$ for each $t$ is a square symmetric and positive definite matrix, i.e. to say not-singular, i.e. to say invertible, so we should be able to calculate $\sum^{-1}$ for Cholesky decomposition (first paper say that $\sum^{-1}=C^TC$) and then, I suppose, to use the spectral theorem to obtain the diagonal matrix in the condition above.

Is my reasoning correct?

Thanks in advance just for reading.

  • 1
    $\begingroup$ Did not read all. The matrix norm condition looks suspiciously like a predefined maximum probability range. Think of it as saying „the realized, normalized, total variation of the vector must be within a certain range“. That range is an ellipsis in 2d-normal probability space . $\endgroup$ Commented Dec 5, 2020 at 10:27
  • $\begingroup$ @Kermittfrog Thanks for your answer! I have tried to understand your comment but... Could you be more specific? I am really stuck. $\endgroup$
    – user51121
    Commented Dec 7, 2020 at 8:32

1 Answer 1


Here, I try to help a bit with the matrix norm question.

Assume an $M$-dimensional multivariate normally distributed return vector $\widetilde{R}$. The covariance of these returns is $\Sigma$, and the expected returns are $\check{R}$.

The probability density function of $R$ is

$$ f(R;\check{R},\Sigma)=\left(2\pi\right)^{-\frac{M}{2}}\left|\Sigma\right|^{-\frac{1}{2}}e^{-\frac{1}{2}\left(R-\check{R}\right)^T\Sigma^{-1}\left(R-\check{R}\right)} $$

We know that the contours of the multivariate density form (higher dimensional) ellipsoids; i.e. an ellipsis in the two-dimensional case of $M=2$. The smaller these ellipsoids, the nearer we are, in some sense, to the expected value vector.

In your case, this can be shown by first decomposing the covariance matrix $\Sigma$ via the Cholesky decomposition into $\Sigma=CC^T$. The inverse of $\Sigma$ then becomes $\Sigma^{-1}=\left(C^{-1}\right)^TC^{-1}$ and with obvious misuse of notation we now write this as $\Sigma^{-1}=C^TC$. now we can rewrite the exponential of the density as

$$ \left(R-\check{R}\right)^T\Sigma^{-1}\left(R-\check{R}\right)=\left(R-\check{R}\right)^TC^TC\left(R-\check{R}\right) $$ And since this is a quadratic form, we can rewrite $$ =\left(R-\check{R}\right)^TC^TC\left(R-\check{R}\right)=\|C(R-\check{R})\|_2 $$

Hence, the condition $\|C(R-\check{R})\|\leq \Gamma $ can be interpreted as a condition on the density of the returns, i.e. on the size of the $M$-dimensional ellipsoid, around the expected return vector.

  • $\begingroup$ That made me a little misty. Really thanks! Now I studied it and let you know! PS: I will upvote but I need 15 reputation... $\endgroup$
    – user51121
    Commented Dec 7, 2020 at 9:31

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